Question

$\text{If}\sqrt{2}=1.414,\sqrt{3}=1.732,\text{then find the value of}\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}.$

• A.
  2
• B.
  3
• C.
  5
• D.
  7

Answer 1

The correct option is A

2
$\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}=\frac{4}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}\times \frac{3\sqrt{3}-2\sqrt{2}}{3\sqrt{3}-2\sqrt{2}}=\frac{12\sqrt{3}+8\sqrt{2}}{(3\sqrt{3}{)}^2-(2\sqrt{2}{)}^2}+\frac{9\sqrt{3}-6\sqrt{2}}{(3\sqrt{3}{)}^2-(2\sqrt{2}{)}^2}=\frac{12\sqrt{3}+8\sqrt{2}}{19}+\frac{9\sqrt{3}-6\sqrt{2}}{19}=\frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{19}=\frac{21\sqrt{3}+2\sqrt{2}}{19}=\frac{21\times 1.732+2\times 1.414}{19}=\frac{36372+2.828}{19}=\frac{39.2}{19}=2.06=2(approx.)$

Hence, the right answer is Option A.