Question : If $x^2-2\sqrt{10}x+1=0$, what is the value of $(x-\frac{1}{x})$?
Option 1: $4$
Option 2: $6$
Option 3: $3$
Option 4: $5$
Correct Answer: $6$
Solution : Given: $x^2-2\sqrt{10}x+1=0$ Dividing both sides by $x$, we get, $⇒x-2\sqrt{10}+\frac{1}{x}=0$ $⇒x+\frac{1}{x}=2\sqrt{10}$ Squaring both sides, we get, $⇒x^2+\frac{1}{x^2}+2×x×\frac{1}{x}=(2\sqrt{10})^2$ $⇒x^2+\frac{1}{x^2}=38$ Subtracting 2 from both sides, we get, $⇒x^2+\frac{1}{x^2}-2=38-2$ $⇒x^2+\frac{1}{x^2}-2×x×\frac{1}{x}=36$ $⇒(x-\frac{1}{x})^2=36$ $\therefore x-\frac{1}{x}=\sqrt{36}=6$ Hence, the correct answer is $6$.
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Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
Question : If $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$, then the value of $(a^{2}-ax)$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
Question : If $x + \frac{1}{x} = \sqrt{3}$, then the value of $x^{18} + x^{12} + x^{6} + 1$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3
Question : If for a non-zero $x$, $3x^{2}+5x+3=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Option 1: $\frac{10}{27}$
Option 2: $-\frac{10}{27}$
Option 3: $\frac{2}{3}$
Option 4: $-\frac{2}{3}$
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