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).
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