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# Find the value of the square root of $\sqrt{20+\sqrt{3}+\sqrt{28-10\sqrt{3}}}$ is:(a) zero(b) 25(c) 49(d) None of the above

Last updated date: 17th Sep 2024
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Hint: To solve this question, we will start from the right of the given expression and then try to eliminate the square root by taking a square inside the square root. To do so, we will use 28 = 25 + 3 and $10\sqrt{3}$ as $2\times 5\times \sqrt{3}$ and then finally use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to get the result.

Complete step-by-step solution
Let us assume a variable as $y=\sqrt{20+\sqrt{3}+\sqrt{28-10\sqrt{3}}}.$ Let us write 28 as 25 + 3 and then write $10\sqrt{3}$ as $2\times 5\times \sqrt{3}.$ So, putting all these substitutions in y, we get,
$y=\sqrt{20+\sqrt{3}+\sqrt{25+3-2\times 5\times \sqrt{3}}}......\left( i \right)$
Consider the identity given as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.$ Let a = 5 and $b=\sqrt{3}.$ Using this above stated identity using a = 5 and $b=\sqrt{3}.$ We get,
${{\left( 5-\sqrt{3} \right)}^{2}}={{\left( 5 \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2\times 5\times \sqrt{3}$
$\Rightarrow {{\left( 5-\sqrt{3} \right)}^{2}}=25+3-2\times 5\times \sqrt{3}......\left( ii \right)$
Comparing equation (i) and (ii), we see that $25+3-2\times 5\times \sqrt{3}$ can be replaced by ${{\left( 5-\sqrt{3} \right)}^{2}}.$
Replacing $25+3-2\times 5\times \sqrt{3}$ by ${{\left( 5-\sqrt{3} \right)}^{2}}$ in equation (i), we get,
$y=\sqrt{20+\sqrt{3}+\sqrt{{{\left( 5-\sqrt{3} \right)}^{2}}}}$
Cancelling the square root and square, we have,
$\Rightarrow y=\sqrt{20+\sqrt{3}+\left( 5-\sqrt{3} \right)}$
$\Rightarrow y=\sqrt{20+5+\sqrt{3}-\sqrt{3}}$
$\Rightarrow y=\sqrt{25}$
$\Rightarrow y=5$
Hence, the value of the term $\sqrt{20+\sqrt{3}+\sqrt{28-10\sqrt{3}}}$ is 5.
Hence, the right option is (d).

Note: Do not proceed with these types of questions by assuming some variable to $\sqrt{3},$ let $x=\sqrt{3},$ then y would become $y=\sqrt{20+x+\sqrt{28-10x}}.$ Then for further processing, take the square on both the sides, ${{y}^{2}}=20+x+\sqrt{28-10x}.$ Then the equation will be of two variables and would have the complex square and square root. Then the equation would become difficult to solve. So, it is better to start by eliminating the square root by forming the perfect square inside.