MCQ
If $\text{x}+\frac{1}{\text{x}}=3,$ than the value of $\text{x}^2+\frac{1}{\text{x}^2}$ is:
- A$0$
- B$1$
- C$9$
- ✓$7$
$\text{x}+\frac{1}{\text{x}}=3,$
Squaring both sides, we get
$\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}=9$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}+2=9$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7$
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