Question

Find the roots of equation $x+\frac{1}{x}=3.$

Answer 1

First we will rewrite the given equation,

$\begin{array}{cc}x+\frac{1}{x}& =3\\ x^2+1& =3x\\ x^2-3x+1& =0\end{array}$

Now, by using quadratic formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=1,b=-3,c=1$

$x=\frac{3\pm \sqrt{(-3{)}^2-4\times 1\times 1}}{2\times 1}$

$x=\frac{3\pm \sqrt{9-4}}{2}$

$x=\frac{3\pm \sqrt{5}}{2}$

Hence the roots are $x=\frac{3+\sqrt{5}}{2}$ or $x=\frac{3-\sqrt{5}}{2}.$