Class 10

### Topic to be coverd

♦ Solution of a Quadratic Equation by Completing the Square

### Solution of a Quadratic Equation by Completing the Square

Let's Consider the following situation:
The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?

To answer this, let her present age (in years) be x. Then the product of her ages two years ago and four years from now is (x – 2)(x + 4).

Therefore, (x – 2)(x + 4) = 2x + 1
i.e., x^2 + 2x – 8 = 2x + 1
i.e., x^2 – 9 = 0

So, Sunita’s present age satisfies the quadratic equation x^2 – 9 = 0.
We can write this as x^2 = 9. Taking square roots, we get x = 3 or x = – 3. Since the age is a positive number, x = 3.

So, Sunita’s present age is 3 years. Now consider the quadratic equation (x + 2)^2 – 9 = 0. To solve it, we can write
it as (x + 2)^2 = 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
Therefore, x = 1 or x = –5

So, the roots of the equation (x + 2)^2 – 9 = 0 are 1 and – 5.
In both the examples above, the term containing x is completely inside a square, and we found the roots easily by taking the square roots.

But, what happens if we are asked to solve the equation x^2 + 4x – 5 = 0 We would probably apply factorisation to do so, unless we realise (somehow!) that x^2 + 4x – 5 = (x + 2)^2 – 9.

So, solving x^2 + 4x – 5 = 0 is equivalent to solving (x + 2)^2 – 9 = 0, which we have seen is very quick to do. In fact, we can convert any quadratic equation to the form (x + a)^2 – b^2 = 0 and then we can easily find its roots. Let us see if this is possible.
Look at Fig. 4.2.

In this figure, we can see how x^2 + 4x is being converted to (x + 2)^2 – 4.

The process is as follows:

x^2+4x = ( x^2+4/2 x ) +4/2 x

= x^2 + 2x + 2x
= (x + 2) x + 2 × x
= (x + 2) x + 2 × x + 2 × 2 – 2 × 2
= (x + 2) x + (x + 2) × 2 – 2 × 2
= (x + 2) (x + 2) – 2^2
= (x + 2)^2 – 4
So, x^2 + 4x – 5 = (x + 2)^2 – 4 – 5 = (x + 2)^2 – 9
So, x^2 + 4x – 5 = 0 can be written as (x + 2)^2 – 9 = 0 by this process of completing the square. This is known as the method of completing the square.

In brief, this can be shown as follows:

x^2+4x= (x+4/2)^2 -(4/2)^2 = (x+4/2)^2-4

So, x^2 + 4x – 5 = 0 can be rewritten as

(x+4/2)^2-4-5 = 0

(x+2)^2-9 = 0

Consider now the equation 3x^2 – 5x + 2 = 0. Note that the coefficient of x^2 is not
a perfect square. So, we multiply the equation throughout by 3 to get

9x^2 – 15x + 6 = 0

Now 9x^2-15x+6 = (3x)^2-2xx3x xx 5/2 +6

= (3x)^2-2xx3x xx 5/2 +(5/2)^2 - (5/2)^2 +6

 = (3x-5/2)^2 -25/4 +6 = ( 3x -5/2)^2 - 1/4

So, 9x^2 – 15x + 6 = 0 can be written as

( 3x - 5/2)^2 - 1/4 = 0

(3x - 5/2)^2 = 1/4

So, the solutions of 9x^2 – 15x + 6 = 0 are the same as those of ( 3x - 5/2)^2 = 1/4.

3x - 5/2 = 1/2  or 3x - 5/2 = -1/2

(We can also write this as 3x - 5/2 = pm 1/2 , where ‘±’ denotes ‘plus minus’.)

