Question

Simplify :- root3 - root 2/ root 3 + root 2

Answer 1

Answer:

see this picture for answer

Answer 2

Given:

$\large \sf \dfrac {\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

Solution:

$\implies \sf \dfrac {\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

Multiply the given fraction by $\bf \dfrac {\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

We get,

$\implies \sf \dfrac {(\sqrt{3} - \sqrt{2}) \times (\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2}) \times (\sqrt{3} - \sqrt{2})}$

Using $\bf (a+b)(a-b) = a^2 - b^2$

$\implies \sf \dfrac {(\sqrt{3} - \sqrt{2}) \times (\sqrt{3} - \sqrt{2})}{3 - 2}$

$\implies \sf \dfrac {(\sqrt{3} - \sqrt{2}) \times (\sqrt{3} - \sqrt{2})}{1}$

$\implies \sf (\sqrt{3} - \sqrt{2}) \times (\sqrt{3} - \sqrt{2})$

$\implies \sf (\sqrt{3} - \sqrt{2})^2$

Using $\bf a^2 - b^2 = a^2 - 2ab + b^2$

$\implies \sf 3 - 2 \sqrt{6} + 2$

$\implies \sf 5 - 2 \sqrt{6}$

$\large{\underline{\boxed{\underline{\bf \therefore 5 - 2 \sqrt{6}}}}}$