Question 40

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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: 

$\begin{align}  & \frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}=\frac{[(p-r)-r][(p-r)+r]}{2p(p-2r)} \\ & \frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}=\frac{(p-2r)p}{2p(p-2r)} \\ & \frac{{{(p-r)}^{2}}-{{r}^{2}}}{2{{p}^{2}}-4pr}=\frac{1}{2} \\\end{align}$

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