Question
If $\text{x}+\frac{1}{\text{x}}=5,$ then $\text{x}^2+\frac{1}{\text{x}^2}=$

Answer

By using identity $(a + b)^2 = a^2 + b^2 + 2ab.$
we have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(5)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big\{\text{x}+\frac{1}{\text{x}}=5\text{ given}\Big\}$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=25-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$

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