Question

Find a and b
$\frac{ \sqrt{7} - 2 }{ \sqrt{7} + 2 } = a \sqrt{7} + b$
​

Answer 1

Given

$\sf \dfrac{\sqrt{7}-2 }{\sqrt{7}+2 } = a\sqrt{7}+b$

Solving the LHS side, we get,

$\implies\sf \dfrac{\sqrt{7}-2 }{\sqrt{7}+2 }$

$\implies\sf \dfrac{\sqrt{7}-2 }{\sqrt{7}+2 } \times \dfrac{\sqrt{7}-2 }{\sqrt{7}-2 }$

$\implies\sf \dfrac{(\sqrt{7}-2 )^{2} }{(\sqrt{7})^{2}-(2)^{2} }$

$\implies \sf \dfrac{7-4\sqrt{7}+4 }{7-4}$

$\implies \sf \dfrac{11-4\sqrt{7} }{3}$

$\implies \sf \dfrac{11}{3} -\dfrac{4\sqrt{7} }{3}$

$\implies \boxed{\boxed{\bold{ b = \dfrac{11}{3} }}}}$

$\implies \boxed{\boxed{\bold{a = \dfrac{-4}{3} }}}}}}}}}$

                                                         

Answer 2

$\huge\mathfrak{Answer:}$

Given:.

• We have been given that $\sf{\dfrac{ \sqrt{7} - 2 }{ \sqrt{7} + 2 } = a \sqrt{7} + b}$

To Find:

• We need to find the value of a and b.

Solution:

As it is given that

t $\sf{\dfrac{ \sqrt{7} - 2 }{ \sqrt{7} + 2 } = a \sqrt{7} + b}$

Taking LHS, we have

$\sf{ \dfrac{ \sqrt{7} - 2}{ \sqrt{7} + 2} }$

On rationalizing the denominator, we have

$\implies\sf{ \dfrac{ \sqrt{7} - 2 }{ \sqrt{7} + 2} \times \dfrac{ \sqrt{7} - 2 }{ \sqrt{7} - 2 }}$

$\implies\sf{\dfrac{(\sqrt{7}-2)^{2} }{(\sqrt{7} )^{2} -(2)^{2} } }$

$\implies\sf{ \dfrac{7 - \sqrt[4]{7} + 4 }{7 - 4}}$

$\implies\sf{ \dfrac{11 - \sqrt[4]{7} }{3}}$

$\implies\sf{ \dfrac{11}{3} - \dfrac{ 4\sqrt{7} }{3} }$

$\sf{Hence, \: a \: = \dfrac{ - 4}{3} and \: b \: = \dfrac{11}{3} }$