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 $6$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Question : If $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$, then the value of $(a^{2}-ax)$ is:
Question : If $x + \frac{1}{x} = \sqrt{3}$, then the value of $x^{18} + x^{12} + x^{6} + 1$ is:
Question : If for a non-zero $x$, $3x^{2}+5x+3=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile