Question : If $x=\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}$, then what is the value of $(x^{2}+x-9)$?
Option 1: 0
Option 2: $3\sqrt2$
Option 3: $3\sqrt3$
Option 4: $5\sqrt3$
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Correct Answer: $5\sqrt3$
Solution :
Given: $x=\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}$
Using the rationalisation method, we get,
⇒ $x=\sqrt{\frac{2+\sqrt3}{2-\sqrt3} \times \frac{2+\sqrt3}{2+\sqrt3}}$
⇒ $x=\sqrt{\frac{(2+\sqrt3)^2}{2^2-\sqrt3^2}}$
⇒ $x=\sqrt{\frac{(2+\sqrt3)^2}{4-3}}$
⇒ $x=(2+\sqrt3)$
Putting $x=(2+\sqrt3)$ in $x^2+x-9$, we get,
= $(2+\sqrt3)^2+(2+\sqrt3)-9$
= $(4+3+4\sqrt3)+(2+\sqrt3)-9$
= $5\sqrt3$
Hence, the correct answer is $5\sqrt3$.
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