Question

if sq root of 3 + sq root of 2 / sq root of 3 - sq root of 2 = a+b* sq root of 6 find the value of a and b

Answer 1

Answer:

$\dfrac{\sqrt3 + \sqrt2}{\sqrt3 -\sqrt2} = a + b\sqrt6$

Rationalizing denominator :

$\implies \dfrac{\sqrt3 + \sqrt2 }{\sqrt3 - \sqrt2}\times \dfrac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = a + b\sqrt6$

$\implies \dfrac{(\sqrt3 +\sqrt2)^2}{(\sqrt3)^2 -(\sqrt2)^2} = a + b\sqrt6$

$\implies \dfrac{3 + 2 + 2\sqrt6}{3 - 2} = a + b\sqrt6$

$\implies 5 + 2\sqrt6 = a + b\sqrt6$

Solving for equation:

$\boxed{\implies{\boxed{a = 5}}}$

$\implies b\sqrt6 = 2 \sqrt6$

$\boxed{\implies{\boxed{b = 2}}}$

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