Mathematics Tricks & Tips of Quadratic Equation for NDA

Nature of roots

See examples
Q 2731367222

Let `alpha ` and ` beta ` be the roots of the equation

`x^2 - (1 - 2a ^2) x + (1 - 2 a^2 ) = 0`
Under what condition does the above equation have real roots ?
NDA Paper 1 2016
(A)

`a^2 < 1/2`

(B)

`a^2 > 1/2`

(C)

`a^2 <= 1/2`

(D)

`a^2 >= 1/2`

Solution:

If equation has real roots

`D ge 0`

`= b^2 -4 ac ge 0`

` [ - (1-2a^2) ]^2 -4 xx 1 xx (1- 2a^2) ge 0`

`( 1-2a^2)^2 -4 (1-2a^2) ge 0`

`(1-2a^2) [ 1- 2a^2 4] ge 0`

` (1-2a^2) (-3-2a^2) ge 0`

`(2a^2 -1) (2a^2 +3) ge 0`

`[a^2 le (-3)/2] text (rejected) ` & `a^2 ge 1/2`
Correct Answer is `=>` (D) `a^2 >= 1/2`
Q 1712280139

The roots of the equation `2a^2 x^2 - 2abx + b^2 = 0`, when
`a < 0` and `b > 0` are
NDA Paper 1 2014
(A)

sometimes complex

(B)

always irrational

(C)

always complex

(D)

always real

Solution:

Given equation,

`2a^2x^2 - 2abx + b = 0`

When, `a< 0` and `b > 0`

`:. x = ( - ( -2ab) pm sqrt((-2ab)^2 - 4.2 a^2 . b^2) )/( 2.2 a^2)`(by quadratic formula)

` = (2ab pm sqrt(4a^2b^2 - 8a^2b^2))/(4a^2)`

` = (2ab pm sqrt(- 4a^2b^2) )/(4a^2)`

` = (2ab pm i 2ab)/(4a^2) = (2ab (1 pm i))/(4a^2)`

` = b/(2a) (1 pm i)`

which shows that the roots of the given equation is always

complex.
Correct Answer is `=>` (C) always complex
Q 1752334234

Every quadratic equation `ax^2 + bx + c = 0`, where
`a, b, c in R, a != 0` has
NDA Paper 1 2014
(A)

exactly one real root

(B)

at least one real root

(C)

at least two real roots

(D)

at most two real roots

Solution:

Every quadratic equation

`ax^2 + bx + c = 0`, where `a, b, c in R, a != 0`

has at most two real roots.
Correct Answer is `=>` (D) at most two real roots
Q 2338101902

How many real roots does the quadratic equation
`f(x) = x^2 + 3 | x | + 2 = 0` have ?
NDA Paper 1 2013
(A)

One

(B)

Two

(C)

Four

(D)

No real root

Solution:

Given quadratic equation is

`f(x) = x^2 + 3 | x | + 2 = 0`

Case I `f(x) = x^2 + 3x + 2 = 0`(when, x > 0)

` => x^2 + 2x + x + 2 = 0`

` => x (x + 2) + 1 (x + 2) = 0`

`=> (x + 2) (x + 1) = 0`

` :. x = -2 , -1` (but x > 0)

So, here no real roots exist.

Case II `f(x) = x^2 - 3x + 2 = 0` (when, x < 0)

`=> x = (3 ± sqrt(9 - 8))/2 = ( 3 ± 1)/2`

`=> x = 2 , 1` (but x < 0)

So, here also no real roots exist.

Hence, given quadratic equation has no real roots.
Correct Answer is `=>` (D) No real root
Q 2308101908

The roots of the equation `x^2 - 8x + 16 = 0`
NDA Paper 1 2013
(A)

are imaginary

(B)

are distinct and real

(C)

are equal and real

(D)

Cannot be determined

Solution:

Given equation is

`x^2 - 8x + 16 = 0`

`=> (x- 4)^2 = 0`

`=> x = 4, 4`

Also, discriminant `(D) = b^2 - 4ac = 0`

So, the roots of the equation are equal and real.
Correct Answer is `=>` (C) are equal and real
Q 2388212107

`(x + 1)^2 - 1 = 0` has
NDA Paper 1 2013
(A)

one real root

(B)

two real roots

(C)

two imaginary roots

(D)

four real roots

Solution:

Given that,

`(x + 1)^2 - 1^2 = 0`

`=> (x + 1)^2 (1)^2 = 0`

` => ( x + 1 + 1) ( x + 1 - 1) = 0 [ ∵ a^2 - b^2 = (a -b) (a +b) ]`

`=> (x+2) (x) = 0`

` => x = 0, -2`

Hence, `(x + 1)^2 - 1 = 0` has two real roots.
Correct Answer is `=>` (B) two real roots
Q 2318012809

If the roots of the quadratic equation
`3x^2 - 5x + p = 0` are real and unequal, then which
one of the following is correct?
NDA Paper 1 2012
(A)

`p = 25//12`

(B)

`p < 25//12`

(C)

`p >25//12`

(D)

`p <= 25//12`

Solution:

Since, the roots of the quadratic equation

`3x^2 - 5x + p = 0` are real and unequal.

`:.` Discriminant `> 0`

`=> b^2 - 4ac > 0`

`=> (-5)^2 - 4(3)(p) > 0` (here, `b = - 5,a = 3, c = p`)

` => 25 - 12p > 0 => 25 > 12p`

`=> 12 p < 25 => p < (25)/(12)`
Correct Answer is `=>` (B) `p < 25//12`
Q 1602245138

The number of real roots of the equation

`x ^(2) - 3 | x | + 2 = 0` is
DSSB Paper 1 2015
(A)

`4`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

We have, `x^(2) -3|x| +2=0 => |x|^(2) -3|x| + 2=0`

`=> | x |^(2) - 2| x | - | x | + 2 = 0`

`=> (| x | - 2) (| x | - 1) = 0`

` => | x | = 2` or `| x | = 1`

`=> x = pm 2` or `x = pm 1`

There are four real roots of the equation.
Correct Answer is `=>` (A) `4`
Q 2416156070

If `p, q` and `r` are positive and are in AP, the
roots of the quadratic equation
`px^2 + qx + r = 0` are real for
UPSEE 2013
(A)

`| r/p -7 | ge 4 sqrt (3)`

(B)

`|p/r -7 | < 4 sqrt (3)`

(C)

all `p` and `r`

(D)

no `p` and `r`

Solution:

Given `p, q` and rare in AP

`:. 2q = p + r`

The roots of `px^2 + qx + r = 0` are real, if

`q^2 - 4pr ge 0`

`=> ((p+r)/2)^2 -4pr ge 0`

`:. (r/p)^2 -14 (r/p) +1 ge 0`

`=> | r/p - 7 | ge 4 sqrt (3)`
Correct Answer is `=>` (A) `| r/p -7 | ge 4 sqrt (3)`
Q 2416412379

If the roots of the equation `ax^2 + bx + c = 0` are
real and distinct then
UPSEE 2010
(A)

both roots are greater than `-b/(2a)`

(B)

both roots are less than `- b/(2a)`

(C)

one of the roots exceeds `-b/(2a)`

(D)

None of the above

Solution:

The roots of the given equation are

`alpha=(-b -sqrt (b^2-4ac))/(2a)`

and `beta =(-b + sqrt(b^2 -4ac))/(2a)`

Since, `alpha , beta` are real and distinct, therefore
`b^2- 4ac > 0`. Now, if `a > 0`, then `beta > -b/(2a)`
and if `a < 0`, then `a > - b/(2a)`. Thus, one of the roots
exceeds ` b/(2a)`.
Correct Answer is `=>` (C) one of the roots exceeds `-b/(2a)`
Q 2304423358

If `r ne 0` and the equation

`lambda/(2x) = alpha/(x+r) +beta/(x-r) `has equal roots,

then one of the value of `lambda` is
BITSAT Mock
(A)

`alpha +beta`

(B)

`alpha - beta`

(C)

`(sqrt(alpha) + sqrt (beta))^2`

(D)

`(alpha - beta)^2`

Solution:

Given equation is `lambda/(2x) = alpha/(x+r) +beta/(x-r) `

or `lambda (x^2 - r^2) = {alpha (x-r) + beta (x+r)} 2x`

or `x^2 (2 alpha +2 beta - lambda) + 2x (beta- alpha) r + lambda r^2 =0`

As this equation has equal roots,
discriminant `= 0`

`=> [2(beta - alpha) r]^2 - 4 lambda r^2 (2 alpha +2 beta - lambda) = 0`

`=> lambda^2 -2 lambda (alpha + beta) + (beta -alpha)^2 =0`

`=> lambda = (2 (alpha + beta) pm sqrt (4 (alpha + beta)^2 -4 (beta - alpha)^2) )/2`

`= alpha + beta pm sqrt (4 alpha beta)`

`= (sqrt (alpha) pm sqrt (beta ))^2`
Correct Answer is `=>` (C) `(sqrt(alpha) + sqrt (beta))^2`
Q 2339301212

For what value of `k`, are the roots of the quadratic
equation `(k + 1)x^2 - 2(k - 1)x + 1 = 0` real and
equal?
NDA Paper 1 2007
(A)

k = 0 only

(B)

k = - 3 only

(C)

k = 0 or k = 3

(D)

k = 0 or k = - 3

Solution:

Since, the roots of the equation

`(k + 1)x^2 - 2(k - 1)x + 1 = 0` are real and equal,

`:. {-2(k -1)}^2 - 4(k + 1) = 0 ( ∵ B^2 - 4AC = 0)`

`=> 4(k^2 - 2k + 1) - 4(k + 1) = 0`

`=> k^2 - 2k + 1 - k - 1 = 0 => k^2 - 3k = 0`

`:. k = 0 , k = 3`
Correct Answer is `=>` (C) k = 0 or k = 3
Q 2349501413

If the equation `x^2 + k^2 = 2(k + 1)x` has equal roots,
then what is the value of `k`?
NDA Paper 1 2007
(A)

` - 1/3`

(B)

`- 1/2`

(C)

`0`

(D)

`1`

Solution:

The given equation is

`x^2 + k^2 =2(k + 1)x`

`=> x^2 - 2(k + 1)x + k^2 = 0`

This equation has equal roots.

`:. { 2 (k + 1)}^2 - 4 xx 1 xx k^2 = 0 ( ∵ B^2 - 4AC = 0)`

`=> 4(k^2 + 2k + 1) - 4k^2 = 0`

`=> k^2 + 2k + 1 - k^2 = 0 => k = - 1/2`
Correct Answer is `=>` (B) `- 1/2`
Q 2343280143

If `p_1p_2 = 2(q_1 + q_2)`, then at least one

of the equations `x^2 + p_1 x + q_1 = 0` and

`x^2 + p_2 x + q_2 = 0` has
BITSAT Mock
(A)

real roots

(B)

non-real roots

(C)

purely imaginary roots

(D)

none of these

Solution:

The given equations are

`x^2 + p_1 x + q_1 = 0` ... (1)

and `x^2 + p_2 x + q_2 = 0` ... (2)

Discriminant of `(1) = p_(1)^2 -4 q_1`

and discriminant of `(2) = p_(2)^2 - 4q_2`

Now, sum of the discriminants

`= p_(1)^2 +p_(2)^2 -4 (q_1 +q_2)`

`= p_(1)^2 + p_(2)^2 -2p_1 p_2`

`= (p_1 -p_2)^2 ge 0`


Hence, at least one of the discriminants is non-negative. This implies
at least one of the equation has real roots.
Correct Answer is `=>` (A) real roots
Q 2318078800

A quadratic polynomial with two distinct roots
has one real root. Then, the other root is
NDA Paper 1 2008
(A)

not necessarily real, if the coefficients are real

(B)

always imaginary

(C)

always real

(D)

real, if the coefficients are real

Solution:

If a quadratic polynomial with two distinct roots, has

one root real, then

`b^2 - 4ac > 0`

So, the other root is also real.
Correct Answer is `=>` (C) always real
Q 2388278107

The roots of the equation `(x - p) (x - q) = r^2`
, where `p, q` and `r` are real, are
NDA Paper 1 2009
(A)

always complex

(B)

always real

(C)

always purely imaginary

(D)

None of the above

Solution:

Given, `(x - p)(x - q)= r^2`

`=> x^2 - (p + q )x + pq - r^2 = 0`

Now, `D = (p+ q)^2 - 4(pq- r^2) ( ∵ D = B^2 - 4AC)`

`= (p - q)^2 + 4r^2 >= 0`

Hence, roots are always real.
Correct Answer is `=>` (B) always real
Q 2318378209

