Question

Find the roots of $x+\frac{1}{x}=3;x\ne 0$.

Answer 1

Finding the roots of the equation by Sreedhar acharya formula

Given equation $x+\frac{1}{x}=3$
$$\begin{gathered} \Rightarrow x+\frac{1}{x}=3 \\ \Rightarrow x^2-3x+1=0 \end{gathered}$$
We know that Sreedhar acharya formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Here $a=1,b=-3\&c=1$, Putting this in above formula
$$\begin{gathered} \Rightarrow x=\frac{3\pm \sqrt{{(-3)}^2-4.1.1}}{2.1} \\ \Rightarrow x=\frac{3\pm \sqrt{5}}{2} \\ \Rightarrow x=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2} \end{gathered}$$

Hence, the required roots are $\frac{3+\sqrt{5}}{2}\&\frac{3-\sqrt{5}}{2}$.