MCQ
The value of $\frac{\sqrt{32}+\sqrt{68}}{\sqrt{8}+\sqrt{12}}$ is
- A$\sqrt{2}$
- ✓2
- C4
- D8
✓
Answer
Correct option: B.
2
(b)
$ \frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}}=\frac{\sqrt{16} \sqrt{2}+\sqrt{16} \sqrt{3}}{\sqrt{4} \sqrt{2} \times \sqrt{4} \sqrt{3}}=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}=2 $
$ \frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}}=\frac{\sqrt{16} \sqrt{2}+\sqrt{16} \sqrt{3}}{\sqrt{4} \sqrt{2} \times \sqrt{4} \sqrt{3}}=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}=2 $
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