Question

If the square root of $\sqrt{35-12\sqrt{6}-4\sqrt{15}+6\sqrt{10}}$ = $m\sqrt{2}-k\sqrt{3}+\sqrt{5}$ . Find k+m

Answer 1

It is given that $\sqrt{35-12\sqrt{6}-4\sqrt{15}+6\sqrt{10}}=m\sqrt{2}-k\sqrt{3}+\sqrt{5}$. Squaring both sides, we get:

${\left(\sqrt{35-12\sqrt{6}-4\sqrt{15}+6\sqrt{10}}\right)}^2={(m\sqrt{2}-k\sqrt{3}+\sqrt{5})}^2$

$\Rightarrow 35-12\sqrt{6}-4\sqrt{15}+6\sqrt{10}={\left(m\sqrt{2}\right)}^2+{(-k\sqrt{3})}^2+{\left(\sqrt{5}\right)}^2$

$+(2\times m\sqrt{2}\times -k\sqrt{3})+(2\times -k\sqrt{3}\times \sqrt{5})+(2\times \sqrt{5}\times m\sqrt{2})$

$(\because (a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca)$

$\Rightarrow 35-12\sqrt{6}-4\sqrt{15}+6\sqrt{10}={2m}^2+{3k}^2+5^2-2mk\sqrt{6}-2k\sqrt{15}+2m\sqrt{10}$

Now, by comparing the coefficients of $\sqrt{6}$, $\sqrt{10}$ and $\sqrt{15}$, we have:

$-2mk=-12\Rightarrow mk=\frac{-12}{-2}\Rightarrow mk=6$

$-2k=-4\Rightarrow k=\frac{-4}{-2}\Rightarrow k=2$

$2m=6\Rightarrow m=\frac{6}{2}\Rightarrow m=3$

Therefore, $k+m=2+3=5$

Hence, $k+m=5$.