PP1 (Shorter) Quant Section 2 (Easy) Q13

<p><span style="color:#27ae60;"><span style="font-size:20px;">Solving the Problem</span></span></p> <p>We can start by looking at the first equation:&nbsp;$(x+2)(x-3) = 0$</p> <p>We know from this that either $x+2=0$ or $x-3=0$</p> <p>So, $x$ could either be $-2$ or $3.$</p> <p>Since we are told that $x&gt;\frac{1}{2}$, we know that $x$ must be $3$.</p> <p>Now, we can solve $x^{-2}$, which is the same as $\frac{1}{x^2}$</p> <p>If we plug in $x=3$, we get the result of $\frac{1}{3^2}$ which is $\frac{1}{9}$.</p> <p>So, the answer must be <span style="color:#27ae60;">D</span>.</p>