Question

Which constant must be added and subtracted to solve the quadratic equation 9x2 + 3/4x– 2 = 0 by the method of completing the square?
(A) 1/8 (B) 1/64 (C) 1/4 (D) 9/64

NCERT Class X
Mathematics - Exemplar Problems

Chapter _QUADRATIC EQUATIONS

Answer 1

Hi,

Given quadratic equation

9x^2 - 3x/4 - 2 = 0

( 3x )^2 - 2 × 3x x 1/8 - 2 = 0

(3x)^2-2 × (3x) x(1/8) + (1/8)^2 - (1/8)^2

= 2 -----------( 1 )

( 3x - 1/8 )^2 = 2 - ( 1/8 )^2

( 3x - 1/8 )^2 = 2 - 1/ 64

( 3x - 1/8 )^2 = (128 - 1)/ 64

( 3x - 1/8)^2 = 127/ 64

From ( 1 )

Here we added and

subtracted

(1/8)^2 = 1/ 64 to solve the

quadratic

equation by the completing square.

Option B is correct.

I hope this will useful to you.

******

Answer 2

The constant which needs to be added and subtracted is (B) 1/64.

Solution:

Given quadratic equation:

$9 x^{2}+\frac{3}{4} x-2=0$

$(3 x)^{2}-(2 \times 3 x) \times\left(\frac{1}{8}\right)-2=0$

By adding and subtracting, $\left(\frac{1}{8}\right)^{2}$ on the above equation,

$(3 x)^{2}-(2 \times 3 x) \times\left(\frac{1}{8}\right)+\left(\frac{1}{8}\right)^{2}-\left(\frac{1}{8}\right)^{2}=2 \rightarrow(i)$

By using the formula, $(a-b)^{2}$

$\left(3 x-\frac{1}{8}\right)^{2}+\left(\frac{1}{8}\right)^{2}=2$

$\left(3 x-\frac{1}{8}\right)^{2} = 2-\left(\frac{1}{8}\right)^{2}$

$\left(3 x-\frac{1}{8}\right)^{2} = 2-\left(\frac{1}{64}\right)$

$\left(3 x-\frac{1}{8}\right)^{2} = \frac{128-1}{64}$

$\left(3 x-\frac{1}{8}\right)^{2} = \frac{127}{64}$

From (i)

Here we added and subtracted

$\left(\frac{1}{8}\right)^{2} = \frac{1}{64}$ is used to resolve the quadratic equation by the completing square.