Question

If \(x+\frac{1}{x}=4\), find the value of \(x^2+\frac{1}{x^2}\).

• A. 12
• B. 14
• C. 16
• D. 18

Hint:

You do not have to solve the quadratic for x explicitly, but if you do, notice how the surd terms are designed to cancel out when you add $x^2$ and $\frac{1}{x^2}$.

Answer 1

The Correct Option is B

Step 1: Instead of squaring the given expression, solve directly for $x$ using the equation $x+\frac{1}{x}=4$.
Step 2: Multiply both sides by $x$: $x^2+1=4x \Rightarrow x^2-4x+1=0$.
Step 3: Apply the quadratic formula: $x=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}$.
Step 4: Take $x=2+\sqrt{3}$. Then $x^2=(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}$.
Step 5: Find $\frac{1}{x^2}$ by rationalising: $\frac{1}{x^2}=\frac{1}{7+4\sqrt{3}}=\frac{7-4\sqrt{3}}{(7+4\sqrt{3})(7-4\sqrt{3})}=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$.
Step 6: Add: $x^2+\frac{1}{x^2}=(7+4\sqrt{3})+(7-4\sqrt{3})=14$. \[\boxed{14}\]