Thus 3x= 5/2+1/2  or 3x = 5/2-1/2

So x = 5/6 +1/6 or x = 5/6 - 1/6

Therefore x = 1 or x = 4/6

x = 1 or x = 4/6

i.e. x = 1 or x = 2/3

Therefore, the roots of the given equation are 1 and 2/3

Remark : Another way of showing this process is as follows :

The equation 3x^2 - 5x+2 = 0

is the same as x^2 - 5/3 x +2/3 = 0

Now x^2 - 5/3 x +2/3 = { x - 1/2 (5/3)}^2 - { 1/2 (5/3)}^2 +2/3

 = ( x-5/6)^2 +2/3 - 25/36

 = ( x-5/6)^2 - 1/36 = ( x-5/6)^2 - ( 1/6)^2

So, the solutions of 3x^2 – 5x + 2 = 0 are the same as those of ( x-5/6)^2 - (1/6)^2 = 0

which are x - 5/6 = pm 1/6
i.e x = 5/6 +1/6 = 1 and x = 5/6-1/6 = 2/3

Let us consider some examples to illustrate the above process.
Q 3119578419

Solve the equation given in Example 3 by the method of completing the square.
Class 10 Chapter 4 Example 7
Solution:

The equation 2x^2 – 5x + 3 = 0 is the same as x^2 -5/2x +3/2 = 0

Now x^2 - 5/2 x +3/2 = ( x-5/4)^2 - ( 5/4)^2 +3/2 = ( x-5/4)^2 - 1/16

Therefore, 2x^2 – 5x + 3 = 0 can be written as (x -5/4)^2 - 1/16 = 0

So, the roots of the equation 2x^2 – 5x + 3 = 0 are exactly the same as those of ( x-5/4)^2 - 1/16 = 0 . Now (x-5/4)^2 - 1/16 = 0 is the same as ( x-5/4)^2 = 1/16

Therefore x- 5/4 = pm 1/4

i.e. x = 5/4 pm 1/4

i.e. x = 5/4+1/4 or x = 5/4-1/4

i.e. x = 3/2 or x = 1

Therefore, the solutions of the equations are x = 3/2 and 1

Let us verify our solutions .
Putting x = 3/2 in 2x^2 - 5x +3 = 0 we get 2 ( 3/2)^2-5(3/2) +3 = 0
which is
correct. Similarly, you can verify that x = 1 also satisfies the given equation.
In Example 7, we divided the equation 2x^2 – 5x + 3 = 0 throughout by 2 to get
x^2-5/2 x +3/2 = 0 to make the first term a perfect square and then completed the square. Instead, we can multiply throughout by 2 to make the first term as 4x^2 = (2x)^2 and then complete the square.
This method is illustrated in the next example.
Q 3139678512

Find the roots of the equation 5x^2 -6x-2 = 0 by the method of completing the square.
Class 10 Chapter 4 Example 8
Solution:

Multiplying the equation throughout by 5, we get

25x^2 – 30x – 10 = 0
This is the same as
(5x)^2 – 2 × (5x) × 3 + 32 – 32 – 10 = 0
i.e., (5x – 3)^2 – 9 – 10 = 0
i.e., (5x – 3)^2 – 19 = 0
i.e., (5x – 3)^2 = 19

i.e. 5x-3 = pm sqrt(19)

i.e 5x = 3 pm sqrt(19)

x = ( 3 pm sqrt(19))/5

Therefore, the roots are ( 3+sqrt(19))/5 and (3-sqrt(19))/5

Verify that the roots are ( 3+sqrt(19))/5 and (3-sqrt(19))/5
Q 3129078811

Find the roots of 4x^2 + 3x + 5 = 0 by the method of completing the
Class 10 Chapter 4 Example 9
Solution:

Note that 4x^2 + 3x + 5 = 0 is the same as

(2x)^2+2 xx (2x) xx 3/4 + ( 3/4)^2 - (3/4)^2+5 = 0

i.e. (2x+3/4)^2 - 9/16 +5 = 0

i.e. ( 2x+3/4)^2 +71/16 = 0
i.e. (2x+3/4)^2 = (-71)/6 < 0

But ( 2x+3/4)^2 cannot be negative for any real value of x (Why?). So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.

