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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Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Question : If $x=2+\sqrt3$, then the value of $\frac{x^{2}-x+1}{x^{2}+x+1}$ is:
Question : If $2 x^2-7 x+5=0$, then what is the value of $x^2+\frac{25}{4 x^2} ?$
Question : If $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$, what is the value of $x$?
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
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