Question

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

$\red{\bold{\underline{\underline{❥Question᎓}}}}$

Q:-Find the value of

$\sqrt{a - b}$

if

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

$\huge\tt\underline\blue{⛶Answer⛶</p><p> }$

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

$⟹ \frac{(8 + 3 \sqrt{7})(8 + 3 \sqrt{7}) - [(8 - 3 \sqrt{7} )(8 - 3 \sqrt{7}) ]}{(8 + 3 \sqrt{7} )(8 - 3 \sqrt{7} )}$=a+b√7

∵${(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$⟹ \frac{4 \times 8 \sqrt{7} }{ {(8)}^{2} - {(3 \sqrt{7}) }^{2} } = a + b \sqrt{7}$

$⟹ \frac{32 \sqrt{7} }{64 - 63} = a + b \sqrt{7}$

$⟹ \frac{32 \sqrt{7} }{1} = a + b \sqrt{7}$

$⟹32 \sqrt{7} = a + b \sqrt{7}$

$⟹0 + 32 \sqrt{7} = a + b \sqrt{7}$

On comparing both sides:-

$⟹a = 0 \: and \: b = 32$

$⟹now \: \sqrt{a - b} = \sqrt{0 - 32} = \sqrt{ - 32 } = \sqrt{ - 2 \times 16} = 4 \sqrt{ - 2} = 4 \sqrt{2} i \: [( {i}^{2} = - 1)$]

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Answer 2

$\huge \blue {Answer}$

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

$\sf \frac{(8 + 3 \sqrt{7})(8 + 3 \sqrt{7}) - [(8 - 3 \sqrt{7} )(8 - 3 \sqrt{7}) ]}{(8 + 3 \sqrt{7} )(8 - 3 \sqrt{7} )}$

$\sf Because, \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab(a+b)$

$\sf \frac{4 \times 8 \sqrt{7} }{ {(8)}^{2} - {(3 \sqrt{7}) }^{2} } = a + b \sqrt{7}$

$\sf \frac{32 \sqrt{7} }{64 - 63} = a + b \sqrt{7}$

$\sf \frac{32 \sqrt{7} }{1} = a + b \sqrt{7}$

$\sf 32 \sqrt{7} = a + b \sqrt{7}$

$\sf 0 + 32 \sqrt{7} = a + b \sqrt{7}$

$\sf \red{Compare \: both \: sides:-}$

$\boxed {\sf a = 0}$

$\boxed {\sf b = 32}$

$\sf ✯ \sqrt{a - b}$

$\sf = \sqrt{0 - 32}$

$\sf = \sqrt{ - 32 }$

$\sf = \sqrt{ - 2 \times 16}$

$\sf = 4 \sqrt{ - 2}$