Question
If $\text{x}^2+\frac{1}{\text{x}^2}=102,$ then $\text{x}-\frac{1}{\text{x}}=$
  1. 8
  2. 10
  3. 12
  4. 13

Answer

  1. 10
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$=102-2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=102\Big\}$
$=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=10$
Hence, correct option is (b).

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