Question

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Answer 1

Question:

Find values of a and b if $\frac{3+2 \sqrt{2}}{2-\sqrt{2}}- \frac{3-2 \sqrt{2}}{2+\sqrt{2}} = a+b\sqrt{2}$

Answer:

a=0

b=7

Explanation:

First of all rationalise both $\frac{3+2 \sqrt{2}}{2-\sqrt{2}}$ and $\frac{3-2 \sqrt{2}}{2+\sqrt{2}}$ separately.

$\tt \frac{3+2 \sqrt{2}}{2-\sqrt{2}} = \frac{3+2 \sqrt{2}}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}} \\\\ \tt \frac{6+3\sqrt{2}+4\sqrt{2}+4}{2} \\\\ \tt \frac{10+7\sqrt{2}}{2}$

Now, rationalise other one

$\tt \frac{3-2 \sqrt{2}}{2+\sqrt{2}} = \frac{3-2 \sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} \\\\ \tt \frac{6-3\sqrt{2}-4\sqrt{2}+4}{2} \\\\ \tt \frac{10-7\sqrt{2}}{2}$

A/q

$\tt \frac{3+2 \sqrt{2}}{2-\sqrt{2}}- \frac{3-2 \sqrt{2}}{2+\sqrt{2}} \\\\ \to\tt \frac{10+7\sqrt{2}}{2} - \tt \frac{10-7\sqrt{2}}{2} \\\\ \tt\to \frac{10+7\sqrt{2}-10+7\sqrt{2}} {2} \\\\ \tt\to \frac{\cancel{10} +7\sqrt{2}\cancel{-10}+7\sqrt{2}} {2} \\\\ \tt \to\frac {2 \times 7\sqrt{2}} {2} \\\\ \tt\to \frac {\cancel{2} \times 7\sqrt{2}} {\cancel{2}} \\\\\to 7\sqrt{2} \\\\ \tt \to 0+7\sqrt{2}$

But answer is $a+b\sqrt{2}$

Comparing both the two, we get a=0 and b=7