Question
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
- $\sqrt{2}$
- 2
- 4
- 8
✓
Answer
- 4
Solution:
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{4\times8}+\sqrt{4\times12}}{\sqrt{8}+\sqrt{12}}$
$=\frac{2\sqrt{8}+2\sqrt{12}}{\sqrt{8}+\sqrt{12}}=\frac{2\big(\sqrt{8}+\sqrt{12}\big)}{\sqrt{8}+\sqrt{12}}=2$
Hence, the correct answer is option (b).
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