MCQ
If $\text{x}=3+\sqrt{8},$ then the value of $(\text{x}^2+\frac{1}{\text{x}^2})$ is:
- ✓$34$
- B$32$
- C$12$
- D$6$
✓
Answer
Correct option: A.
$34$
Given, $(\text{3}+\sqrt{8})$
$\frac{1}{\text{x}}=\frac{1}{(3+\sqrt{8})}=\frac{1}{(3+\sqrt{8})}\times\frac{(3-\sqrt{8})}{(3-\sqrt{8})}$
$=\frac{(3-\sqrt{8})}{(3^2-(\sqrt{8})^2)}=\frac{3+\sqrt{8}}{9-8}=(3-\sqrt{8})$
$(\text{x}+\frac{1}{\text{x}})=(3+\sqrt{8})+(3-\sqrt{8})=6$
$\Rightarrow(\text{x}+\frac{1}{\text{x}})^2=6^2=36$
$\Rightarrow(\text{x}^2+\frac{1}{\text{x}^2})+2\times\text{x}\times\frac{1}{\text{x}}=36$
$\Rightarrow(\text{x}^2+\frac{1}{\text{x}^2})+2=36$
$\Rightarrow(\text{x}+\frac{1}{\text{x}^2})=36-2=34$
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