Question : If $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$, then the value of $x$ is equal to:
Option 1: 3
Option 2: 4
Option 3: 1
Option 4: 2
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Correct Answer: 2
Solution : Given: $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$. Use the algebraic identity, $(a+b)^2=a^2+b^2+2ab$. $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{4+3+4 \sqrt{3}}}}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{(2+\sqrt{3}}})^2}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+8 (2+\sqrt{3}})}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{3+16+8\sqrt{3}}}$ $⇒x=\sqrt{–\sqrt{3}+\sqrt{(4+\sqrt{3}})^2}$ $⇒x=\sqrt{–\sqrt{3}+4+\sqrt{3}}$ $\therefore x=\sqrt4=2$ Hence, the correct answer is 2.
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