If `a, b` and `c` are real numbers, then the roots of the
equation `(x- a)(x- b)+ (x- b)(x- c) + (x- c)(x- a) = 0` are
always
NDA Paper 1 2009
(A)

real

(B)

imaginary

(C)

positive

(D)

negative

Solution:

Given,

`(x - a) (x - b)+ (x - b )(x -c) + (x - c )(x - a) = 0`

` => 3x^2 - 2(b +a+ c)x + ab +be+ ca = 0`

Now, `D = 4(a + b + c)^2 - 12(ab +bc+ ca)`

`= 4a^2 + b^2 + c^2 - ab - bc - ca`

`= 2 1/2 = {(a-b)^2 +(b-c)^2 +(c-a)^2 } > 0`

Hence, the roots are always real.
Correct Answer is `=>` (A) real
Q 2328067801

If the equation `x^2 - bx + 1 = 0` does not possess
real roots, then which one of the following is
correct?
NDA Paper 1 2010
(A)

`- 3 < b < 3`

(B)

`- 2 < b < 2`

(C)

`b > 2`

(D)

`b < - 2`

Solution:

Since, equation `x^2 - bx + 1 = 0` has no real roots i.e.,

it has imaginary roots which is possible only, if

`b^2 - 4 < 0 ( ∵ B^2 - 4AC < 0)`

` => b^2 < 4 => -2 < b < 2`
Correct Answer is `=>` (B) `- 2 < b < 2`
Q 2348167003

Consider the equation `(x - p) (x - 6) + 1 = 0` having
integral coefficients. If the equation has integral
roots, then what values can `p` have?
NDA Paper 1 2010
(A)

4 or 8

(B)

5 or 10

(C)

6 or 12

(D)

3 or 6

Solution:

The given equation can be rewritten as

`x^2 - (p + 6)x + (6p + 1) = 0`

Now, discriminant, `D = 0 => b^2 - 4ac = 0` for integral roots.

`=> (p + 6)^2 - 4(6p + 1) = 0`

`=> p^2 - 12p + 32 = 0`

` => (p- 4) (p- 8) = 0`

` :. P = 4, 8`
Correct Answer is `=>` (A) 4 or 8
Q 2378356206

If the roots of the equation
` (a^2 + b^2 ) x^2 -2b(a + c) x + (b^2 + c^2) = 0`
are equal, then which one of the following is
correct?
NDA Paper 1 2010
(A)

`2b = a + c`

(B)

`b^2 = ac`

(C)

`b + c = 2a`

(D)

`b = ac`

Solution:

Since, the roots of given equation

`(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0` are equal.

`:. 4b^2 (a + c)^2 = 4(a^2 + b^2) (b^2 + c^2) ( ∵ B^2 - 4AC = 0)`

`=> b^2(a^2 + c^2 + 2ac) = a^2b^2 + b^2c^2 + a^2c^2 + b^4`

`=> a^2b^2 + b^2c^2 + 2acb^2 = a^2b^2 + b^2c^2 + a^2c^2 + b^4`

`=> b^4 + a^2c^2 - 2acb^2 = 0`

`=> (b^2 - ac )^2 = 0`

`=> b^2 = ac`
Correct Answer is `=>` (B) `b^2 = ac`
Q 2348845703

One of the roots of the quadratic equation
`ax^2 + bx + c = 0, a != 0` is positive and the other
root is negative. The condition for this to happen
is
NDA Paper 1 2011
(A)

`a > 0 , b > 0 ,c > 0`

(B)

`a > 0, b < 0, c > 0`

(C)

`a < 0, b > 0, c < 0`

(D)

`a < 0 ,c > 0`

Solution:

Since, one root of `ax^2 + bx + c = 0, a != 0` is positive

and another root is negative which is possible only if `a < 0` and

`c > 0`.

`∵` Product of roots `= c/a < 0` (since, the roots are opposite signs)

So, `c` and `a` are also opposite signs.

i.e., `(c > 0, a < 0` or `c < 0, a < 0)`
Correct Answer is `=>` (D) `a < 0 ,c > 0`
Q 2348745603

If `p, q` and `r` are rational numbers, then the roots of
the equation `x^2 - 2px + p^2 - q^2 + 2qr - r^2 = 0` are
NDA Paper 1 2011
(A)

complex

(B)

pure imaginary

(C)

irrational

(D)

rational

Solution:

The given equation is

`x^2 - 2 px + p^2 - q^2 + 2qr - r^2 = 0`.

Now, `B^2 - 4AC = (-2p)^2 - 4(1)(p^2 - q^2 + 2q r - r^2)`

`= 4p^2 - 4p^2 + 4(q - r)^2 = 4(q - r)^2`

which is always greater than zero.

Therefore, the roots of the given equation are rational.
Correct Answer is `=>` (D) rational
Q 2378023806

If the equation `x^2 - 4x + 29 = 0` has one root
`2 + 5i`, then what is the other root (where, `i = sqrt(-1)`)?
NDA Paper 1 2011
(A)

`2`

(B)

`5`

(C)

`2 + 5i`

(D)

`2 - 5i`

Solution:

We know that, if one root of the quadratic equation is

complex, then its other root is its conjugate.

Now, `x^2 - 4x + 29 = 0` has one root `alpha = 2 + 5i`, then its

conjugate `bar alpha = 2 - 5i` is the second root of this equation.
Correct Answer is `=>` (D) `2 - 5i`
Q 2318212109

What is the degree of the equation

` 1/(x - 3) = 1/( x - 2) - 1/2 ` ?
NDA Paper 1 2013
(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

Given equation is

`1/(x-3) = 1/(x+2) - 1/2`

`=> 1/(x-3) = (2 - x - 2)/(2(x+2)) => 2(x + 2) = - x (x - 3)`

` => 2x + 4 = - x^2 + 3x => x^2 - x + 4 = 0`

which is a quadratic equation and degree of the above equation

is `2`.
Correct Answer is `=>` (C) `2`
Q 2239278112

If `a, b` are odd integers, then the roots of
the equation `2ax^2 + (2a + b) x + b = 0`,
`a ≠ 0` is :
BITSAT Mock
(A)

non-real

(B)

equal

(C)

irrational

(D)

rational

Solution:

Equation is `2ax^2 + (2a + b) x + b = 0`

Discriminant `= (2a + b)^2 − 4 . 2a . b`

`= (2a + b)^2 − 8ab`

`> 0`

`∴ ` Roots are rational.
Correct Answer is `=>` (D) rational
Q 1754691554

If for real values of `x, cos theta = x + 1/x`, then
NCERT Exemplar
(A)

`theta` is an acute angle

(B)

`theta` is right angle

(C)

`theta` is an obtuse angle

(D)

No value of `theta` is possible

Solution:

Here, `cos theta = x + 1/x`

` => cos theta = (x^2 + 1)/x`

` x^2 - xcos theta + 1 = 0`

For real value of `x , (-cos theta)^2 - 4 xx 1 xx 1 = 0`

` cos^2 theta = 4`

` cos theta = pm 2`

which is not possible.
Correct Answer is `=>` (D) No value of `theta` is possible

Relation between Roots and Coefficients

1. Quadratic Roots

If `alpha` and `beta` are the roots of quadratic equation `ax^2 + b x +c = 0` , then

sum of roots `= alpha + beta = -b/a`

and product of roots ` = alpha beta = c/a`

2. Cubic Roots

If `alpha, beta ` and `gamma` are the roots of cubic equation `ax^2 + bx^2 +cx + d = 0 ; a != 0` ,then

`alpha + beta + gamma = -b/a`

`beta gamma + gamma alpha + alpha beta = c/a` and `alpha beta gamma = - d/a`
Q 2713080840

If cot `alpha` and cot `beta` are the roots of the equation `x^2 + bx + c = 0` 'with `b != 0`, then the value of cot `(alpha + beta)` is
NDA Paper 1 2017
(A)

`(c - 1)/b`

(B)

`(1 - c)/b`

(C)

`b/(c - 1)`

(D)

`b/(1 - c)`

Solution:

`cot (alpha+beta) =(cot alpha cot beta-1)/(cot alpha+ cot beta)`

`=(c-1)/(-b)`

`=(1-c)/b`
Correct Answer is `=>` (B) `(1 - c)/b`
Q 2703080848

The sum of the roots of the equation `ax^2 + x + c = 0`. (where a and c are non-zero) is equal to the sum of the reciprocals of their squares. Then `a , ca^2 , c^2` are in
NDA Paper 1 2017
(A)

AP

(B)

GP

(C)

HP

(D)

None of the above

Solution:

`alpha+ beta =1/(alpha^2) + 1/(beta^2)`

`alpha+ beta =((alpha+ beta)^2 - 2 alpha beta)/((alpha beta)^2)`

`(-1)/a =((-1/a)^2 - 2(c/a))/((c/a)^2)`

`2(c/a) = 1/(a^2) + c^2/a`

`2 ac = 1+ c^2 a`
Correct Answer is `=>` (D) None of the above
Q 1608423308

If m and n are roots of the equation
`(x + p )(x + q) - k = 0`, then roots of the equation
`(x- m)(x- n) + k =0` are
NDA Paper 1 2015
(A)

`p` and `q`

(B)

`1/p` and `1/q`

(C)

`-p` and `-q`

(D)

`p + q` and `p- q`

Solution:

Since, m and n are roots of the equation

`(x + p)(x + q) - k = 0`

`=> x^ 2 + qx + px + pq - k = 0`

`=> x^ 2 +(p+q) x - t - pq - k = 0 ...... (i)`

For `m` and `n` to be roots the equation should be

`x^ 2 - (m+n) x + m . n = 0 ...... (ii)`

On comparing Eqs. (i) and (ii), we get

`=> p+q = - m - n`

` => p + q = - (m+ n) .... (iii)`

and `pq - k = m . n ........ (iv)`

Also `(x - m)(x - n) + k = 0`

`=> x^ 2- nx - mx + mn + k = 0`

`=> x^ 2 - (m + n) x + mn + k = 0`

`=> x^ 2 + (p + q)x + pq - k + k = 0`

[using Eqs. (iii) and (iv)]

`=> x^ 2 + (p + q)x + pq = 0`

`=> x^ 2 - [(-p) -(-q)] + [(-p)(-q)] = 0`

Hence. `-p` and `- q` are the required roots.
Correct Answer is `=>` (C) `-p` and `-q`
Q 2106123978

Let `alpha` and `beta (alpha < beta)` be the roots of the equation `x^(2) + bx + c = 0`,
where `b > 0` and `c < 0`.

Consider the following
1. `alpha + beta + alpha > 0 ` quad 2. quad `alpha ^(2) beta + beta^(2) alpha > 0`
Which of the above is/are correct?
NDA Paper 1 2016
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given, `alpha` and `beta` are the roots of equation `x^(2) + bx + c = 0`.

`:. alpha + beta = - b` and `alpha beta = c`

As `b > 0`

So, `alpha + beta < 0`

`=> beta < -alpha`

and given that `alpha < beta => alpha < 0` and ` beta > 0`

As,` alpha+beta = 0` and `alpha beta < 0`

:. Statement 1 is not correct.

Now, `alpha^(2) beta + alpha beta (alpha+beta)`

As, `alpha + beta < 0` and `alpha beta < 0`

`=> alpha beta ( alpha+beta ) > 0`

:. Statement 2 is correct.
Correct Answer is `=>` (B) Only 2
Q 2388578407

Let `alpha , gamma ` be the roots of `Ax^2 - 4x + 1 = 0` and `beta , delta` be
the roots of `Bx^2 - 6x + 1 = 0`. If `alpha , beta , gamma` and `delta` are in
HP, then what are the values of `A` and `B`, respectively?
NDA Paper 1 2009
(A)

3, 8

(B)

-3, -8

(C)

3, -8

(D)

-3, 8

Solution:

Since, `alpha` and `gamma` be the roots of `Ax^2 - 4x + 1 = 0`

`:. alpha + gamma = 4/A` and `alpha gamma = 1/A`

and `beta` and `delta` be the roots of `Bx^2 - 6x + 1 = 0`.

`:. beta + delta = 6/B` and `beta delta = 1/B`

Also, `alpha ,beta , gamma` and `delta` are in HP.

Hence `1/alpha , 1/beta , 1/ gamma` and `1/delta` are in AP.

