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Question:

The square root of $1+x^2+\sqrt{1+x^2+x^4}$1

Explanation:

Instead of solving algebraically, solve the question by first substituting x = 1. You will get one original value and three values in the options. Square each value from the options and match it with the original value of the expression.

Original expression:$1+x^2+\sqrt{(1+x^2+x^4)}$

$=1+{\left(1\right)}^2+\sqrt{[1+{\left(1\right)}^2+{\left(1\right)}^4]}$

$=1+1+\sqrt{3}=2+\sqrt{3}$

Option 1:

$\frac{1}{\sqrt{2}}\left[\sqrt{1+x+x^2}+\sqrt{1-x+x^2}\right]$

$=\frac{1}{\sqrt{2}}\left[\sqrt{1+1+1^2}+\sqrt{1-1+1^2}\right]$

$=\frac{\sqrt{3}+1}{\sqrt{2}}$

Now, square this value.

$=\frac{3+1+2\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$

Hence, option (a).