MCQ
If $\text{x}=3+\sqrt{8}$ then $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=?$
- ✓$34$
- B$56$
- C$28$
- D$63$
✓
Answer
Correct option: A.
$34$
$\text{x}=3+\sqrt{8}$
$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$
Now, $\frac{1}{\text{x}}=\frac{1}{3+\sqrt{8}}$
$=\frac{1}{3+\sqrt{8}}\times\frac{3-\sqrt{8}}{3-\sqrt{8}}$
$=\frac{3-\sqrt{8}}{3^2-\big(\sqrt{8}\big)^2}$
$=\frac{3-\sqrt{8}}{9-8}$
$=3-\sqrt{8}$
$\Rightarrow\Big(\frac{1}{\text{x}}\Big)^2=\big(3-\sqrt{8}\big)^2=9+8-6\sqrt{8}=17-6\sqrt{8}$
Then, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=17+6\sqrt{8}+17-6\sqrt{8}=34$
Hence, the correct option is $(a).$
$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$
Now, $\frac{1}{\text{x}}=\frac{1}{3+\sqrt{8}}$
$=\frac{1}{3+\sqrt{8}}\times\frac{3-\sqrt{8}}{3-\sqrt{8}}$
$=\frac{3-\sqrt{8}}{3^2-\big(\sqrt{8}\big)^2}$
$=\frac{3-\sqrt{8}}{9-8}$
$=3-\sqrt{8}$
$\Rightarrow\Big(\frac{1}{\text{x}}\Big)^2=\big(3-\sqrt{8}\big)^2=9+8-6\sqrt{8}=17-6\sqrt{8}$
Then, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=17+6\sqrt{8}+17-6\sqrt{8}=34$
Hence, the correct option is $(a).$
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