Question

If $x+\frac{1}{x}=3$, then find the value of

$x^2+\frac{1}{x^2}$.

• A. $9$
• B. $8$
• C. $7$
• D. $11$

Answer 1

The correct option is C $7$

Given: $x+\frac{1}{x}=3$

We need to find the value of $x^2+\frac{1}{x^2}$.

Squaring on both the sides of the equation $x+\frac{1}{x}=3$:
$(x+\frac{1}{x}{)}^2=3^2$

$x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=9$

$[\because (x+y{)}^2=x^2+y^2+2xy]$

$x^2+\frac{1}{x^2}+2=9$

$(\because x\times \frac{1}{x}=1)$

Subtracting $2$ from both the sides of the above equation:
$x^2+\frac{1}{x^2}+2-2=9-2$

$x^2+\frac{1}{x^2}=7$

The value of $x^2+\frac{1}{x^2}$ is $7$.

Therefore, option (c.) is the correct answer.