Question

. Find the value of a and b if :
(i) root7−1/root7+1−√7+1/root7−1= a + b root7

Answer 1

Required answer:

The value of 'a' and 'b' in a + b√7 is 0 and – ⅔.

Step-by-step explanation:

$\tt{ : \implies \: \: \dfrac{ \sqrt{7} - 1}{ \sqrt{7} + 1} - \dfrac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} = a + b \sqrt{7} }$

$\:$

Firstly, rationalising the first part of the LHS :

$\tt{ : \implies \: \: \dfrac{ \sqrt{7} - 1 }{ \sqrt{7} + 1} \times \bigg \{ \dfrac{ \sqrt{7} - 1}{ \sqrt{7} - 1 } \bigg \}}$

$\tt{ : \implies \: \: \dfrac{ {( \sqrt{7} - 1) }^{2} }{ {( \sqrt{7} )}^{2} - {(1)}^{2} } }$

$\tt{ : \implies \: \: \dfrac{ ( \sqrt{7})^{2} + {(1)}^{2} - 2 \: ( \sqrt{7} ) \: (1)}{7 - 1} }$

$\tt{ : \implies \: \: \dfrac{7 + 1 - 2 \sqrt{7} }{6} }$

$\tt{ : \implies \: \: \dfrac{8 - 2 \sqrt{7} }{6} }$

$\tt{ : \implies \: \: \dfrac{ \cancel2 \: (4 - \sqrt{7} )}{ \cancel6 } }$

$\tt{ : \implies \: \: \dfrac{4 - \sqrt{7} }{3}\qquad\dots\dots eq^n.1 }$

$\:$

Now, rationalising the denominator of the second part of the LHS :

$\tt { : \implies \: \: \dfrac{ \sqrt{7} + 1 }{\sqrt{7} - 1} \times \bigg \{ \dfrac{ \sqrt{7} + 1 }{ \sqrt{7} + 1 } \bigg \}}$

$\tt{ : \implies \: \: \dfrac{ {( \sqrt{7} + 1)}^{2} }{ {( \sqrt{7} )}^{2} - {(1)}^{2} } }$

$\tt{ : \implies \: \: \dfrac{ ( \sqrt{7} )^{2} + {(1)}^{2} + 2 \: ( \sqrt{7}) \: (1) }{7 - 1} }$

$\tt{ : \implies \: \: \dfrac{7 + 1 + 2 \sqrt{7} }{6} }$

$\tt{ : \implies \: \: \dfrac{8 + 2 \sqrt{7} }{6} }$

$\tt{ : \implies \: \: \dfrac{ \cancel2 \: (4 + \sqrt{7} )}{ \cancel6} }$

$\tt{ : \implies \: \: \dfrac{4 + \sqrt{7} }{3}\qquad\dots\dots eq^n.2 }$

$\:$

Now, let's subtract eq. ⓶ from eq. ⓵ :

$\tt{ : \implies \: \: \dfrac{4 - \sqrt{7} }{3} - \bigg \{ \dfrac{4 + \sqrt{7} }{3} \bigg \}}$

$\tt{ : \implies \: \: \dfrac{4 - 4 - \sqrt{7} - \sqrt{7} }{3} }$

$\tt{ : \implies \: \: \dfrac{ - 2 \sqrt{7} }{3} }$

$\:$

On comparing this to RHS, we get the values :

$\tt{ \bigstar \: \: \: a = 0 \: and \: b = \dfrac{ - 2}{3} \: \: \: \bigstar}$