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Tricks & Tips of Quadratic Equation for NDA

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Nature of roots

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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`

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