# How do you solve 3x^2-5x+1=0 by completing the square?

##### 1 Answer
Oct 12, 2016

x=(5+sqrt13)/6 or
$x = \frac{5 - \sqrt{13}}{6}$

#### Explanation:

To solve this equation we have to factorize $3 {x}^{2} - 5 x + 1$
Since we can not use any of the polynomial identities so let us
compute $\textcolor{b l u e}{\delta}$

$\textcolor{b l u e}{\delta = {b}^{2} - 4 a c}$
$\delta = {\left(- 5\right)}^{2} - 4 \left(3\right) \left(1\right)$
$\delta = 25 - 12 = 13$

The roots are :
${x}_{1} = \frac{- b + \sqrt{\delta}}{2 a} = \textcolor{red}{\frac{5 + \sqrt{13}}{6}}$
${x}_{2} = \frac{- b + \sqrt{\delta}}{2 a} = \textcolor{red}{\frac{5 - \sqrt{13}}{6}}$

Now let us solve the equation:
$3 {x}^{2} - 5 x + 1 = 0$
$\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = 0$
$\left(x - \textcolor{red}{\frac{5 + \sqrt{13}}{6}}\right) \left(x - \textcolor{red}{\frac{5 - \sqrt{13}}{6}}\right) = 0$
x-(5+sqrt13)/6=0 rArr x=(5+sqrt13)/6 or
$x - \frac{5 - \sqrt{13}}{6} = 0 \Rightarrow x = \frac{5 - \sqrt{13}}{6}$