quare. So, let us give this method in general.
Consider the quadratic equation ax^2 + bx + c = 0 (a ≠ 0). Dividing throughout by

a we get x^2 +b/a x +c/a = 0

This is the same as (x+b/(2a))^2 - (b/(2a))^2+c/a = 0

i.e. (x+b/(2a))^2 - (b^2-4ac)/(4a^2) = 0

So, the roots of the given equation are the same as those of (x +b/(2a))^2 - (b^2-4ac)/(4a^2) = 0

i.e. those of (x+b/(2a))^2 = (b^2-4ac)/(4a^2) ........(1)

If b^2 – 4ac ≥ 0, then by taking the square roots in (1), we get

x+b/(2a) = ( pm sqrt(b^2-4ac))/(2a)

Therefore x = (- b pm sqrt(b^2-4ac))/(2a)

So, the roots of ax^2 + bx + c = 0 are  ( -b + sqrt(b^2-4ac))/(2a) and (-b - sqrt(b^2-4ac))/(2a) , if b^2 – 4ac ≥ 0. If b^2 – 4ac < 0, the equation will have no real roots. (Why?)
Thus, if b2^ – 4ac ≥ 0, then the roots of the quadratic equation.

ax^2 + bx + c = 0 are given by (-b pm sqrt(b^2-4ac))/(2a)

This formula for finding the roots of a quadratic equation is known as the quadratic formula.
Let us consider some examples for illustrating the use of the quadratic formula.
Q 3139078812

The area of a rectangular plot is 528 m^2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Class 10 Chapter 4 Example 10
Solution:

Let the breadth of the plot be x metres. Then the length is (2x + 1) metres.
Then we are given that x(2x + 1) = 528, i.e., 2x^2 + x – 528 = 0.
This is of the form ax^2 + bx + c = 0, where a = 2, b = 1, c = – 528.
So, the quadratic formula gives us the solution as

x = (-1 pm sqrt(1+4(2)(528)))/4 = (-1 pm sqrt(4225)/4 = (-1 pm 65)/4

 x = 64/4 or x = (-66)/4

x = 16 or x = - 33/2

Since x cannot be negative, being a dimension, the breadth of the plot is 16 metres and hence, the length of the plot is 33m.
You should verify that these values satisfy the conditions of the problem.
Q 3159078814

Find two consecutive odd positive integers, sum of whose squares is 290.
Class 10 Chapter 4 Example 11
Solution:

Let the smaller of the two consecutive odd positive integers be x. Then, the
second integer will be x + 2. According to the question,
x^2 + (x + 2)^2 = 290
i.e., x^2 + x^2 + 4x + 4 = 290
i.e., 2x^2 + 4x – 286 = 0
i.e., x^2 + 2x – 143 = 0
which is a quadratic equation in x.
Using the quadratic formula, we get

x = (-2 pmsqrt(4+572))/2 = (-2 pm sqrt(576))/2 = (-2 pm 24)/2

i.e., x = 11 or x = – 13
But x is given to be an odd positive integer. Therefore, x ≠ – 13, x = 11.
Thus, the two consecutive odd integers are 11 and 13.
Check : 11^2 + 13^2 = 121 + 169 = 290.
Q 3169078815

A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig. ). Find its length and breadth.
Class 10 Chapter 4 Example 12
Solution:

Let the breadth of the rectangular park be x m.
So, its length = (x + 3) m.
Therefore, the area of the rectangular park = x(x + 3) m^2 = (x^2 + 3x) m^2.
Now, base of the isosceles triangle = x m.
Therefore, its area = 1/2 × x × 12 = 6 x m^2.
According to our requirements, x^2 + 3x = 6x + 4
i.e., x^2 – 3x – 4 = 0
Using the quadratic formula, we get x = (3 pm sqrt(25))/2 = (3 pm 5)/2 = 4 or -1

