Question

^{If $9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right)$,} then the value of b is
(a) 0

(b) $\frac{1}{\sqrt{2}}$

(c) $\frac{1}{4}$

(d) $\frac{1}{2}$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right) \\ 9x^2-b=\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right) \\ \Rightarrow 9x^2-b={\left(3x\right)}^2-{\left(\frac{1}{2}\right)}^2\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:a^2-b^2=\left(a+b\right)\left(a-b\right)\right) \\ \Rightarrow 9x^2-b=9x^2-\frac{1}{4} \\ \Rightarrow -b=-\frac{1}{4} \\ \Rightarrow b=\frac{1}{4} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$