Math, asked by ramneek3, 1 year ago

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

Answers

Answered by DaIncredible
553
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

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Answered by boffeemadrid
166

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}

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