Question

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Q:-Find the value of

$\sqrt{a - b}$




if $\frac{8 + 3 \sqrt{7} }{8 - 3 \sqrt{7} } - \frac{8 - 3 \sqrt{7} }{8 + 3 \sqrt{7} } = a + b \sqrt{7}$


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Answer 1

Solution

Given

$\rm \implies \dfrac{8 + 3 \sqrt{7} }{8 -3 \sqrt{7} } - \dfrac{8 - 3 \sqrt{7} }{8 + 3 \sqrt{7} } = a + b \sqrt{7}$

To find

$\rm \implies \: \sqrt{a - b}$

Now take

$\rm \implies \dfrac{8 + 3 \sqrt{7} }{8 -3 \sqrt{7} } - \dfrac{8 - 3 \sqrt{7} }{8 + 3 \sqrt{7} } = a + b \sqrt{7}$

By taking lcm we get

$\rm \implies \dfrac{(8 + 3 \sqrt{7})(8 + 3 \sqrt{7} ) - (8 - 3 \sqrt{7} )(8 - 3 \sqrt{7} ) }{(8 - 3 \sqrt{7} )(8 + 3 \sqrt{7} )}$

$\rm \implies \: \dfrac{(8 + 3 \sqrt{7} ) {}^{2} - (8 - 3 \sqrt{7} ) ^{2} }{(8) {}^{2} - (3 \sqrt{7} ) {}^{2} }$

We using this identity

$\rm \to \: (a - b)(a + b) = {a}^{2} - {b}^{2}$

$\rm \to(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab$

$\rm \to(a - b) ^{2} = {a}^{2} + {b}^{2} - 2ab$

Apply this identity

$\rm \implies \: \dfrac{ {8}^{2} + (3 \sqrt{7} ) {}^{2} + 2 \times 8 \times 3 \sqrt{7} - ( {8}^{2} + (3 \sqrt{7} ) {}^{2} - 2 \times 8 \times 3 \sqrt{7} }{64 - 63}$

$\rm \implies \: \dfrac{ {8}^{2} + (3 \sqrt{7} ) {}^{2} +48 \sqrt{7} - 8 {}^{2} - (3 \sqrt{7} ) {}^{2} + 48 \sqrt{7} }{1}$

$\rm \implies \: \cancel{8}^{2} + \cancel{ (3 \sqrt{7} )} {}^{2} +48 \sqrt{7} - \cancel{8 {}^{2} } - \cancel{ (3 \sqrt{7} }) {}^{2} + 48 \sqrt{7}$

$\rm \implies \: 96 \sqrt{7}$

So value of a and b is

$\rm \implies \: a = 0 \: \: and \: b \: = 96$

We have find the value of

$\rm \implies \sqrt{a - b}$

Put the value on

$\rm \implies \sqrt{0 - 96} = \sqrt{ - 96}$

value of √-96 is doesn't exist