Question

Guys please help me with this question

Answer 1

Answer:

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

Step-by-step explanation:

Given that:

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

To find:

The value of a and b.

Solution:

We have,

$\frac{ \sqrt{7} - 1}{ \sqrt{7} + 1 } - \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1}$

By Rationalising the denominator, we get

${\huge{\Rightarrow}} \: \frac{ \sqrt{7} - 1}{ \sqrt{7 + 1} } \times \frac{ \sqrt{7 } - 1}{ \sqrt{7} - 1 } - \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 } \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1 }$

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

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

${\huge{\Rightarrow}} \: \frac{8 - 2 \sqrt{7} }{7 - 1} - \frac{8 + 2 \sqrt{7} }{7 - 1}$

${\huge{\Rightarrow}} \: \frac{1}{6}(8 - 2 \sqrt{7} - 8 - 2 \sqrt{7} )$

${\huge{\Rightarrow}} \:0 - \frac{4 \sqrt{7} }{6 } = 0 + ( \frac{ - 2}{3} ) \sqrt{7}$

$∴ \: 0 + ( \frac{ - 2}{3} ) = a + b \sqrt{7}$

According to the question we find,the value of

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