Question

if root 7 + 1 upon root 7 minus 1 equals to a minus b root 7 then find the value of a and b

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 } = a - b \sqrt{7} \\ \\ on \: rationalizing \: we \: get \\ \\ = \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1 } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ we \: get \\ \\ = \frac{ {( \sqrt{7} )}^{2} + {(1)}^{2} + 2 \times \sqrt{7} \times 1 }{ {( \sqrt{7}) }^{2} - {(1)}^{2} } \\ \\ = \frac{7 + 1 + 2 \sqrt{7} }{7 - 1} \\ \\ = \frac{8 + 2 \sqrt{7} }{6} \\ \\ = \frac{4 + \sqrt{7} }{6} \\ \\ \frac{4 + \sqrt{7} }{6} = a - b \sqrt{7} \\ \\ a = \frac{4}{6} \\ \\ a = \frac{2}{3} \\ \\ b = - \frac{1}{6}$

Hope this helps!!!!!

@Mahak24

Thanks....
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Answer 2

\frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 } = a - b \sqrt{7} \\ \\ on \: rationalizing \: we \: get \\ \\ = \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1 } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ we \: get \\ \\ = \frac{ {( \sqrt{7} )}^{2} + {(1)}^{2} + 2 \times \sqrt{7} \times 1 }{ {( \sqrt{7}) }^{2} - {(1)}^{2} } \\ \\ = \frac{7 + 1 + 2 \sqrt{7} }{7 - 1} \\ \\ = \frac{8 + 2 \sqrt{7} }{6} \\ \\ = \frac{4 + \sqrt{7} }{6} \\ \\ \frac{4 + \sqrt{7} }{6} = a - b \sqrt{7} \\ \\ a = \frac{4}{6} \\ \\ a = \frac{2}{3} \\ \\ b = - \frac{1}{6}