` => 1/beta - 1/alpha = 1/ delta - 1/ gamma => 1/beta - 1/delta = 1/alpha - 1/gamma`

` => ( delta - beta)/( beta delta) = ( gamma - alpha)/(alpha gamma) = sqrt((delta + beta)^2 - 4 beta delta)/(beta delta) = sqrt((gamma + alpha)^2 - 4 alpha gamma )/(alpha gamma)`

` => sqrt((36)/B^2 - 4/B) /(1 /B) = sqrt((16)/A^2 - 4/A)/(1/A) => sqrt(36 - 4B) =sqrt(16 - 4A)`

`=> 36 - 48 = 16 - 4A => 4A - 48 = - 20`

It is possible, when `A = + 3` and `B = 8`.
Correct Answer is `=>` (A) 3, 8
Q 2359801714

If `alpha , beta` are the roots of the equation
`ax^2 + bx + c = 0`, then what is the value of
`(a alpha + b)^(-1) + (a beta + b)^(-1)` ?
NDA Paper 1 2007
(A)

`a/(bc)`

(B)

`b/(ac)`

(C)

`(-b)/(ac)`

(D)

`(-a)/(bc)`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`ax^2 + bx + c = 0`, then

`alpha + beta = - b/a` and `alpha beta = c/a` ... (i)

Now, `(a alpha + b)^(-1) + (a beta + b)^(-1)`

`= 1/(a alpha + b) + 1/(a beta + b) = (a beta + b + a alpha + b)/((a alpha + b )(a beta + b))`

` = (a(alpha + beta) + 2b)/(a^2 alpha beta + ab (alpha + beta) + b^2)`

`= (a(-b//a) + 2b)/( a^2(c//a) + ab(-b//a) + b^2) = ( - b + 2b)/( ac- b^2 + b^2)` [from Eq. (i)]

`= b/(ac)`
Correct Answer is `=>` (B) `b/(ac)`
Q 2751067824

`2 x^2 + 3 x - alpha = 0` has roots -2 and `beta` while the equation `x^2 - 3mx + 2m^2 = 0` has both roots positive , where `alpha > 0` and `beta > 0`.
What is the value of `alpha` ?
NDA Paper 1 2016
(A)

`1/2`

(B)

`1`

(C)

`2`

(D)

`4`

Solution:

`2x^2 +3x - alpha = 0`

`=> -2 + beta = -3/2` ............(i)

`=> -2 beta = - alpha `..................(ii)

Putting the value of `beta` from (i) in (ii)

`-2 (-3/2 +2) = - alpha`

`alpha = -3 +4 =1`
Correct Answer is `=>` (B) `1`
Q 2166780675

Given that `tan alpha` ard `tan beta` are
the roots of the equation ` x^(2) + bx + c = 0` with `b != 0`.

What is `tan (alpha + beta)` equal to?

NDA Paper 1 2016
(A)

`b(c- 1)`

(B)

`c(b- 1)`

(C)

`c(b- 1)^(-1)`

(D)

`b(c - 1)^(-1)`

Solution:

Given,` x^(2) + bx + c = 0 , b != 0`

`tanalpha + tanbeta =-b`

and `tan alpha tan beta = c`

Now ` tan (alpha + beta ) =(tanalpha + tanbeta)/(1 - tan alpha tan beta) = (-b) /(1-c) = b(c-1)^(-1)`
Correct Answer is `=>` (D) `b(c - 1)^(-1)`
Q 2146880773

Given that `tan alpha` ard `tan beta` are the roots of the equation ` x^(2) + bx + c = 0` with `b != 0`.

What is `sin (alpha + beta ) sec alpha sec beta` equal to?
NDA Paper 1 2016
(A)

`b`

(B)

`-b`

(C)

`c`

(D)

`-c`

Solution:

` ∵ tan alpha + tan beta =-b`

` => (sin alpha)/(cos alpha) + (sin beta)/(cos beta)= -b`

` => ( sin alpha cos beta + cos alpha sin beta)/(cos alpha cos beta) = - b`

` => (sin (alpha + beta))/(cos alpha cos beta) = - b`

`=> sin( alpha + beta ) sec alpha sec beta = -b`
Correct Answer is `=>` (B) `-b`
Q 2308001808

If `alpha` and `beta` are the roots of the equation
`x^2 + x + 2 = 0`, then what is `(alpha^(10) + beta^(10))/(alpha^(-10) + beta^(-10))` equal to'?

NDA Paper 1 2013
(A)

`4096`

(B)

`2048`

(C)

`1024`

(D)

`512`

Solution:

Here, `alpha` and `beta` are the roots of the equation

`x^2 + x + 2 = 0`, then

`alpha + beta = - 1` ... (i)

and `alpha · beta = 2` ... (ii)

Now, `( alpha^(10) + beta^(10) )/( alpha^(-10) + beta^(-10) ) = ( alpha . beta )^(10) = (2)^(10)` [fromEq. (ii)]

`= 1024`
Correct Answer is `=>` (C) `1024`
Q 2127001881

If one root of the equation `( l - m) x^2 + lx + 1 = 0`
is double the other and `l` is real, then what is the greatest
value of `m`?
NDA Paper 1 2016
(A)

` - 9/8`

(B)

`9/8`

(C)

`- 8/9`

(D)

`8/9`

Solution:

Given equation is

`(l - m) x^2 + lx + 1 = 0`

Let `alpha` and `2alpha` be the roots of given equation. Then, we
have

`alpha + 2alpha = 3alpha = (-l)/( l - m)`

and `alpha = (-l)/(3 (l - m)) ` and `2 alpha^2 = 1/(l - m)`

` => 2*( (-l)/(3 ( l - m))) = 1/(l - m)`

` => (2l^2)/(9(l - m)^2) = 1/(l - m)`

` => 2l^2 = 9(l - m) quad [ ∵l!= m]`

` => 2l^2 - 9l + 9m = 0`

` => l = (9 pm sqrt(81 - 72m))/4`

`∵ l` is real.

`:. D >= 0 => 81 - 72m >= 0`

` => 81 >= 72m`

` => m <= (81)/(72) => m <= 9/8`

Hence, greatest value of `m` is `9/8`.
Correct Answer is `=>` (B) `9/8`
Q 2116023879

Let `alpha` and `beta (alpha < beta)` be the roots of the equation `x^(2) + bx + c = 0`,
where `b > 0` and `c < 0`.

Consider the following
1. `beta < - alpha quad`
2. `quad beta < |alpha|`
Which of the above is/are correct?
NDA Paper 1 2016
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given, `alpha` and `beta` are the roots of equation `x^(2) + bx + c = 0`.

`:. alpha + beta = - b` and `alpha beta = c`

As `b > 0`

So, `alpha + beta < 0`

`=> beta < -alpha`

and given that `alpha < beta => alpha < 0` and ` beta > 0`

` alpha + beta < 0 => < - alpha`...............(i)

:. Statement 1 is correct.

As, `alpha < beta => -alpha > beta`..........(ii)

From Eqs. (i) and (ii),

` | alpha| > |beta|`

:. Statement 2 is correct.
Correct Answer is `=>` (C) Both 1 and 2
Q 2318112000

What is the difference in the roots of the equation
`x^2 - 10x + 9 = 0`?
NDA Paper 1 2013
(A)

`2`

(B)

`3`

(C)

`5`

(D)

`8`

Solution:

Given equation,

`x^2 - 10x + 9 = 0`

Let `(alpha , beta)` be the roots of the given equation.

Then, `alpha + beta = 10` .......(i)

and `alpha · beta = 9` .........(ii)

Now, we use the identity

`(alpha - beta)^2 = (alpha + beta)^2 - 4alpha beta = (10)^2 - 4 (9)= 100- 36 = 64`

` => alpha - beta = ± 8`

`:. | alpha - beta | = 8`
Correct Answer is `=>` (D) `8`
Q 2368567405

If `alpha, beta` are the roots of the quadratic equation
`x^2 - x + 1= 0`, then which one of the following is
correct?
NDA Paper 1 2010
(A)

`(alpha^4 - beta^4 )` is real

(B)

`2(alpha^5 + beta^5 ) = (alpha beta )^5`

(C)

`(alpha^6 - beta^6) = 0`

(D)

`(alpha^8 + beta^8) = (alpha beta)^8`

Solution:

Since, `alpha` and `beta` be the roots of the equation

`x^2 - x + 1 = 0`.

`:. alpha + beta = 1` and `alpha beta = 1`

Now, `alpha - beta = sqrt((alpha + beta )^2 - 4 alpha beta ) = sqrt(3)i`

` => alpha = (1 + i sqrt3)/2 ` and `(1 - i sqrt3)/2 `

Now `alpha = cos pi/3 + i sin pi/3`

and ` beta = cos pi/3 - i sin pi/3`

(a) `alpha^4 - beta^4 = cos (4 pi)/3 + i sin (4 pi)/3 - cos (4 pi)/3 + i sin (4 pi)/3`

`= 2 i sin (4 pi)/3`

Hence, `alpha^4 - beta^4 ` is not real.

(b) `2(alpha^5 + beta^5 ) = 2 ( cos (5 pi)/3 + i sin (5 pi)/3 + cos (5 pi)/3 - i sin (5 pi)/3 )`

`= 2 . 2 cos (5 pi)/3 = 4 . 1/2 = 2`

Now, `(alpha beta )^5 = 1 => 2 (alpha^5 + beta^5) != (alpha beta )^5`

(c ) `alpha^6 - beta^6 = cos (6 pi)/3 + i sin (6 pi)/3 - cos (6 pi)/3 + i sin (6 pi)/3`

` = 2i sin 2 pi = 0`

(d) `alpha^8 + beta^8 = cos (8 pi)/3 + sin (8 pi)/3 + cos (8 pi)/3 - i sin (8 pi)/3`

`= 2 cos (8 pi)/3 = 2 ( - 1/2) = -1`

Now, `(alpha beta )^8 = (1)^8 = 1 => (alpha^8 + beta^8) != (alpha beta)^8`
Correct Answer is `=>` (C) `(alpha^6 - beta^6) = 0`
Q 2368101905

If `alpha` and `beta` are the roots of the equation
`a.x^2 + bx + b = 0`, then what is the value of
` sqrt(alpha/ beta) + sqrt(beta/alpha) + sqrt(b/a) `?
NDA Paper 1 2013
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Given quadratic equation is `ax^2 + b x + b = 0`

Let `alpha` and `beta` be the roots of given equation.

` :. alpha + beta = - b/a ` and ` alpha beta = b/a`

` :. sqrt(alpha/ beta) + sqrt(beta/alpha) + sqrt(b/a) = ( alpha + beta )/sqrt(alpha beta) + sqrt(b/a)`

` = (-b)/a . sqrt(a/b) + sqrt(b/a) = - sqrt(b/a) + sqrt(b/a) = 0`
Correct Answer is `=>` (B) `0`
Q 2329001811

If `alpha , beta` are the roots of the equation `x^2 - 2x -1 = 0`,
then what is the value of `alpha^2 beta^(-2) + alpha^(-2)beta^(2)` ?
NDA Paper 1 2007
(A)

`-2`

(B)

`0`

(C)

`30`

(D)

`34`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`x^2 - 2x - 1 = 0`, then

`alpha + beta = 2` and `alpha beta = -1`

`(alpha + beta)^2 = alpha^2 + beta^2 + 2 alpha beta`

`=> 4 = alpha^2 + beta^2 - 2 => alpha^2 + beta^2 = 6`

`=> (alpha^2 + beta^2)^2 = 6^2`

`=> alpha^4 + beta^4 + 2 = 36`

`=> alpha^4 + beta^4 = 34`

Now, `alpha^2 beta^(-2) + alpha^(-2) beta^2 = alpha^2/beta^2 + beta^2 /alpha^2 = ( alpha^4 + beta^4)/(alpha beta)^2 = (34)/(-1)^2 = 34`
Correct Answer is `=>` (D) `34`
Q 2319701619

If `alpha , beta` are the roots of `ax^2 + 2bx + c = 0` and `alpha + delta`,
`beta + delta` are the roots of `Ax^2 + 2Bx + C = 0`, then what
is `(b^2 - ac)//(B^2 - AC)` equal to
NDA Paper 1 2007
(A)

`(b/B)^2`

(B)

`(a/A)^2`

(C)

`(a^2b^2)/(A^2B^2)`

(D)

`(ab)/(AB)`

Solution:

