Question

Find the value of $${\displaystyle \frac{26-15\sqrt{3}}{{[5\sqrt{2}-\sqrt{38+5\sqrt{3}}]}^2}}+{\displaystyle \frac{\sqrt{10}+\sqrt{18}}{\sqrt{8}+\sqrt{(3-\sqrt{5})}}}$$
(a) ${\displaystyle \frac{1}{3}}$
(b) ${\displaystyle \frac{2}{3}}$
(c) ${\displaystyle \frac{3}{5}}$
(d) ${\displaystyle \frac{7}{3}}$

Answer 1

Hint: Initially, we should try to simplify $$\sqrt{38+5\sqrt{3}}$$ by expressing it in such a way that we can use the identity $$a^2+b^2+2ab={(a+b)}^2$$ to get rid of square root sign. Similarly, $$\sqrt{3-\sqrt{5}}$$ can be simplified by expressing it in such a way that we can use the identity $$a^2+b^2-2ab={(a-b)}^2$$and after that the fraction can be rationalised which will lead to final answer.

Complete step-by-step solution:
Let us first consider $$\sqrt{38+5\sqrt{3}}$$ and multiplying $\sqrt{2}$ in numerator and denominator, we get
$$\Rightarrow \sqrt{38+5\sqrt{3}}={\displaystyle \frac{1}{\sqrt{2}}}\sqrt{2\times (38+5\sqrt{3}}$$
$$\Rightarrow \sqrt{38+5\sqrt{3}}={\displaystyle \frac{\sqrt{76+10\sqrt{3}}}{\sqrt{2}}}$$
$$\Rightarrow \sqrt{38+5\sqrt{3}}={\displaystyle \frac{\sqrt{75+1+2\times 1\times 5\sqrt{3}}}{\sqrt{2}}}$$
We can put $$75\text{=}{\left(5\sqrt{3}\right)}^2$$ in the above equation
$$\Rightarrow \sqrt{38+5\sqrt{3}}={\displaystyle \frac{\sqrt{{\left(5\sqrt{3}\right)}^2+1^2+2\times 1\times 5\sqrt{3}}}{\sqrt{2}}}$$
Applying $$a^2+b^2+2ab={(a+b)}^2$$ , we get
$$\Rightarrow \sqrt{38+5\sqrt{3}}={\displaystyle \frac{\sqrt{{(5\sqrt{3}+1)}^2}}{\sqrt{2}}}$$
We can neutralize square root by the square sign over the expression
$$\Rightarrow \sqrt{38+5\sqrt{3}}={\displaystyle \frac{5\sqrt{3}+1}{\sqrt{2}}}..............(1)$$
Now, let us consider $$\sqrt{3-\sqrt{5}}$$ and multiplying $\sqrt{2}$ in numerator and denominator, we get
$$\Rightarrow \sqrt{3-\sqrt{5}}={\displaystyle \frac{\sqrt{2\times (3-\sqrt{5})}}{\sqrt{2}}}$$
$$\Rightarrow \sqrt{3-\sqrt{5}}={\displaystyle \frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}}$$
$$\Rightarrow \sqrt{3-\sqrt{5}}={\displaystyle \frac{\sqrt{5+1-2\sqrt{5}}}{\sqrt{2}}}$$
$$\Rightarrow \sqrt{3-\sqrt{5}}={\displaystyle \frac{\sqrt{{\left(\sqrt{5}\right)}^2+1^2-2\sqrt{5}}}{\sqrt{2}}}$$
Applying $$a^2+b^2-2ab={(a-b)}^2$$ , we get
$$\Rightarrow \sqrt{3-\sqrt{5}}={\displaystyle \frac{\sqrt{{(\sqrt{5}-1)}^2}}{\sqrt{2}}}$$
We can neutralize square root by the square sign over the expression
$$\Rightarrow \sqrt{3-\sqrt{5}}={\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}..............(2)$$

Now substituting equation (1) and (2) in the question we get,
$$\Rightarrow {\displaystyle \frac{26-15\sqrt{3}}{{[5\sqrt{2}-\sqrt{38+5\sqrt{3}}]}^2}}+{\displaystyle \frac{\sqrt{10}+\sqrt{18}}{\sqrt{8}+\sqrt{(3-\sqrt{5})}}}={\displaystyle \frac{26-15\sqrt{3}}{{[5\sqrt{2}-{\displaystyle \frac{5\sqrt{3}+1}{\sqrt{2}}}]}^2}}+{\displaystyle \frac{\sqrt{10}+\sqrt{18}}{\sqrt{8}+{\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}}}$$
We can take $\sqrt{2}$ common from $\sqrt{10}$ and $\sqrt{18}$ in the second expression
$$={\displaystyle \frac{26-15\sqrt{3}}{{\left[{\displaystyle \frac{9-5\sqrt{3}}{\sqrt{2}}}\right]}^2}}+{\displaystyle \frac{\sqrt{2}(\sqrt{5}+\sqrt{9})}{{\displaystyle \frac{3+\sqrt{5}}{\sqrt{2}}}}}$$
$$={\displaystyle \frac{2(26-15\sqrt{3})}{156-90\sqrt{3}}}+{\displaystyle \frac{2(\sqrt{5}+3)}{3+\sqrt{5}}}$$
$$={\displaystyle \frac{2(26-15\sqrt{3})}{6(26-15\sqrt{3})}}+2$$
$$={\displaystyle \frac{1}{3}}+2$$
$$={\displaystyle \frac{7}{3}}$$
Therefore, the given expression can be simplified to $${\displaystyle \frac{7}{3}}$$
Hence, the correct option is (d).

Note: Sometimes we need to multiply the expression with some number (like we multiplied the expression with $\sqrt{2}$ in this question) to achieve the form of $$a^2+b^2+2ab$$ or $$a^2+b^2-2ab$$ . Now you might think that why only $\sqrt{2}$ and not any other number is multiplied to achieve the desired form in this question, the logic behind it is that we first focused on getting $$2ab$$ term and it cannot be achieved by multiplying the expression with any other number except $\sqrt{2}$. After making $$2ab$$ term, we can easily write the remaining expression in the form of $$a^2+b^2$$ , as we can easily predict $$a$$ and $$b$$ from the $$2ab$$ term.