Question:
If $49 x^{2}-b=\left(7 x+\frac{1}{2}\right)\left(7 x-\frac{1}{2}\right)$, then the value of $b$ is
(a) 0
(b) $\frac{1}{\sqrt{2}}$
(c) $\frac{1}{4}$
(d) $\frac{1}{2}$
Solution:
(c) Given,
$\left(49 x^{2}-b\right)=\left(7 x+\frac{1}{2}\right)\left(7 x-\frac{1}{2}\right)$
$\Rightarrow \quad\left[49 x^{2}-(\sqrt{b})^{2}\right]=\left[(7 x)^{2}-\left(\frac{1}{2}\right)^{2}\right]\left[\right.$ [using identity, $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$
$\Rightarrow \quad 49 x^{2}-(\sqrt{b})^{2}=49 x^{2}-\left(\frac{1}{2}\right)^{2}$
$\Rightarrow \quad-(\sqrt{b})^{2}=-\left(\frac{1}{2}\right)^{2}$
$\Rightarrow \quad(\sqrt{b})^{2}=\left(\frac{1}{2}\right)^{2}$ [multiplying both sides by $-1$ ]
$\therefore$ $b=\frac{1}{4}$