Question

3/√3-√2=a√3-b√2.Find the value of a and b?​

Answer 1

Given:

• $\tt\dfrac{3}{\sqrt 3-\sqrt2}=a\sqrt3-b\sqrt2$

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To find:

• Value of a and b

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Solution:

Rationalising the given equation by multiplying LHS with the conjugate of its denominator.

$\implies\tt\dfrac{3}{ \sqrt 3-\sqrt 2 }\times \dfrac{\sqrt 3+\sqrt 2}{\sqrt 3+\sqrt 2}=a\sqrt3-b\sqrt2$

Now solve it

$\tt\implies \dfrac{3(\sqrt 3+\sqrt 2)}{( \sqrt 3-\sqrt 2 )(\sqrt 3+\sqrt 2)}=a\sqrt3-b\sqrt2$

Applying algebraic identity in denominator: (x+y)(x-y)  =x²-y²   where x=√3 and y=√2

$\tt \implies\dfrac{3\sqrt 3+3\sqrt 2}{( \sqrt 3)^2-(\sqrt 2 )^2}=a\sqrt3-b\sqrt2$

$\tt\implies \dfrac{3\sqrt 3+3\sqrt 2}{( 3-2)}=a\sqrt3-b\sqrt2$

$\tt \implies\dfrac{3\sqrt 3+3\sqrt 2}{1}=a\sqrt3-b\sqrt2$

$\tt \implies{3\sqrt 3+3\sqrt 2}=a\sqrt3-b\sqrt2$

Now by comparing LHS and RHS, we get:

• a=√3

• b=-√3

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