Solve for $x$:
$x+\frac{1}{x}=3, x≠0$

Given:

Given quadratic equation is $x+\frac{1}{x}=3, x≠0$.

To do:

We have to solve the given quadratic equation.

Solution:

$x+\frac{1}{x}=3, x≠0$

$\frac{x(x)+1}{x}=3$ 

$\frac{x^2+1}{x}=3$

$x^2+1=3(x)$   (On cross multiplication)

$x^2+1=3x$

$x^2-3x+1=0$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=1, b=-3$ and $c=1$.

Therefore, the roots of the given equation are

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

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

$x=\frac{3\pm \sqrt{9-4}}{2}$ 

$x=\frac{3\pm \sqrt{5}}{2}$  

$x=\frac{3+\sqrt5}{2}$ or $x=\frac{3-\sqrt5}{2}$

The values of $x$ are $\frac{3+\sqrt5}{2}$ and $\frac{3-\sqrt5}{2}$. 

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Updated on: 10-Oct-2022

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