Question

Evaluate: $\sqrt{12-\sqrt{68+48\sqrt{2}}}$

• A. $\sqrt{2}-3$
• B. $2+\sqrt{2}$
• C. $2-\sqrt{2}$
• D. $6–2\sqrt{2}$

Answer 1

The correct option is C

$2-\sqrt{2}$

Explanation for the correct answer:

Given:

$\sqrt{12-\sqrt{68+48\sqrt{2}}}$

Simplifying the given expression:

By using the algebraic identity ${\left(a+b\right)}^2=a^2+b^2+2ab$.

$\begin{array}{rcl}\sqrt{12-\sqrt{68+48\sqrt{2}}}& =& \sqrt{12-\sqrt{36+32+48\sqrt{2}}}\\ & =& \sqrt{12-\sqrt{{\left(6\right)}^2+{\left(4\sqrt{2}\right)}^2+2\left(6\times 4\sqrt{2}\right)}}\end{array}$

Here, we get $a=6,b=4\sqrt{2}$, then with the algebraic identity we get

$\begin{array}{rcl}\sqrt{12-\sqrt{{\left(6+4\sqrt{2}\right)}^2}}& =& \sqrt{12-\left(6+4\sqrt{2}\right)}\\ & =& \sqrt{12-6-4\sqrt{2}}\\ & =& \sqrt{6-4\sqrt{2}}\end{array}$

Now we can split 6 as 4 = 2, and by using the algebraic identity ${\left(a+b\right)}^2=a^2+b^2+2ab$.

$\Rightarrow \sqrt{4+2-4\sqrt{2}}=\sqrt{{\left(2\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(2\sqrt{2}\right)}$

Here, we get $a=2,b=\sqrt{2}$, then with the algebraic identity we get

$\Rightarrow \sqrt{{\left(2-\sqrt{2}\right)}^2}=2-\sqrt{2}$

Therefore, The value of $\sqrt{12-\sqrt{68+48\sqrt{2}}}$ is $2-\sqrt{2}$.

Hence, the correct answer is option (C).