MCQ
If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ than $\text{x}+\frac{1}{\text{x}}=$
- A$27$
- ✓$3\sqrt{3}$
- C$25$
- D$-3\sqrt{3}$
✓
Answer
Correct option: B.
$3\sqrt{3}$
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=623$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}^2\times\frac{1}{\text{x}^2}=623+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=625$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{625}=25$
Now,
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=25+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=27$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{27}=3\sqrt{3}$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}^2\times\frac{1}{\text{x}^2}=623+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=625$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{625}=25$
Now,
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=25+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=27$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{27}=3\sqrt{3}$
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