Question

Question 20
If $49x^2-b=(7x+\frac{1}{2})(7x-\frac{1}{2})$,
then the value of b is

A) $0$
B) $\frac{1}{\sqrt{2}}$
C) $\frac{1}{4}$
D) $\frac{1}{2}$

Answer 1

Given,
$(49x^2-b)=(7x+\frac{1}{2})(7x-\frac{1}{2})$
$\Rightarrow [49x^2-(\sqrt{b}{)}^2]=\left[\right(7x{)}^2-{\left(\frac{1}{2}\right)}^2][usingidentity,(a+b)(a-b)=a^2-b^2]$
$\Rightarrow 49x^2-(\sqrt{b}{)}^2=49x^2-{\left(\frac{1}{2}\right)}^2$
$\Rightarrow -(\sqrt{b}{)}^2=-{\left(\frac{1}{2}\right)}^2$
$\Rightarrow (\sqrt{b}{)}^2=-{\left(\frac{1}{2}\right)}^2$ [multiplying both sides by -1]
$\therefore b=\frac{1}{4}$
The answer is C.