Question

$\dfrac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=........$
A.$\sqrt{2}$
B.$\sqrt{3}$
C.$\sqrt{3}+\sqrt{2}$
D.$\sqrt{3}-\sqrt{2}$

Answer 1

Hint: In such type of question we first simplify the surds in simplest form by using formula
$\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}\text{ ; a}\ge \text{0,b}\ge \text{0}$ , try to write the number in square root, in prime factorization and then simplify the expression by cancelling the common terms.

Complete step-by-step answer:
Here we have to simplify the given expression. So we can write it
\[\Rightarrow \dfrac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=\dfrac{3+\sqrt{2\times 3}}{\sqrt{2\times 5\times 5}-\sqrt{2\times 2\times 2\times 2\times 3}-\sqrt{2\times 2\times 2\times 2}+\sqrt{5\times 5\times 2}}\]
Here we factorize the terms in square roots in prime factor, now we can write the expression as\[\begin{align}
  & \Rightarrow \dfrac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=\dfrac{\sqrt{3}\times \sqrt{3}+\sqrt{2}\times \sqrt{3}}{\sqrt{5\times 5}\times \sqrt{3}-\sqrt{2\times 2\times 2\times 2}\times \sqrt{3}-\sqrt{2\times 2\times 2\times 2}\times \sqrt{2}+\sqrt{5\times 5}\times \sqrt{2}} \\
 & \\
\end{align}\]
Now we take$\sqrt{3}$ common in numerator so we can write\[\Rightarrow \dfrac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=\dfrac{\sqrt{3}\left( \sqrt{3}+\sqrt{2} \right)}{5\times \sqrt{3}-4\times \sqrt{3}-4\times \sqrt{2}+5\times \sqrt{2}}\]
Now we can write the above expression as
\[\Rightarrow \dfrac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=\dfrac{\sqrt{3}\left( \sqrt{3}+\sqrt{2} \right)}{\sqrt{3}+\sqrt{2}}\]
As we see here $\sqrt{3}+\sqrt{2}$ is both in numerator and denominator so we can cancel it so we have
\[\Rightarrow \dfrac{3+\sqrt{6}}{\sqrt{75}-\sqrt{48}-\sqrt{32}+\sqrt{50}}=\dfrac{\sqrt{3}}{1}\]
So the value of given question is $\sqrt{3}$
Hence option B is correct

Note: When we have an expression we contain irrational numbers in such case we simplify it in such a way that we get common irrational terms by using rule of surds and prime factorization so that common terms cancel each other. Once we do prime factorization of surds we get a look how to proceed further.