Question

If $y=\sqrt{\sqrt{7}+7+\sqrt{8+2\sqrt{7}}}-\sqrt{7},$ then the value of $y$ will be

• A. $1$
• B. $\sqrt{7}$
• C. $8-2\sqrt{7}$
• D. $\sqrt{7}-1$

Answer 1

The correct option is A

$1$

We have,

$y=\sqrt{\sqrt{7}+7+\sqrt{8+2\sqrt{7}}}-\sqrt{7}$

$\Rightarrow y=\sqrt{\sqrt{7}+7+\sqrt{7+1+2\sqrt{7}}}-\sqrt{7}$

$\Rightarrow y=\sqrt{\sqrt{7}+7+\sqrt{{\left(\sqrt{7}\right)}^2+1^2+2\times 1\times \sqrt{7}}}-\sqrt{7}$ $\left[\because {\left(\sqrt{a}\right)}^2=a\right]$

$\Rightarrow y=\sqrt{\sqrt{7}+7+\sqrt{{\left(\sqrt{7}+1\right)}^2}}-\sqrt{7}$ $\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]$

$\Rightarrow y=\sqrt{\sqrt{7}+7+\sqrt{7}+1}-\sqrt{7}$

$\Rightarrow y=\sqrt{2\sqrt{7}+7+1}-\sqrt{7}$

$\Rightarrow y=\sqrt{{\left(\sqrt{7}\right)}^2+1^2+2\sqrt{7}}-\sqrt{7}$

$\Rightarrow y=\sqrt{{\left(\sqrt{7}+1\right)}^2}-\sqrt{7}$

$\Rightarrow y=\sqrt{7}+1-\sqrt{7}$

$\Rightarrow y=1$

Therefore, the value of $y$ is $1.$

Hence, option (A) is correct.