Prove the following
Question:

If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, then find the value of

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

Solution:

We have, $\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}$

$=\frac{4(3 \sqrt{3}+2 \sqrt{2})+3(3 \sqrt{3}-2 \sqrt{2})}{(3 \sqrt{3}-2 \sqrt{2})(3 \sqrt{3}+2 \sqrt{2})}$

$=\frac{12 \sqrt{3}+8 \sqrt{2}+9 \sqrt{3}-6 \sqrt{2}}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}}$            [using identity, $(a+b)(a-b)=a^{2}-b^{2}$ ]

$=\frac{21 \sqrt{3}+2 \sqrt{2}}{27-8}=\frac{21 \sqrt{3}+2 \sqrt{2}}{19}$

$=\frac{21 \times 1.732+2 \times 1.414}{19}$            [put $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$ ]

$=\frac{36.372+2.828}{19}=\frac{39.2}{19}=2.06316 \approx 2.063$

Administrator

Leave a comment

Please enter comment.
Please enter your name.