Question

If a and b rational numbers and $\frac{6}{3\sqrt{2}-2\sqrt{3}}=a\sqrt{2}-b\sqrt{3}$ then find the value of a and b.

Answer 1

Given that,

$\frac{6}{3\sqrt{2}-2\sqrt{3}}=a\sqrt{2}-b\sqrt{3}$

$\Rightarrow \frac{6}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=a\sqrt{2}-b\sqrt{3}$

$\Rightarrow \frac{6(3\sqrt{2}-2\sqrt{3})}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2}=a\sqrt{2}-b\sqrt{3}$

$\Rightarrow \frac{6(3\sqrt{2}-2\sqrt{3})}{18-12}=a\sqrt{2}-b\sqrt{3}$

$\Rightarrow 3\sqrt{2}-2\sqrt{3}=a\sqrt{2}-b\sqrt{3}$

On comparing that and we get,

$\Rightarrow a=3andb=2$

Hence, this is the answer.