MCQ
If $\text{x}+\frac{1}{\text{x}}=4,$ then $\text{x}^4+\frac{1}{\text{x}^4}=$
- A$196$
- ✓$194$
- C$192$
- D$190$
✓
Answer
Correct option: B.
$194$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(4)^2-2=16-2=14\ ...(1)$
Squaring equation $(1)$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2(\text{x}^2)\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=194$
Hence, correct option is $(b)$.
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(4)^2-2=16-2=14\ ...(1)$
Squaring equation $(1)$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2(\text{x}^2)\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=194$
Hence, correct option is $(b)$.
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