Question

Given that $\sqrt{3}=1.732;(\sqrt{6}+\sqrt{2})÷(\sqrt{6}-\sqrt{2})$ is equal to

• A. $5$
• B. $2$
• C. $3.732$
• D. $0$

Answer 1

Answer: (C)

$3.732$

$\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3\times 2}+\sqrt{2}}{\sqrt{3\times 2}-\sqrt{2}}$

$=\frac{\sqrt{3}.\sqrt{2}+\sqrt{2}}{\sqrt{3}.\sqrt{2}-\sqrt{2}}$

$=\frac{\sqrt{2}(\sqrt{3}+1)}{\sqrt{2}(\sqrt{3}-1)}$

$=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$=\frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

$=\frac{(\sqrt{3}{)}^2+2\sqrt{2}+1}{3-1}$

$=\frac{3+2\sqrt{3}+1}{2}$

$=\frac{4+2\sqrt{3}}{2}$

$=2+\sqrt{3}$

$=2+1.732=3.732$