MCQ
If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ then $\text{x}+\frac{1}{\text{x}}=$
- A$27$
- B$25$
- C$3\sqrt3$
- D$-3\sqrt3$
✓
Answer
- $3\sqrt3$
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2.\text{x}.\frac{1}{\text{x}}=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}$
Squaring both sides.
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2.\text{x}^2.\frac{1}{\text{x}}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2=(623)+2$
$\Rightarrow623+2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2\ \Big\{\text{x}^4+\frac{1}{\text{x}^4}=623\Big\}$
$\Rightarrow625=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=\sqrt{625}=25$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=25+2=27$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{27}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=3\sqrt{3}$
Hence, correct option is (c).
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