Since, `alpha` and `beta` are the roots of `ax^2 + 2bx + c = 0`

`:. alpha + beta = - (2b)/a` and `alpha beta = c/a` ......(i)

Also, `alpha + delta` and `beta + delta` are the roots of

`Ax^2 + 2 Bx + C = 0`

`(alpha + beta) + 2 delta = -(2B)/ A` ......(ii)

and `(alpha + delta )(beta + delta) = C/A` ......(iii)

` => - (2b)/a + 2 delta = - (2B)/ A` [from Eqs. (ii) and (i)]

`=> delta = b/a - B/A` ... (iv)

and `(alpha + delta )(beta + delta) = C/A` [from Eq. (iii)]

`=> alpha beta + (alpha + beta) delta + delta^2 = C/A`

`=> c/a - (2b)/a (b/a - B/A) + (b/a - B/A)^2 = C/A`

[from Eqs. (i) and (iv)]

`=> c/a - (2b^2) /a^2 + (2bB)/(aA) + (b/a)^2 = C/A`

` => c/a - (b/a)^2 + (B/A)^2 = C/A`

`=> B^2/A^2 - C/A = b^2/a^2 => (B^2 - AC)/A^2 = (b^2 - ac^2 )/(a^2)`

` :. (b^2 - ac)/(b^2 - AC) = (a/A)^2`
Correct Answer is `=>` (B) `(a/A)^2`
Q 2318178000

If the roots of `ax^2 + bx + c = 0` are `sin alpha` and `cos alpha`
for some `alpha`, then which one of the following is
correct?
NDA Paper 1 2009
(A)

`a^2 + b^2 = 2ac`

(B)

`b^2 - c^2 = 2ab`

(C)

`b^2 - a^2 = 2ac`

(D)

`b^2 + c^2 = 2ab`

Solution:

Since, `sin alpha` and `cos alpha` are the roots of

`ax^2 + bx + c = 0`

`:. sin alpha + cos alpha = (-b)/a` ......(i)

and `sin alpha cos alpha = c/a` ... (ii)

`=> sin^2 alpha + cos^2 alpha + 2 sin alpha cos alpha = b^2/a^2` [from Eq. (i)]

` => 1 + ( 2c)/a = b^2/a^2` [from Eq. (ii)]

` => b^2 - a^2 = 2ac`
Correct Answer is `=>` (C) `b^2 - a^2 = 2ac`
Q 2349201113

If `alpha` and `beta` are the roots of the equation
`x^2 + 6x + 1 = 0`, then what is `| alpha - beta |` equal to?
NDA Paper 1 2007
(A)

`6`

(B)

`3 sqrt(2)`

(C)

`4 sqrt(2)`

(D)

`12`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`x^2 + 6x + 1 = 0`.

`:. alpha + beta = - 6` and `alpha beta = 1`

Now, `(alpha - beta)^2 = (alpha + beta)^2 - 4 alpha beta`

`= ( -6)^2 - 4 = 36 - 4 = 32`

`=> | alpha - beta | = sqrt(32) = 4sqrt(2)`
Correct Answer is `=>` (C) `4 sqrt(2)`
Q 2318878700

If `alpha , beta` are the roots of the equation
`2x^2 - 2(1 + n^2 )x + (1 + n^2 + n^4) = 0`, then what is
the value of `alpha^2 + beta^2` ?
NDA Paper 1 2009
(A)

`2n^2`

(B)

`2n^4`

(C)

`2`

(D)

`n^2`

Solution:

Since, `alpha` and `beta` be the roots of

`2x^2 - 2(1 + n^2)x + (1 + n^2 + n^4) = 0`.

`:. alpha + beta = (n^2 + 1)` and `alpha beta = (1 + n^2 + n^4)/2`

Now, `alpha^2 + beta^2 = (alpha + beta )^2 - 2 alpha beta = (n^2 + 1)^2 - (1 + n^2 + n^4)`

`= n^4 + 1 + 2n^2 - 1 - n^2 - n^4 = n^2`
Correct Answer is `=>` (D) `n^2`
Q 2318191909

If sum of the roots of `3x^2 + (3p + 1)x- (p + 5) = 0`
is equal to their product, then what is the value of
`p`?
NDA Paper 1 2008
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`9`

Solution:

Let `alpha` and `beta` are the roots of the given equation.

`:. alpha + beta = (- (3p + 1))/3`

and ` alpha beta = (- (p+ 5))/3`

Now , `(- (3p+ 1))/3 = (- (p+ 5))/3`(according to the question)

` => 3p + 1 = p + 5`

`=> 2p = 4`

`=> p = 2`
Correct Answer is `=>` (A) `2`
Q 2388191907

If `alpha` and `beta` are the roots of `x^2 + 4x + 6 = 0`, then
what is the value of `alpha^3 + beta^3` ?
NDA Paper 1 2008
(A)

` - 2/3`

(B)

`2/3`

(C)

`4`

(D)

`8`

Solution:

Since, `alpha` and `beta` are the roots of `x^2 + 4x + 6 = 0`.

`:. alpha + beta = - 4` and `alpha beta = 6`

Now, `alpha^3 + beta^3 = (alpha + beta)^3 - 3alpha beta (alpha + beta) = (-4)^3 - 3 xx 6 (-4)`

`= -64 + 72 = 8`
Correct Answer is `=>` (D) `8`
Q 2378191906

If `r` and `s` are roots of `x^2 + px + q = 0`, then what is
the value of `(1// r^2) + (1// s^2 )`?
NDA Paper 1 2008
(A)

`p^2 - 4q`

(B)

`(p^2 - 4q)/2`

(C)

`(p^2 - 4q)/q^2`

(D)

`(p^2 - 2q)/q^2`

Solution:

Since, `r` and `s` are the roots of the equation

`x^2 + px + q = 0`.

`:. r + s = - p` and `rs = q`

Now, `1/r^2 + 1/s^2 = (r^2 + s^2)/(rs)^2 = ((r + s )^2 - 2rs)/(rs)^2`

` = ((-p)^2- 2q)/q^2 = ( p^2 - 2q)/q^2`
Correct Answer is `=>` (D) `(p^2 - 2q)/q^2`
Q 2368191905

If the roots of the equation `x^2 - bx + c = 0` are two
consecutive integers, then what is the value of
`b^2 - 4c`?
NDA Paper 1 2008
(A)

`1`

(B)

`2`

(C)

`-2`

(D)

`3`

Solution:

Let the roots of:the equation `x^2 - bx + c = 0` be `alpha`

and `alpha + 1`.

`:. alpha + (alpha + 1) = b`

`=> 2 alpha + 1 = b` and `alpha (alpha + 1 ) = c`

Now, `b^2 - 4c = (2alpha + 1)^2 - 4[alpha (alpha + 1)]`

`= 4 alpha ^2 + 1 + 4 alpha - 4alpha ^2 - 4alpha = 1`
Correct Answer is `=>` (A) `1`
Q 2358191904

If `sin alpha` and `cos alpha` are the roots of the equation
`px^2 + qx + r = 0`, then which one of the following
is correct?
NDA Paper 1 2008
(A)

`p^2 + q^2 -2pr = 0`

(B)

`p^2 - q^2 + 2pr = 0`

(C)

`(p + r)^2 = 2(p^2 + r^2 )`

(D)

`(p- r)^2 = q^2 + r^2`

Solution:

Since, `sin alpha` and `cos alpha` are the roots of the equation

`px^2 + qx + r = 0`.

`:. sin alpha + cos alpha = (-q)/p` .....(i)

and `sin alpha cos alpha = r/p` .........(ii)

From Eq. (i),

`(sin alpha + cos alpha)^2 = q^2/p^2`

`=> sin^2 alpha + cos^2 alpha + 2 sin alpha cos alpha = q^2/p^2`

`=> 1 + 2.r/p = q^2/p^2` [from Eq. (ii)]

`=> p^2 + 2rp = q^2`

`=> p^2 - q^2 + 2 r p = 0`
Correct Answer is `=>` (B) `p^2 - q^2 + 2pr = 0`
Q 2368878705

The roots of `Ax^2 + Bx + C = 0` are `r` and `s`. For the
roots of `x^2 + px + q = 0` to be `r^2` and `s^2`, what must
be the value of `p`?
NDA Paper 1 2009
(A)

`((B^2 - 4AC))/A^2`

(B)

`((B^2 - 4AC))/A^2`

(C)

`((2AC - B^2))/A^2`

(D)

`B^2 - 2C`

Solution:

Since, r and s are the roots of `Ax^2 + Bx + C = 0`, then

`r + s = - B/A` and `rs = C/A`

Now, roots of `x^2 + px + q = 0` be `r^2` and `s^2`

`:. r^2 + s^2 = - p` and `r^2s^2 = q`.

`=> (r + s)^2 - 2rs = - p`

` => B^2/A^2 - (2C)/A = - p => ( B^2 - 2AC)/A^2 = - p`

` => p = ((2AC - B^2))/A^2`
Correct Answer is `=>` (C) `((2AC - B^2))/A^2`
Q 2378867706

If the product of the roots of the equation
`x^2 - 5x + k = 15` is `- 3`, then what is the value of `k`?
NDA Paper 1 2010
(A)

`12`

(B)

`15`

(C)

`16`

(D)

`18`

Solution:

Let `alpha` and `beta` be the roots of the equation

`x^2 - 5x + k - 15 = 0`.

`:. alpha beta = k - 15`

But `alpha beta = - 3`

` => -3 = k - 15 => k = 15 - 3 = 12`
Correct Answer is `=>` (A) `12`
Q 2378823706

If the equations
`x^2 - px + q = 0` and `x^2 - ax + b = 0`
have a common root and the roots of the second
equation are equal, then which one of the
following is correct?
NDA Paper 1 2011
(A)

`aq = 2(b + p)`

(B)

`aq = b + p`

(C)

`ap = 2(b + q)`

(D)

`ap= b + q`

Solution:

Let the roots of the equation `x^2 - px + q = 0` is `(alpha, beta)`

and the roots of the equation `x^2 - ax + b = 0` is (`alpha , alpha`).

Then, `alpha + beta = p` and `alpha beta = q` ... (i)

`alpha + alpha = a` and `alpha ·alpha = b`

` => alpha = a/2` and `alpha^2 = b`

`=> (a/2)^2 = b => a^2 = 4b` ... (ii)

From Eq. (i), `alpha beta = q`

`=> a/2 . beta = q => beta = (2q)/a`

Now, put the value of `alpha` and `beta` in the following equation

`alpha + beta = p`

`=> a//2 + 2q//a = p`

`=> a^2 + 4q = 2ap`

`=> 4b + 4q = 2ap` (from Eq. (ii)]

`=> 2(b+q) = ap`
Correct Answer is `=>` (C) `ap = 2(b + q)`
Q 2308745608

What is .the sum of the roots of the equation
`(2 - sqrt(3)) x^2 - (7 - 4 sqrt3) x + (2 + sqrt(3)) = 0` ?
NDA Paper 1 2011
(A)

`2 - sqrt(3)`

(B)

`2 + sqrt(3)`

(C)

`7 - 4sqrt(3)`

(D)

`4`

Solution:

The given equation is

`(2 - sqrt3)x^2 - (7 - 4 sqrt(3) )x + (2 + sqrt(3)) = 0`

`:.` Sum of roots `= (7 - 4sqrt(3))/(2 - sqrt3) = ( 2- sqrt3)^2/( 2- sqrt3) = 2 - sqrt(3)`
Correct Answer is `=>` (A) `2 - sqrt(3)`
Q 2318045800

What is the condition that one root of the
equation `ax^2 + bx + c = 0`, where `a != 0` should be
double the other?
NDA Paper 1 2011
(A)

`2a^2 = 9bc`

(B)

`2b^2 = 9ac`

(C)

`2c^2 = 9ab`

(D)

None of these

Solution:

Let the roots of the equation `ax^2 + bx + c = 0` be `alpha`

and `2 alpha`.

` :. alpha + 2 alpha = (-b)/a` and `alpha . 2 alpha = c/a`

` => alpha = (-b)/(3a) ` and ` alpha^2 = c/(2a)`

`=> ( (-b)/(3a) )^2 - c/(2a) => b^2/(9a^2) = c/(2a)`

` => 2b^2 = 9ac`
Correct Answer is `=>` (B) `2b^2 = 9ac`
Q 2318045809

If sum of squares of the roots of the equation
`x^2 + k x - b = 0` is `2b`, then what is `k` equal to?
NDA Paper 1 2011
(A)

`1`

(B)

`b`

(C)

`-b`

(D)

`0`

Solution:

Let the roots of the equation `x^2 + kx - b = 0` be `alpha` and

`beta`.

