# What are the zero(s) of: 3x^2 - 5x -4 = 0?

May 22, 2018

$x = \frac{5}{6} \pm \frac{\sqrt{73}}{6}$

#### Explanation:

Given:

$3 {x}^{2} - 5 x - 4 = 0$

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

Complete the square and use this with $A = \left(6 x - 5\right)$ and $B = \sqrt{73}$ as follows:

$0 = 12 \left(3 {x}^{2} - 5 x - 4\right)$

$\textcolor{w h i t e}{0} = 36 {x}^{2} - 60 x - 48$

$\textcolor{w h i t e}{0} = {\left(6 x\right)}^{2} - 2 \left(6 x\right) \left(5\right) + 25 - 73$

$\textcolor{w h i t e}{0} = {\left(6 x - 5\right)}^{2} - {\left(\sqrt{73}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(6 x - 5\right) - \sqrt{73}\right) \left(\left(6 x - 5\right) + \sqrt{73}\right)$

$\textcolor{w h i t e}{0} = \left(6 x - 5 - \sqrt{73}\right) \left(6 x - 5 + \sqrt{73}\right)$

Hence:

$6 x = 5 \pm \sqrt{73}$

Hence:

$x = \frac{5}{6} \pm \frac{\sqrt{73}}{6}$