Question

find the value of a and b if root 7 -1/root7+1-root7+1/root7-1 =a+b root7

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{ \sqrt{7} - 1}{ \sqrt{7} + 1} - \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} = a + b \sqrt{7} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1} \times \frac{ \sqrt{7} - 1 }{ \sqrt{7} - 1} - \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1 }{ \sqrt{7} + 1} \\ \\ using \: the \:identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{7}) }^{2} + {(1)}^{2} - 2( \sqrt{7} )(1)}{ {( \sqrt{7} )}^{2} - {(1)}^{2} } - \frac{ {( \sqrt{7}) }^{2} + {(1)}^{2} + 2( \sqrt{7} )(1)}{ {( \sqrt{7} )}^{2} - {(1)}^{2} } \\ \\ = \frac{7 + 1 - 2 \sqrt{7} }{7 - 1} - \frac{7 + 1 + 2 \sqrt{7} }{7 - 1} \\ \\ = \frac{8 - 2 \sqrt{7} }{6} - \frac{8 + 2 \sqrt{7} }{6} \\ \\ = \frac{8 - 2 \sqrt{7} - 8 - 2 \sqrt{7} }{6} \\ \\ = \frac{ - 2 \sqrt{7} - 2 \sqrt{7} }{6} \\ \\ = \frac{ - 4 \sqrt{7} }{6} \\ \\ \frac{ - 2 \sqrt{7} }{3} = a + b \sqrt{7} \\ \\ a = 0 \\ \\ b = - \frac{2}{3}$


Hope this helps!!!

@Mahak24

Thanks...
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Answer 2

Answer:

a=0 and b=$\frac{-2}{3}$

Step-by-step explanation:

The given equation is:

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

Solving the LHS of the above equation, we get

$\frac{(\sqrt{7}-1)^{2}-(\sqrt{7}+1)^{2}}{7-1}$

=$\frac{7+1-2\sqrt{7}-7-1-2\sqrt{7}}{6}$

=$\frac{-4}{6}\sqrt{7}$

=$\frac{-2}{3} \sqrt{7}$

Now, comparing with the RHS of the given equation, we get

a=0 and b=$\frac{-2}{3}$