`=> alpha + beta = - k` and `alpha beta = -b`.

According to the question,

`alpha^2 + beta^2 = 2b`

`=> (alpha + beta)^2 - alpha beta = 2b => k^2 + 2b = 2b`

` :. k = 0`
Correct Answer is `=>` (D) `0`
Q 2328412301

If `alpha` and `beta` are the roots of the equation
`x^2 + bx + c = 0`, then what is the value of
`alpha^(-1) + beta^(-1)`?
NDA Paper 1 2013
(A)

` - b/a`

(B)

`b/c`

(C)

`c/b`

(D)

` - c/b`

Solution:

Given that, `alpha` and `beta` are the roots of the equation

`x^2 + bx + c = 0`.

Then, sum of the roots `= alpha + beta = (-b)/1 = - b` ... (i)

and product of roots `= alpha . beta = c/1 = c ` .........(ii)

` :. alpha^(-1) + beta^(-1) = 1/alpha + 1/beta = ( alpha + beta)/(alpha beta) = (-b)/c`
Correct Answer is `=>` (A) ` - b/a`
Q 2338812702

If `alpha , beta` are the roots of `x^2 + px - q = 0` and `gamma , delta` are the
roots of `x^2 - px + r = 0`, then what is the value of `( beta + gamma) (beta + delta)`?
NDA Paper 1 2012
(A)

`q + r`

(B)

`p + q`

(C)

`q + r`

(D)

`p + q`

Solution:

Since , `alpha` and `beta` are the roots of `x^2 + px - q = 0`.

`:. alpha + beta = - p, alpha beta = - q`

Since, `gamma ` and `delta` are the roots of `x^2 - px + r = 0`.

`:. gamma + delta = p , gamma delta = r`

`:. (beta + gamma) (beta + delta ) = beta^2 + beta delta + gamma beta + gamma delta = beta^2 + beta (gamma + delta ) + gamma delta`

`= beta^ 2 + beta (P) + gamma delta ( ∵ gamma + delta = p` and ` gamma delta = r)`

`= beta^2 + beta (- alpha - beta ) + r [ ∵ p = -(alpha + beta)]`

`= beta^2 + (- beta ) (alpha + beta ) + r`

`= beta^2 - alpha beta - beta^2 + r = - alpha beta + r`

`= - (-q) + r = q + r `
Correct Answer is `=>` (C) `q + r`
Q 2358645504

If `alpha` and `beta` are the roots of the equation
`4x^2 + 3x + 7 = 0`, then what is the value of
`(alpha^(-2) + beta^(-2))`?
NDA Paper 1 2011
(A)

`47//49`

(B)

`49//47`

(C)

`-47 //49`

(D)

`- 49//47`

Solution:

Since, `alpha` and `beta` are the roots of the equation

`4x^2 + 3x + 7 = 0`.

Now,

`:. alpha + beta = - 3/4` and `alpha beta = 7/4`

Now, `alpha^(-2) + beta^(-2) = 1/alpha^2 + 1/beta^2 = ( alpha^2 + beta^2 )/(alpha beta)^2`

` = (( alpha + beta )^2 - 2 alpha beta )/( alpha beta )^2 = (9/(16) - 7/2)/( (49)/(16))`

` = ( (9 - 56)/(16) )/( (49)/(16)) = ( -47)/(16) . (16)/(49) = - (47)/(49)`
Correct Answer is `=>` (C) `-47 //49`
Q 2328534401

If the roots of the equation `3x^2 - 5x + q = 0` are
equal, then what is the value of `q`?
NDA Paper 1 2011
(A)

`2`

(B)

`5//12`

(C)

`12//25`

(D)

`25//12`

Solution:

Let the roots of the equation,

`3x^2 - 5x + q = 0` is `(alpha , alpha)`.

Then, sum of the roots `= - (- 5/3)`

`=> alpha + alpha = 5/3 => alpha = 5/6`

Since, this roots satisfy the given equation.

`:. 3 (5/6)^2 - 5 (5/6 ) + q = 0 => (75)/(36) - (25)/6 + q = 0`

` => 75 - 150 - 36q = 0 => 36q = 75`

`:. q = (25)/(12)`
Correct Answer is `=>` (D) `25//12`
Q 2348123903

If one of the roots of the equation
`a(b-c)x^2 + b(c-a)x + c(a-b) = 0` is `1`, then
what is the second root?
NDA Paper 1 2011
(A)

` - ( b (c - a))/(a(b - c))`

(B)

` ( b (c - a))/(a(b - c))`

(C)

` ( c (a - b))/(a(b - c))`

(D)

` - ( c (a - b))/(a(b - c))`

Solution:

Given quadratic equation is

`a(b -c)x^2 + b(c- a)x + c(a- b)= 0`

Given that, one root is `1`.

Let the other root be `alpha`.

Then , ` alpha + 1 = - (b (c- a))/(a (b -c))`

`:. a = - 1 - ((bc - ab))/((ab - ac)) = ((-ab + ac- bc + ab))/(a(b - c) ) = ( c (a- b))/(a(b - c))`
Correct Answer is `=>` (C) ` ( c (a - b))/(a(b - c))`
Q 2328123001

What is the sum of the squares of the roots of the
equation `x^2 + 2x - 143 = 0`?
NDA Paper 1 2012
(A)

`170`

(B)

`180`

(C)

`190`

(D)

`290`

Solution:

Given quadratic equation is

`x^2 + 2x- 143 = 0`

Let `(alpha, beta)` be the roots of this equation.

Then, `alpha + beta = -2` and `alpha · beta = -143`

Now, `alpha^2 + beta^2 =(alpha + beta )^2 -2 alpha beta`

`= (-2)^2 - 2 (-143) = 4 + 286= 290`
Correct Answer is `=>` (D) `290`
Q 2338323202

If the difference between the roots of
`ax^2 + bx + c = 0` is `1`, then which one of the
following is correct?
NDA Paper 1 2012
(A)

`b^2 = a(a + 4c)`

(B)

`a^2 = b(b + 4c)`

(C)

`a^2 = c(a + 4c)`

(D)

`b^2 = a(b + 4c)`

Solution:

Let the roots of the equation `ax^2 + bx + c = 0` are `alpha`

and `(alpha - 1)` by given condition.

Then, sum of the roots `= - b/a`

`=> alpha + (alpha - 1) = - b/a`

`=> 2alpha = 1 - b/a => alpha = (a - b)/(2a)`

and product of roots `= c/a`

` => alpha (alpha - 1 ) = c/a`

`=> (a- b)/(2a) { (a - b)/(2a) - 1 } = c/a`

` => (a-b)/(2a) . (-b-a)/(2a) = c/a`

`=> -(a^2 - b^2) = 4ac`

`=> b^2 - a^2 = 4ac`

`:. b^2 = a(a + 4c)`
Correct Answer is `=>` (A) `b^2 = a(a + 4c)`
Q 2338423302

If one of the roots of the equation `x^2 + ax - b = 0`
is `1`, then what is the value of `(a-b)`?
NDA Paper 1 2012
(A)

`-1`

(B)

`1`

(C)

`2`

(D)

`-2`

Solution:

Let the roots of the given equation `x^2 + ax - b = 0` are

`1` and `alpha` by given condition.

`:.` Sum of the roots `= - a`

i.e., `alpha + 1 = -a`

`alpha = -1 -a` ......(i)

and product of the roots `= - b`

i.e., `alpha ·1 = -b`

`alpha = -b` ......(ii)

From Eqs. (i) and (ii),

`-b = - 1 - a => a - b = - 1`

Alternate Method

Since, one of the roots of the equation is `1`.

Hence, satisfy the given equation

`(1)^2 + a(1) - b = 0`

`=> a+a-b = 0 => a - b = -1`
Correct Answer is `=>` (A) `-1`
Q 2318523400

If `alpha` and `beta` are the roots of the equation
`x^2 - q(1+x)- r = 0`, then what is the value of
`(1 + alpha) (1 + beta)`?
NDA Paper 1 2012
(A)

`1 - r`

(B)

`q- r`

(C)

`1 + r`

(D)

`q + r`

Solution:

Given that, `alpha` and `beta` be the roots of the equation

`x^2 - q(1 + x) - r = 0`

i.e., `x^2 - qx - (q + r) = 0`.

Then, sum of the roots `= q`

i.e., `alpha + beta = q` ... (i)

and product of the roots `= - (q + r)`

i.e., `alpha beta = -(q + r)` ... (ii)

Now, `(1 + alpha ) (1 + beta ) = 1 + (alpha + beta ) + alpha beta = 1 + q - (q + r)`

`= 1- r`
Correct Answer is `=>` (A) `1 - r`
Q 2260467315

If the sum of the roots of the equation `ax ^(2) + bx + c = 0` is
equal to the sum of their squares, then
NDA Paper 1 2015
(A)

`a^(2) + b^(2) = c^(2)`

(B)

`a^(2) + b^(2) = a +b`

(C)

`ab + b^(2) = 2ac`

(D)

`ab - b^(2) = 2ac`

Solution:

We have `ax ^(2) + bx + c = 0`

Sum of roots, i.e. `alpha + beta = - b/a`

and product of roots, i.e ` alpha beta = c/a `

It is given that, `alpha + beta = alpha^(2) + beta^(2)`

`= alpha + beta = (alpha + beta)^(2) -2alpha beta`

` => - b/a =( - b/a)^(2) - 2 (c/a)`

` => - b/a =( b)^(2)/(a)^(2) -(2 c)/a`

` => - b/a = ( b^(2) - 2ac)/a^(2) `

` => -ab = b^(2) - 2ac => b^(2) = 2ac - ab `

`=> b^(2) + ab = 2ac`
Correct Answer is `=>` (C) `ab + b^(2) = 2ac`
Q 1732434332

If `alpha , beta ` are the roots of `ax^2 + bx + c = 0` and `alpha + h, beta + h`
are the roots of `px^2 + qx + r = 0`, then what is `h` equal to?
NDA Paper 1 2014
(A)

` 1/2 (b/a - q/p)`

(B)

` 1/2 (- b/a + q/p)`

(C)

` 1/2 (b/p + q/a)`

(D)

` 1/2 (- b/p + q/a)`

Solution:

` ∵ alpha + beta = -b/a ` and ` alpha beta =c/a `

Also ` alpha + h + beta + h = - q/p`

`=> alpha + beta + 2h = - q/p`

` => 2h = -q/p + b/a quad ( ∵ alpha + beta = - b/a)`

` => h = 1/2 [b/a - q/p]`
Correct Answer is `=>` (A) ` 1/2 (b/a - q/p)`
Q 1782280137

If `alpha` and `beta` are the roots of the equation `ax^2 + bx + c = 0`,
where `a != 0`, then `(a alpha +b) (a beta + b)` is equal to
NDA Paper 1 2014
(A)

`ab`

(B)

`bc`

(C)

`ca`

(D)

`abc`

Solution:

Given that, `alpha` and `beta` are the roots of the

equation `ax^2 + bx + c = 0`, where `a != 0`.

