waecmaths question:
Simplify $\frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}$
Option A:
$\frac{1}{2}$
Option B:
p – 2r
Option C:
$\frac{1}{p-2r}$
Option D:
$\frac{2p}{p-2r}$
waecmaths solution:
$\begin{align} & \frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}=\frac{[(p-r)-r][(p-r)+r]}{2p(p-2r)} \\ & \frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}=\frac{(p-2r)p}{2p(p-2r)} \\ & \frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}=\frac{1}{2} \\\end{align}$
maths year:
maths topics: