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: $\sqrt{3}$
Correct Answer: $\frac{\sqrt{15}}{2}$
Solution : Given: $x^2+\frac{1}{x^2}=\frac{7}{4}$ $⇒x^2+\frac{1}{x^2}+2=\frac{7}{4}+2$ [adding 2 on both sides] $⇒x^2+\frac{1}{x^2}+2×x^2×\frac{1}{x^2}=\frac{15}{4}$ $⇒(x+\frac{1}{x})^2=\frac{15}{4}$ $\therefore (x+\frac{1}{x})=\frac{\sqrt{15}}{2}$ Hence, the correct answer is $\frac{\sqrt{15}}{2}$.
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