Then, sum of roots `= alpha + beta = (-b)/a`

and product of roots `=alpha . beta = c/a`

Now, we have

`(aalpha + b) (a beta +b) = a^2 (alpha beta) + ab (alpha+ beta) + b^2`

`=a^2 (c/a) + ab(- b/a) + b^2`

` = ac - b^2 + b^2`

` = ac`
Correct Answer is `=>` (C) `ca`
Q 2250291114

If the roots of the equation `x^( 2) - nx + m = 0` differ by `1`,
then
NDA Paper 1 2015
(A)

`n^(2) - 4m- 1 = 0`

(B)

`n^(2) + 4m- 1 = 0`

(C)

`m^(2) + 4n + 1 = 0`

(D)

`m^(2) - 4n - 1 = 0`

Solution:

We have, `x^(2) - nx + m =0`

`:. alpha + beta = n , alpha beta = m`

and given that, `alpha - beta = 1`

We know that, `(alpha+ beta)^(2) -( alpha - beta ) ^(2) = 4 alpha beta`

`=> n^(2) -1 = 4m`

`=> n^(2) - 4m- 1 = 0`
Correct Answer is `=>` (A) `n^(2) - 4m- 1 = 0`
Q 1543291143

If `\alpha` and `\beta` are the roots of the quadratic equation `4x^2+3x+7=0`, then the value of ` \frac{1}{\alpha}+\frac{1}{\beta}` is:
BITSAT 2012
(A)

`- \frac{3}{4}`

(B)

`- \frac{3}{7}`

(C)

`\frac{3}{7}`

(D)

`\frac{4}{7}`

Solution:

If `\alpha, \beta` are the roots of quadratic `4x^2+3x+7=0`, then

`\alpha+\beta=-\ frac{3}{4}` and `\alpha\beta =\ frac{7}{4}`

Hence `\ frac{1}{\alpha}+\ frac{1}{\beta}=\ frac{\alpha+\beta}{\alpha\beta}=\ frac {-3}{4}\times \ frac {4}{7}`

`=-\ frac{3}{7}`
Correct Answer is `=>` (B) `- \frac{3}{7}`
Q 2529291111

If `alpha` and `beta` are the roots of the equation `x^2 + px - 1/(2p^2) = 0`, where `p in R`. Then, the minimum value of `alpha^4 + beta^4` is
BCECE Mains 2015
(A)

`2`

(B)

`2 + sqrt2`

(C)

`2 sqrt2`

(D)

`2 - sqrt2`

Solution:

Here, ` alpha + beta = -p`

and ` alpha beta = - 1/(2p^2)`

Now , ` alpha^4 + beta^4 = (alpha^2 + beta^2)^2 - 2 alpha^2 beta^2`

` = [ ( alpha + beta)^2 - 2 alpha beta ]^2 - 2(alpha beta)^2`

` = [ p^2 + 1/p^2 ]^2 - 2( 1/(4p^4) )`

` = p^4 + 1/(2p^4) + 2`

` = ( p^2 - 1/(sqrt2 p^2) )^2 + 2 + sqrt2`

` >= 2 + sqrt2`

Thus, minimum value of `alpha^4 + beta^4` is `2 + sqrt2`.
Correct Answer is `=>` (B) `2 + sqrt2`
Q 2469191915

If `alpha` and `beta` are the roots of `x^2 + 5x + 4 = 0`, then equation whose roots are `(alpha +2)/3` and `(beta +2)/3`, is
BCECE Stage 1 2014
(A)

`9x^2+ 3x + 2 = 0`

(B)

`9x^2 - 3x + 2 = 0`

(C)

`9x^2+ 3x - 2 = 0`

(D)

`9x^2 - 3x + 2 = 0`

Solution:

Given. `alpha` and `beta` are the roots of the equation `x^2 + 5x +4 =0`

`:. alpha +beta = -5 ` and `alpha beta = 4`

Now, `(alpha +2 )/3 + (beta +2 )/3 = (alpha +beta +4)/3`

`= (-5 +4)/3 = -1/3`

and `((alpha+2 )/3 ) (( beta +2 )/3) = (alpha beta +2 (alpha +beta) +4)/9`

`= (4 + 2(-5) + 4) /9 = -2/9`

`:.` Required equation is

`x^2` - (sum of roots) x + product of roots =0

`=> x^2 + 1/3 x - 2/9 =0`

`=> 9x^2 + 3x -2 =0`
Correct Answer is `=>` (C) `9x^2+ 3x - 2 = 0`
Q 2580134017

If `alpha`, and `beta` are the roots of `ax^2 + c = bx`, then
the equation `(a + cy)^2 = b^2 y` in `y` has the
roots
BCECE Stage 1 2013
(A)

`alpha^(-1) , beta^(-1)`

(B)

`alpha^2 , beta^2`

(C)

`alpha beta^(-1) , alpha^(-1) beta`

(D)

`alpha^(-2) , beta^(-2)`

Solution:

`ax^2 - bx + c = 0`

`alpha + beta = b/a , alpha beta = c/a`

Also, `(a + cy)^2 = b^2y`

` => c^2y^2 - (b^2 - 2ac)y + a^2 = 0`

`=> (c/a)^2 y^2 - {(b/a)^2 -2 (c/a)} y + 1 = 0`

`=> (alpha beta )^2 y^2 - (alpha^2 + beta^2) y + 1 = 0`

`=> y^2 - (alpha^(-2) + beta^(-2)) y + alpha^(-2) beta^(-2) = 0`

` => (y - alpha^(-2)) (y - beta^(-2)) = 0`

So, the roots are `alpha^(-2)` and `beta^(-2)`.
Correct Answer is `=>` (D) `alpha^(-2) , beta^(-2)`
Q 2773680546

If the roots of the equation `x^2 + px + q = 0` are in the same ratio as those of the equation `x^2 + lx + m = 0`, then which one of the following is correct ?
NDA Paper 1 2017
(A)

`p^2 m = l^2q`

(B)

`m^2p = l^2q`

(C)

`m^2p = q^2l`

(D)

`m^2p^2 = l^2q`

Solution:

If ration is same of given equations , then

`(-p + sqrt(p^2 -4q))/(-p - sqrt(p^2 -4q)) = (-l + sqrt(l^2 -4m))/(-l - sqrt(l^2 -4m))`

by C & D theorem

`(-2p)/(2 sqrt(p^2- 4q)) =(-2l)/(2 sqrt(l^2 - 4m))`

`(p^2)/(p^2- 4q) =(l^2)/(l^2- 4m)`

`p^2m = l^2 q`
Correct Answer is `=>` (A) `p^2 m = l^2q`

Location of roots

See examples
Q 2731245122

If `c > 0` and `4 a +c < 2 b `, then `ax^2 - bx + c = 0` has a root in which one of the following intervals ?
NDA Paper 1 2016
(A)

`(0,2)`

(B)

`(2,3)`

(C)

`(3,4)`

(D)

`(-2, 0)`

Solution:

`f(0) = C >0`

`f(2) = 4a -2b +c < 0`

As `f(x) ` changes sign between `0` to `2` and is continuous

`=>` one root between `(0,2)`
Correct Answer is `=>` (A) `(0,2)`
Q 2781745627

If both the roots of the equation `x^2 - 2 kx + k^2 - 4 = 0` lie between -3 and 5 , then which one of the following is correct ?
NDA Paper 1 2016
(A)

`- 2 < k < 2 `

(B)

`-5 < k < 3`

(C)

`-3 < k < 5`

(D)

`-1 < k < 3`

Solution:

Roots of the equation `x^2 - 2 kx +k^2 - 4 = 0` are

`(-(-2 k ) pm sqrt ((-2 k)^2 - 4 (k^2 -4) ))/(2 xx 1) `

`= (2 k pm sqrt 16 )/2 = k pm 2`

As given roots are lie between `-3` and 5, so

`-3 < k +2 < 5 ` and ` -3 < k -2 < 5`

`=>
Correct Answer is `=>` (A) `- 2 < k < 2 `
Q 2348112003

The quadratic equation `x^2 + bx + 4 = 0` will have
real roots, if
NDA Paper 1 2013
(A)

Only `b <= -4`

(B)

Only `b >= 4`

(C)

`-4 < b < 4`

(D)

`b <= - 4, b >= 4`

Solution:

Given that, the equation `x^2 + bx + 4 = 0` have real

roots, if discriminate `(D) = B^2 - 4AC >= 0`

`=> b^2 - 4 (1) (4) >= 0 => b^2 - 16 >= 0`

`=> (b - 4) (b + 4) >= 0`

` :. b <= -4 , b >= 4`
Correct Answer is `=>` (D) `b <= - 4, b >= 4`
Q 2544134053

Find all values of a for which the equation `4x^2 - 2x + a = 0` has two roots lie in the interval `(- 1, 1)`.

Solution:

Let `f(x) = 4x^2 - 2x + a` as both roots of the equation, `f(x) = 0` are

lie between `(-1, 1)`, we can take `D >= 0, af(-1) > 0, af(1) > 0` and `-1 < 1/4 < 1`.

(i) Consider ` D >= 0`

`(-2)^ 2 - 4·4·a >= 0`

`=> a <= 1/4` ......(i)

(ii) Consider `af (- 1) > 0`

`4(4+2+ a) > 0`

`=> a > - 6`

`=> a in (-6, oo)` ........(ii)

(iii) Consider `a f (1) > 0`

`4(4 - 2 + a) > 0`

`=> a > -2`

` => a in (-2 ,oo)` .......(iii)

Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at the same time are

` a in ( -2, 1/4 ]`
Q 2781745627

If both the roots of the equation `x^2 - 2 kx + k^2 - 4 = 0` lie between -3 and 5 , then which one of the following is correct ?
NDA Paper 1 2016
(A)

`- 2 < k < 2 `

(B)

`-5 < k < 3`

(C)

`-3 < k < 5`

(D)

`-1 < k < 3`

Solution:

Roots of the equation `x^2 - 2 kx +k^2 - 4 = 0` are

`(-(-2 k ) pm sqrt ((-2 k)^2 - 4 (k^2 -4) ))/(2 xx 1) `

`= (2 k pm sqrt 16 )/2 = k pm 2`

As given roots are lie between `-3` and 5, so

`-3 < k +2 < 5 ` and ` -3 < k -2 < 5`

`=>
Correct Answer is `=>` (A) `- 2 < k < 2 `
Q 1467156085

The real values of `A` for which the equation,
`3x^3 + x^ 2- 7 x + lambda= 0`, has two distinct real roots in `[0, 1]`
lie in the intetval (s)

(This question may have multiple correct answers)

(A) `(- 2, 0)`
(B) `[0, 1]`
(C) `[1, 2]`
(D) `(-oo,oo)`
Solution:

Given equation is

`3x^3 + x^ 2- 7 x + lambda = 0` ....................(i)

Let `f(x) = 3x^3 + x^ 2 -7x +lambda`

`:.` `f' (x) = 9x^2 + 2x -7`

`=9(x + 1)(x -7/9)`

For max or min `f' (x) = 0`

`:.` `x=- 1, x = 7/9`

`7/9 in [0,1]`

Hence Eq. (i) has two distinct real roots in `[0, 1`] for all
values of ` lambda`

`:.` `lambda in (-oo,oo)`
Correct Answer is `=>` (A)
Q 2431145022

All the values of `m` for which both the roots of
the equation `x ^2- 2mx + m^2- 1 = 0` are greater
than `-2` but less than `4`, lie in the interval
UPSEE 2015
(A)

`-2 < m < 0`

(B)

`m > 3`

(C)

`-1 < m < 3`

(D)

`1 < m < 4`

Solution:

The given equation is

`x^2 - 2mx + m^2 -1 = 0`

`=> (x-m)^2-1=0`

`=> (x-m-1)(x-m+1)=0`

`=> x=m+1` or `x=m-1`

According to the given condition

`m-1 > =-2` and `m+1 < 4`

`=> m > =-1` and `m < 3`

Hence, `-1 < m < 3`
Correct Answer is `=>` (C) `-1 < m < 3`
Q 1458556404

Let `alpha, beta in R` be the roots of the equation `ax^2 + bx + c = 0`.
Then `k in R` lies between `alpha` and `beta` if ......

(A)

`a^2k + abk + ac < 0`

(B)

`a^2k^2 + ab + ac < 0`

(C)

`a^2k^2 + abk + ac < 0`

(D)

`ak^2 + abk + ac < 0`

Solution:

If ` a > 0` or `a < 0`

Then `af(k) <0`

`a (a k^2 +bk +c) <0`

`=> a^2 k^2 +abk+ac < 0`
Correct Answer is `=>` (C) `a^2k^2 + abk + ac < 0`
Q 1147780683

If α, β are the roots of `x^2 – 3x + a = 0`, `a ⊂ R` and `α < 1 < β` then,

(A)

`a⊂ (−∞,9/4)`

(B)

None of these.

