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

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

$\huge\mathcal{Answer}$

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

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

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

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

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

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

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

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

$\bold{\red{On \:comparing\: both \:sides:-}}$

$⟹\bold{a = 0 \: and \: b = 32}$

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

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Answer 2

❥Question᎓

Q:-Find the value of

$\sqrt{a - b}$

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

$\huge\mathcal{Answer}Answer$

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

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

∵$\bold{ {(a + b)}^{2} - {(a - b)}^{2}$

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

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

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

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

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

On comparing both sides:−

⟹$\bold{a = 0 \: and \: b = 32}$

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

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