waecmaths question:
If $\sqrt{72}+\sqrt{32}-3\sqrt{18}=x\sqrt{8}$ find the value of x
Option A:
1
Option B:
$\tfrac{3}{4}$
Option C:
$\tfrac{1}{2}$
Option D:
$\tfrac{1}{4}$
waecmaths solution:
$\begin{align} & \sqrt{72}+\sqrt{32}-3\sqrt{18}=x\sqrt{8} \\ & \sqrt{36\times 2}+\sqrt{16\times 2}-3\sqrt{9\times 2}=x\sqrt{4\times 2} \\ & 6\sqrt{2}+4\sqrt{2}-9\sqrt{2}=2x\sqrt{2} \\ & \sqrt{2}=2x\sqrt{2} \\ & x=\frac{1}{2} \\\end{align}$
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