(C)

`a ⊂ ( – ∞ , 2 )`

(D)

`(−2,9/4)`

Solution:

According to question, 1 lies between the roots, therefore,

`f (1) < 0`

`⇒ 1 - 3 + a < 0`

`∵ a < 2` or `a ∈ ( - ∞, 2).`

The correct answer is: `a ⊂ ( – ∞ , 2 )`
Correct Answer is `=>` (C) `a ⊂ ( – ∞ , 2 )`

Quadratic graph

See examples
Q 2743880743

If the graph of a quadratic polynomial lies entirely above x-axis, then which one of the following is correct?
NDA Paper 1 2017
(A)

Both the roots are real

(B)

One root is real and the other is complex

(C)

Both the roots are complex

(D)

Cannot say

Solution:

If graph lies entirely above the x-axis mean , it doesn't cut on x-axis , so quadratic equation has no real roots or both roots will be complex.
Correct Answer is `=>` (C) Both the roots are complex
Q 2146478373

If `x^( 2) - px + 4 > 0` for all real values of `x`, then which one

of the following is correct?
NDA Paper 1 2016
(A)

` |P| < 4`

(B)

` |P| <= 4`

(C)

` |P| > 4`

(D)

` |P| >= 4`

Solution:

`:. x^(2) - px + 4 > 0`

Here, `a > 0` and `f(x) > 0`

`:. D < O`

`:. p^( 2) -16 < 0`

`=> p^( 2) < 16`

`=> |p| < 4`
Correct Answer is `=>` (A) ` |P| < 4`
Q 2389401317

If roots of an equation `ax^2 + bx + c = 0` are
positive, then which one of the following is
correct?
NDA Paper 1 2007
(A)

Signs of a and c should be like

(B)

Signs of b and c should be like

(C)

Signs of a and b should be like

(D)

None of the above

Solution:

If roots of an equation `ax^2 + bx + c = 0` are positive,

then signs of `a` and `c` should be same (like).
Correct Answer is `=>` (A) Signs of a and c should be like
Q 2743880743

If the graph of a quadratic polynomial lies entirely above x-axis, then which one of the following is correct?
NDA Paper 1 2017
(A)

Both the roots are real

(B)

One root is real and the other is complex

(C)

Both the roots are complex

(D)

Cannot say

Solution:

If graph lies entirely above the x-axis mean , it doesn't cut on x-axis , so quadratic equation has no real roots or both roots will be complex.
Correct Answer is `=>` (C) Both the roots are complex
Q 2613723640

The graph of the quadratic polynomial `y = ax^2 + bx +c` is as shown in the figure. Then

(A)

`b^2 - 4ac > 0`

(B)

`b < 0`

(C)

`a > 0`

(D)

`c < 0`

Solution:

`f(0) < 0 => c < 0`

Roots are real & distinct ` => D > 0`

product of roots `< 0`

`=> c/a < 0 => a > 0`

Sum of roots ` = - b/a > 0 => b < 0`
Correct Answer is `=>` (A)

Formation of an equation

Q 1628423301

In solving a problem that reduces to a quadratic
equation, one student makes a mistake in the
constant term and obtains `8` and `2` for roots.
Another student makes a mistake only in the
coefficient offirst degree term and finds `-9` and `-1`
for roots. The correct equation is
NDA Paper 1 2015
(A)

`x^ 2 - 10 x + 9 = 0`

(B)

`x^ 2 + 10 x + 9 = 0`

(C)

`x^ 2 - 10 x + 16 = 0`

(D)

`x^ 2 - 8 x - 9 = 0`

Solution:

Given, roots are `8` and `2`, then the equation is

`x^2 -` (sum of roots) `x +` product of roots `= 0`

`=> x^ 2 - (8+2) x + 16 = 0`

`=> x ^2 -10 x + 16 = 0 ....... (i)`

(mistake in the constant term)

For roots `-9` and `-1`, we have

`x ^2 (-9x - 1 ) x + ( -9) x (-1) = 0`

`=> x^ 2 - 9x + 9 = 0 .... (ii)`

(mistake in the coefficient of first degree term)

But as per given information. we have the right

equation as `x ^2 - 10x + 9 = 0`
Correct Answer is `=>` (A) `x^ 2 - 10 x + 9 = 0`
Q 2318101900

If `a` and `b` are rational and `b` is not perfect square,
then the quadratic equation with rational
coefficients whose one root is `3a + sqrt(b)` is
NDA Paper 1 2013
(A)

`x^2 - 6ax + 9a^2 - b = 0`

(B)

`3ax^2 + x - sqrt(b) = 0`

(C)

`x^2 + 3ax + sqrt(b) = 0`

(D)

` sqrt(b)x^2 + x - 3a = 0`

Solution:

If one root of any quadratic equation is in the form

`3a + sqrt(b)`. then other root of this equation should be `3a - sqrt(b)`.

`:.` Required equation is

`x^2 -` (Sum of roots) · x + (Product of roots)` = 0`

`=> x^2 - {(3a + sqrt(b)) + (3a - sqrt(b)} ·x + {( 3a + sqrt(b))`

` (3a - sqrt(b))} = 0`

`=> x^2 - 6ax + 9a^2 - b = 0`
Correct Answer is `=>` (A) `x^2 - 6ax + 9a^2 - b = 0`
Q 2318112009

If the roots of a quadratic equation
`ax^2 + bx + c = 0` are `alpha` and `beta`, then the quadratic
equation having roots `alpha^2` and `beta^2` is
NDA Paper 1 2013
(A)

`x^2 - (b^2 - 2ac) x + c = 0`

(B)

`a^2x^2 - (b^2 - 2ac)x + c = 0`

(C)

`ax^2 - (b^2 - 2ac)x + c^2 = 0`

(D)

`a^2x^2 - (b^2 - 2ac)x + c^2 = 0`

Solution:

Given quadratic equation is `ax^2 + bx + c = 0`.

and its root are `alpha` and `beta`.

`:.` Sum of the roots `= alpha + beta = (-b)/a`

and product of the roots `= alpha . beta = c/a`

` ∵ alpha^2 + beta^2 = ( alpha + beta)^2 - 2 alpha beta = ( - b/a)^2 -2 . c/a`

` = b^2/a^2 - (2c)/a = (b^2 - 2ac)/a^2 `

and ` alpha^2 . beta^2 = ( alpha beta)^2 = (c/a)^2 = c^2/a^2`

` :.` Required quadratic equation whose roots are `alpha^ 2` and `beta^2` is

`x^2 - (alpha^2 + beta^2)x+ alpha^2 beta^2 = 0`

` => x^2 - (b ^2 - 2ac)/a^2 x + c^2/a^2 = 0`

`:. a^2x^2 - (b^2 - 2ac) x + c^2 = 0`
Correct Answer is `=>` (D) `a^2x^2 - (b^2 - 2ac)x + c^2 = 0`
Q 2338312202

If the sum of the roots of a quadratic equation is `3`
and the product is `2`, then the equation is
NDA Paper 1 2013
(A)

`2x^2 - x + 3= 0 `

(B)

`x^2 - 3x + 2 = 0`

(C)

`x^2 + 3x + 2 = 0`

(D)

`x^2 - 3x - 2 = 0`

Solution:

Let the required quadratic equation is

`ax^2 + bx + c= 0` ... (i)

Now, sum of the roots `= (-b)/a = 3 = (-(-3))/1` (given)

and product of the roots `= c/a = 2 = 2/1` (given)

`:. a = 1, b = -3` and `c = 2`

Hence, the required quadratic equation is `x^2 - 3x + 2 = 0`.
Correct Answer is `=>` (B) `x^2 - 3x + 2 = 0`
Q 2328334201

If let `alpha` and `beta` be the roots of the equation
`x^2 + x + 1 = 0`, then the equation whose roots are
`alpha^( 19)` and `beta^7` is
NDA Paper 1 2011
(A)

`x^2 - x - 1 = 0`

(B)

`x^2 - x + 1 = 0`

(C)

`x^2 + x - 1 = 0`

(D)

`x^2 + x + 1 = 0`

Solution:

The given quadratic equation is

`x^2 + x + 1 = 0`

` x = (-1 ± sqrt(1-4))/2 = (-1 ± i sqrt3)/2`

`x = (-1+ i sqrt3)/2 , (-1-i sqrt3)/2`

or `x = omega , omega^2`

i.e., `alpha = omega` and `beta = omega^2`

`=> alpha^(19) = omega^(19) = ( omega^3)^6 · omega = (1)^6 . omega = omega`

`=> beta^7 = ( omega^2)^7 = omega^(14) = ( omega^3)^4. omega^2 = (1)^4. omega^2 = omega^2`

Now, sum of roots, `alpha^( 19) + beta^( 7) = omega + omega^2 = -1 ( ∵ 1 + omega + omega^2 = 0)`

and product of roots , `alpha^(19) - beta^7 = omega . omega^2 = omega^3 = 1 ( ∵ omega^3 = 1)`

So, the required quadratic equation whose roots are `alpha^( 19)` and `beta^7` is

`x^2 - (alpha^(19) + beta^7)x + (alpha^(19) · beta^7) = 0`

` => x^2 - (-1)x + (1) = 0`

` => x^2 + x + 1 = 0`
Correct Answer is `=>` (D) `x^2 + x + 1 = 0`
Q 2388512407

If the roots of a quadratic equation are `m + n` and
`m - n`, then the quadratic equation will be
NDA Paper 1 2012
(A)

`x^2 + 2 mx + m^2 - mn + n^2 = 0`

(B)

`x^2 + 2 mx + (m - n)^2 = 0`

(C)

`x^2 - 2 mx + m^2 - n^2 = 0`

(D)

`x^2 + 2 mx + m^2 - n^2 = 0`

Solution:

Given that, `m + n` and `m- n` be the roots of given

equation

Sum of roots `= (m + n) + (m - n)= 2m`

and product of roots `= (m + n)(m- n)= m^2 - n^2`

So, the required quadratic equation is

`x^2 -` (Sum of roots )x + Product of roots `= 0`

`=> x^2 - 2mx + (m^2 - n^2 ) = 0`
Correct Answer is `=>` (C) `x^2 - 2 mx + m^2 - n^2 = 0`
Q 2318267109

If `1/(2 - sqrt(-2)) ` is one of the roots of `ax^2 + bx + c = 0`,
where, `a, b` and `c` are real, then what are the values
of `a, b` and `c` respectively?
NDA Paper 1 2010
(A)

6, - 4, 1

(B)

4, 6, - 1

(C)

3,- 2, 1

(D)

6, 4, 1

Solution:

Given, root `= 1/(2 - sqrt(-2)) = 1/(2 - sqrt(2i) ) . (2 + sqrt(2) i)/(2 + sqrt(2) i)`

` = (2 + sqrt(2) i)/(4 + 2) = (2 + sqrt(2) i)/6`

So, the another root will be ` (2 - sqrt(2) i)/6`

Thus, sum of roots `= (2 + sqrt(2) i)/6 + (2 - sqrt(2) i)/6 = 4/6`

and product of roots `= ((2 + sqrt(2) i)/6) ( (2 - sqrt(2) i)/6) = (4 + 2)/(36) = 1/6`

`:.` Required equation is

`x^2 -` (sum of roots) x + (product of roots) `= 0`

` => x^2 - 4/6 x + 1/6 = 0`

` => 6x^2 - 4x + 1 = 0`

Thus, the values of `a, b` and `c` are `6,- 4` and `1`, respectively.
Correct Answer is `=>` (A) 6, - 4, 1
Q 2388767607

If `p` and `q` are positive integers, then which one of
the following equations has `p - sqrtq` as one of its
roots?
NDA Paper 1 2010
(A)

`x^2 - 2px - (p^2 - q) = 0`

(B)

`x^2 - 2px + (p^2 - q) = 0`

(C)

`x^2 + 2px - (p^2 - q) = 0`

(D)

`x^2 + 2px + (p^2 - q) = 0`

Solution:

If any equation has `p - sqrt q`. as a root, then the another

root will be `p + sqrt q`.

`:.` Sum of roots `= p - sqrt q + p + sqrt q = 2p`,

and product of roots `= (p - sqrt q) (p + sqrt q ) = p^2 - q`

Now, the required equation is

`x^2 -` (sum of roots)x + (product of roots) `= 0`

` => x^2 - 2 px + (p^2 - q) = 0`
Correct Answer is `=>` (B) `x^2 - 2px + (p^2 - q) = 0`

Solution of quadratic equation

Q 1658156904

Let `alpha` be the root of the
equation `25 cos^2 theta + 5 cos theta - 12 = 0`, where ` pi/2 < alpha < pi`.

What is the tan `alpha` equal to?
NDA Paper 1 2015
(A)

`- 3/4`

(B)

`3/4`

(C)

` - 4/3`

(D)

` -4/5`

Solution:

`25cos^2 theta + 5cos theta -12 = 0`

`=> 25cos^2 theta + 20cos theta -15 cos theta -12 = 0`

` => 5cos theta (5 cos theta + 4)- 3 (5 cos theta + 4 ) = 0`

`=> (5cos theta - 3)(5cos theta + 4) = 0`

`=> cos theta = 3/5 ` or `cos theta = -4/ 5`

` alpha in ( pi/2 , pi)`

Since, `alpha` is a root of the given equation.

