# then the value of p is

Question:

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

(a) 0

(b) $-\frac{1}{4}$

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

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

Solution:

$\left(3 x+\frac{1}{2}\right)\left(3 x-\frac{1}{2}\right)=9 x^{2}-p$

$9 x^{2}-\frac{1}{4}=9 x^{2}-p$

$\left(\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right)$

$\Rightarrow p=\frac{1}{4}$

Hence, the correct answer is option (c).