Question

Solving $x\sqrt{24}=x\sqrt{3}+\sqrt{6}$ we get $x=\frac{a+\sqrt{b}}{7}$. Find $a$ and $b$

Answer 1

$x\sqrt{24}=x\sqrt{3}+\sqrt{6}$

$x\left(\sqrt{8\times 3}\right)=x\sqrt{3}+\sqrt{6}$

$2x\sqrt{6}=x\sqrt{3}+\sqrt{6}$

$2x\sqrt{2}=x+\sqrt{2}$

$x(2\sqrt{2}-1)=\sqrt{2}$

$x=\frac{\sqrt{2}}{2\sqrt{2}-1}\times \frac{2\sqrt{2}+1}{2\sqrt{2}+1}$

$=\frac{\sqrt{2}}{7}(2\sqrt{2}+1)$

$=\left(\frac{4+\sqrt{2}}{7}\right)=\frac{a+\sqrt{b}}{7}$

$a=7$ $b=2$