Question

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

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

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