Question

If $\left(3x+\frac{1}{2}\right)\left(3x-\frac{1}{2}\right)=9x^2-p$ then the value of p is
(a) ⁰
(b) $-\frac{1}{4}$
(c) $\frac{1}{4}$
(d) $\frac{1}{2}$

Answer 1

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