Question

If √2 + √3 upon 2√2 - 2√3 = a-b√6 find the value of a and b.
Pls fast

Answer 1

Answer

a = - ⁵/₂ , b = 1

Given

$\bullet\ \; \rm \dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}-2\sqrt{3}}=a-b\sqrt{6}$

To Find

Values of a and b

Solution

Let's take LHS part ,

$\to \rm \dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}-2\sqrt{3}}$

Now , rationalize the denominator ,

$\to \rm \dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}-2\sqrt{3}}\times \dfrac{2\sqrt{2}+2\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}\\\\\to \rm \dfrac{\sqrt{2}( 2\sqrt{2}+2\sqrt{3} )+\sqrt{3}(2\sqrt{2}+2\sqrt{3})}{(2\sqrt{2})^2-(2\sqrt{3})^2}\\\\ \because\ \; \bf (x+y)(x-y)=x^2-y^2\ \\\\\to \rm \dfrac{4+2\sqrt{6}+2\sqrt{6}+6}{8-12}\\\\\to \rm \dfrac{10+4\sqrt{6}}{-4}\\\\\to \rm \dfrac{2(5+2\sqrt{6})}{2(-2)}\\\\\to \rm \dfrac{-5-2\sqrt{6}}{2}\\\\\to \bf -\dfrac{5}{2}-\sqrt{6}$

Now , compare this part with RHS part , we get ,

$\bf a=-\dfrac{5}{2}\ \; \; \&\ \; \; b=1$