Question

If $\frac{2\sqrt{7}+3\sqrt{5}}{\sqrt{7}+\sqrt{5}}=P\sqrt{35}+Q$, then what is the value of $2P+Q$

Answer 1

Given equation is $\frac{2\sqrt{7}+3\sqrt{5}}{\sqrt{7}+\sqrt{5}}=P\sqrt{35}+Q$

Solving the L.H.S.:

On rationalising, we get:

$\frac{2\sqrt{7}+3\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times$ $\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}$

$=\frac{2\left(7\right)-2\sqrt{35}+3\sqrt{75}-3\left(5\right)}{7-5}$

$=\frac{-1}{2}+\frac{1}{2}\sqrt{35}$

$\Rightarrow \frac{1}{2}\sqrt{35}-\frac{1}{2}$

Equating the L.H.S. with the R.H.S., we get:

$\frac{1}{2}\sqrt{35}-\frac{1}{2}=P\sqrt{35}+Q$

Hence, $P=\frac{1}{2}$ and $Q=-\frac{1}{2}$.

Hence, $2P+Q=2\times \frac{1}{2}-\frac{1}{2}=1-\frac{1}{2}$

$\therefore 2P+Q=\frac{1}{2}$

Hence, the required value of $2P+Q$ is $\frac{1}{2}$.