But x ≠ – 1 (Why?). Therefore, x = 4.
So, the breadth of the park = 4m and its length will be 7m.
Verification : Area of rectangular park = 28 m^2,
area of triangular park = 24 m^2 = (28 – 4) m^2
Q 3179078816

Find the roots of the following quadratic equations, if they exist, using the quadratic formula:
(i) 3x^2 – 5x + 2 = 0
(ii) x^2 + 4x + 5 = 0
(iii) 2x^2 – 2 sqrt2 x + 1 = 0
Class 10 Chapter 4 Example 13
Solution:

(i) 3x^2 – 5x + 2 = 0. Here, a = 3, b = – 5, c = 2. So, b^2 – 4ac = 25 – 24 = 1 > 0.

Therefore, x = ( 5 pm sqrt1)/6 = (5pm1)/6

i.e.  x = 1 or x = 2/3

So the roots are 2/3 and 1

(ii) x^2 + 4x + 5 = 0. Here, a = 1, b = 4, c = 5. So, b^2 – 4ac = 16 – 20 = – 4 < 0.

Since the square of a real number cannot be negative, therefore sqrt(b^2-4ac) will
not have any real value.
So, there are no real roots for the given equation.

(iii) 2x^2 - 2 sqrt2 +1 = 0. Here, a = 2, b = -2 sqrt2 , c = 1.

So, b^2 – 4ac = 8 – 8 = 0

Therefore, x = (2 sqrt2 pm sqrt0)/4 = sqrt2/2 pm 0

i.e. x = 1/sqrt2

So, the roots are 1/sqrt2 , 1/sqrt2
Q 3189078817

Find the roots of the following equations

(i) x +1/x = 3 , x ne 0

(ii) 1/x - 1/(x-2) = 3 , x ne 0 , 2
Class 10 Chapter 4 Example 14
Solution:

(i) x +1/x = 3 Multiplying throughout by x, we get

x^2 + 1 = 3x

i.e. x^2-3x+1= 0 which is a quadratic equation.

Here, a = 1, b = – 3, c = 1

So, b^2 - 4ac = 9-4 = 5 > 0

Therefore, x = ( 3 pm sqrt5)/2 (Why?)

So, the roots are (3+sqrt5)/2 and (3-sqrt5)/2

(ii) 1/x - 1/(x-2) = 3 , x ne 0 , 2

As x ≠ 0, 2, multiplying the equation by x (x – 2), we get

(x – 2) – x = 3x (x – 2)
= 3x^2 – 6x
So, the given equation reduces to 3x^2 – 6x + 2 = 0, which is a quadratic equation.

Here, a = 3, b = – 6, c = 2. So, b^2 – 4ac = 36 – 24 = 12 > 0

Therefore, x = ( 6 pm sqrt(12))/6 = ( 6 pm 2sqrt3)/6 = (3 pm sqrt3)/3

So, the roots are  (3 + sqrt3)/3 and (3-sqrt3)/3
Q 3119078819

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Class 10 Chapter 4 Example 15
Solution:

Let the speed of the stream be x km//h.
Therefore, the speed of the boat upstream = (18 – x) km//h and the speed of the boat
downstream = (18 + x) km//h.

The time taken to go upstream = text(distance)/text(speed) = (24)/(18-x) hours.

Similarly, the time taken to go downstream  = (24)/(18+x) hours.

According to the question, 24/(18-x) - 24/(18+x) = 1

i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)
i.e., x^2 + 48x – 324 = 0
Using the quadratic formula, we get

x = (-48 pm sqrt(48^2+1296))/2 = (-48 pm sqrt(3600))/2

 = (-48 pm 60)/2 = 6 or -54

Since x is the speed of the stream, it cannot be negative. So, we ignore the root
x = – 54. Therefore, x = 6 gives the speed of the stream as 6 km//h.