`:. cos alpha = 3/2` and `cos alpha = -4/5`

:. We take `cos alpha = -4/5 => tan alpha =- 3/4`
Correct Answer is `=>` (A) `- 3/4`
Q 2338512402

The area of a rectangle whose length is five more
than twice its width is `75` sq units. The length is
NDA Paper 1 2013
(A)

`5` units

(B)

`10` units

(C)

`15` units

(D)

`20` units

Solution:

Let the width of the rectangle is x, then by condition its

length `= 5+ 2x`

Given that, Area of the rectangle `= 75`

`=>` Length `xx` Width `=75 => (5 + 2x) xx x = 75`

`=> 2x^2 + 5x - 75 = 0`

` => 2x^2 + 15x - 10x -75 = 0`

`=> x(2x + 15) -5(2x+15) = 0`

`=> (2x + 15) (x - 5) = 0`

`:. x = 5 , (-15)/2 ( ∵ x != -(15)/2)`

`:.` Required length `= 2x + 5 = 2(5) + 5 = 10 + 5 = 15` units
Correct Answer is `=>` (C) `15` units
Q 2318134000

What are the roots of the equation
`2 (y + 2)^2 - 5 (y + 2) = 12 ?`
NDA Paper 1 2011
(A)

`-7//2,2`

(B)

`-3//2,4`

(C)

`-5//3,3`

(D)

`3//2,4`

Solution:

The given quadratic equation is

`2(y + 2)^2 - 5(y + 2) = 12`

`=> 2(y+ 2)^2- 5(y+ 2)-12 = 0`

Let `z =y + 2` .....(i)

`=> 2 z^2 - 5z - 12 = 0`

` => 2z^2 - 8z + 3z- 12 = 0`

` => (2z + 3) (z- 4) = 0`

`:. z = - 3/2 , 4`

`=> y + 2 = - 3/2 , 4` [from Eq. (i)]

`=>y = - 2 - 3/2 , 4 -2`

`:. y = - 7/2 , 2`

So, the required roots are `- 7/2` and `2`.
Correct Answer is `=>` (A) `-7//2,2`
Q 2348434303

What is the value of
` sqrt(8 + 2 sqrt (8 + 2 sqrt(8 + 2 sqrt(8 + .... oo))))` ?
NDA Paper 1 2011
(A)

10

(B)

8

(C)

6

(D)

4

Solution:

Let `y = sqrt(8 + 2 sqrt (8 + 2 sqrt(8 + 2 sqrt(8 + .... oo)))) `

` => y = sqrt(8+2y) => y^2 = 8 + 2y`

`=> y^2 - 2y - 8 = 0 => y^2 - 4y + 2y - 8 = 0`

`=> (y- 4)(y + 2) = 0`

`:. y = - 2, 4`

So, the required value of expression is ` 4`.
Correct Answer is `=>` (D) 4
Q 2318167900

If `p` and `q` are the roots of the equation
`x^2 - px + q = 0`, then what are the values of `p` and `q`
respectively'?
NDA Paper 1 2010
(A)

1, 0

(B)

0, 1

(C)

-2, 0

(D)

-2, 1

Solution:

Since, `p` and `q` are the roots of `x^2 - px + q = 0`.

then, `p + q = p` and `pq = q`

` => q(p- 1) = 0 => q = 0, p = 1`
Correct Answer is `=>` (A) 1, 0
Q 2358778604

Which one of the following is one of the roots of
the equation `(b -c)x^2 + (c -a)x +(a -b) =0`?
NDA Paper 1 2009
(A)

` (c-a)/(b - c)`

(B)

` (a - b)/(b - c)`

(C)

` ( b - c)/(a - b)`

(D)

`(c -a)/(a - b)`

Solution:

`(b- c)x^2 + (c - a)x + (a- b) = 0`

`=> (b- c)x^2 - (b- c - b + a)x +(a- b)= 0`

`=> (b- c)x(x - 1)- (a- b)(x- 1) = 0`

`=> {(b- c)x- (a- b)} (x- 1) = 0`

`:. x = (a - b)/(b - c)` and `x = 1`
Correct Answer is `=>` (B) ` (a - b)/(b - c)`
Q 2309101018

The numerical value of the perimeter of a square
exceeds that of its area by `4`. What is the side of
the square?
NDA Paper 1 2008
(A)

`1` unit

(B)

`2` units

(C)

`3` units

(D)

`4` units

Solution:

Let the side of the square `= x` units

`:.` Area of square `= x^2` units

and perimeter of the square `= 4x` units

According to the question,

`x^2 + 4 = 4x`

`=> x^2 - 4x + 4 = 0 => (x - 2)^2 = 0`

`=> x = 2`

`:.` Side of square `= 2` units
Correct Answer is `=>` (B) `2` units
Q 2250467314

The number of real roots of the equation
`x^2 - 3 | x | + 2 = 0` is
NDA Paper 1 2015
(A)

`4`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

We have, `x^2 - 3 | x |+ 2 = 0 => | x |^2 -3|x| + 2 = 0`

`=> | x|^2 - 2 | x| - | x| + 2 = 0`

`=> (| x | - 2) (| x | - 1) = 0`

` => | x | = 2` or `| x | = 1`

`=> x = ± 2` or `x = ± 1`

`:.` There are four real roots of the equation.
Correct Answer is `=>` (A) `4`
Q 2378723606

If `alpha` and `beta` be the roots of the equation
`(x- a) (x -b) = c, c != 0`. Then, the roots of the
equation `(x -alpha) (x - beta) + c = 0` are
NDA Paper 1 2011
(A)

`(a, c)`

(B)

`(b, c)`

(C)

`(a, b)`

(D)

`(a + b, a + c)`

Solution:

Given quadratic equation is

`(x- a)(x- b) = c. c != 0`

`=> x^2 - (a+ b)x + (ab-c)= 0`

Since, the roots of this equation is `alpha` and `beta`.

Then, `alpha + beta = -{ - (a+b)/1} = a + b` ......(i)

and `alpha beta = ab - c` ........(ii)

Now, consider the equation,

`(x- alpha) (x - beta) + c = 0`

` => x^2 - (alpha + beta )x + (alpha beta +c) = 0`

From Eqs. (i) and (ii),

`x^2 - (a+ b)x + (ab-c +c)= 0`

` => x^2 - (a+ b)x + ab = 0`

So, the roots of this equation is `(a, b)`.
Correct Answer is `=>` (C) `(a, b)`
Q 2348045803

If `3` is the root of the equation `x^2 - 8x + k = 0`,
then what is the value of `k`?
NDA Paper 1 2011
(A)

`- 15`

(B)

`9`

(C)

`15`

(D)

`24`

Solution:

Since, `3` is the root of the equation `x^2 - 8x + k = 0`.

`:. 9 - 24 + k = 0 => k = 15`
Correct Answer is `=>` (C) `15`
Q 2318656509

Which of the following are the two roots of the
equation `(x^2 + 2)^2 + 8x^2 = 6x (x^2 + 2)`?
NDA Paper 1 2010
(A)

`1 ± i`

(B)

`2 ± i `

(C)

`1 ± sqrt(2)`

(D)

`2 ± i sqrt(2)`

Solution:

` ∵ (x^2 + 2)^2 + 8x^2 = 6x(x^2 + 2)`

`=> x^4 + 4x^2 + 4 + 8x^2 = 6x^3 + 12 x`

`=> x 4 - 6x^3 + 12x^2 - 12x + 4 = 0`

Also, this equation is satisfied by `1 ± i`. Hence, required roots are

`1 ± i`.
Correct Answer is `=>` (A) `1 ± i`
Q 2379201116

The number of rows in a lecture hall equals to the
number of seats in a row. If the number of rows is
doubled and the number of seats in every row in
reduced by `10`, the number of seats is increased by
`300`. If `x` denotes the number of rows in the lecture
hall, then what is the value of `x`?
NDA Paper 1 2007
(A)

`10`

(B)

`15`

(C)

`20`

(D)

`30`

Solution:

` ∵` Number of rows `= x`

`:.` Number of seats in each row `= x`

Thus, total number of seats `= x^2`

Now, new number of rows `= 2x`

and new number in each row `= x - 10`

Thus, total number of seats `= 2x(x - 10) = 2x^2 - 20x`

According to the question,

`2x^2 - 20x = 300 + x^2`

`=> x^2 - 20x - 300 = 0`

`=> x^2 - 30x + 10x - 300 = 0`

`=> (x - 30)(x + 10) = 0`

`=> x = 30 ( ∵ x != -10)`
Correct Answer is `=>` (D) `30`
Q 2368156905

If `alpha` and `beta` are the roots of the equation
`x^2 + x + 1 = 0`, then which of the following are the
roots of the equation `x^2 - x + 1 = 0`?
NDA Paper 1 2010
(A)

`alpha^7` and `beta^(13)`

(B)

`alpha^(13)` and `beta^7`

(C)

`alpha^(20)` and `beta^(20)`

(D)

None of these

Solution:

Since, `alpha` and `beta` be the roots of the equation

`x^2 + x + 1 = 0`.

`:. alpha + beta = -1` and `alpha beta = 1`

`=> alpha = omega` and `beta = omega^2`

Now, for equation `x^2 - x + 1 = 0`,

`:.` Sum of roots `= 1`

and product of roots `= 1`

Looking through options,

`alpha^7. beta^(13) = (omega)^7 (omega^2)^(13) = 1 != -1`

`alpha^(13).beta^7 = (omega)^(13) (omega^2) = 1 != -1`

`alpha^(20). beta^(20) = (omega)^(20) (omega^2)^(20) = 1 != -1`

Hence, no one of (a), (b) and (c) are satisfied for the given

equation.
Correct Answer is `=>` (D) None of these
Q 2378778606

What is the value of `x` satisfying the equation

` 16( (a-x)/(a + x))^3 = ( a + x)/(a - x)` ?
NDA Paper 1 2009
(A)

` a/2`

(B)

`a/3`

(C)

`a/4`

(D)

`0`

Solution:

`16( (a-x)/(a + x))^3 = ( a + x)/(a - x)`

` => ( (a-x)/(a + x))^4 = (1/2)^4 => (a - x)/(a +x) = 1/2`

` => 2a - 2x = a + x => a = 3x`

`:. x = a /3`
Correct Answer is `=>` (B) `a/3`
Q 2359101014

What is the value of `sqrt(5sqrt(5sqrt(5sqrt(...oo))))`?
NDA Paper 1 2008
(A)

`5`

(B)

`sqrt5`

(C)

`1`

(D)

`(5)^(1//4)`

Solution:

Let ` x = sqrt(5sqrt(5sqrt(5sqrt(...oo))))`

`=> x = sqrt(5x) => x^2 = 5x`

`=> x^2 - 5x = 0 => x(x - 5) = 0`

`:. x = 0` or `5`
Correct Answer is `=>` (A) `5`

Solution of quadratic Inequation

Q 2349101013

If `x` is real and `x^2 - 3x + 2 > 0, x^2 - 3x- 4 <= 0`, then
which one of the following is correct?
NDA Paper 1 2008
(A)

`-1 <= x <= 4`

(B)

`2 <= x <= 4`

(C)

`-1 < x <= 1`

(D)

`-1 <= x <= 1` or `2 < x <= 4`

Solution:

`∵ x^2 - 3x + 2 > 0`

`=> (x -1) (x -2) > 0`

`=> x < 1 ` or `x > 2` ......(i)

and `x^2 - 3x - 4 <= 0`

`=> (x - 4) (x + 1) <= 0`

`:. -1 <= x <= 4` .......(ii)

`=> -1 <= x < 1` or `2 < x <= 4` [from Eqs. (i) and (ii)]
Correct Answer is `=>` (D) `-1 <= x <= 1` or `2 < x <= 4`
Q 2319112010

If `- x^2 + 3x + 4 > 0`, then which one of the
following is correct?
NDA Paper 1 2007
(A)

`x in (-1, 4)`

(B)

`x in [-1, 4]`

(C)

`x in (-oo, -1) cup (4, oo)`

(D)

`x in (-oo , 1) cup [4, oo)`

Solution:

`- x^ 2 + 3x + 4 > 0`

`=> x^2 - 3x- 4 < 0 => (x- 4) (x + 1) < 0`

`=> x in (-1, 4)`
Correct Answer is `=>` (A) `x in (-1, 4)`

 
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