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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