How do you solve the quadratic 6y^2-13y+6=0 using any method?

Oct 6, 2016

Using the quadratic formula
color(white)("XXX")color(green)(y=2/3color(black)(" or ")y=3/2)

Explanation:

The quadratic formula for the general quadratic (in $y$)
$\textcolor{w h i t e}{\text{XXX}} a {y}^{2} + b y + c = 0$
is
$\textcolor{w h i t e}{\text{XXX}} y = \frac{- b \pm \sqrt{{b}^{2} + 4 a c}}{2 a}$

For the example: $6 {y}^{2} - 13 y + 6 = 0$
$\textcolor{w h i t e}{\text{XXX}} a = 6$
$\textcolor{w h i t e}{\text{XXX}} b = - 13$
$\textcolor{w h i t e}{\text{XXX}} c = 6$
giving
$\textcolor{w h i t e}{\text{XXX}} y = \frac{13 \pm \sqrt{{\left(- 13\right)}^{2} - 4 \left(6\right) \left(6\right)}}{2 \left(6\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{13 \pm \sqrt{169 - 144}}{12}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{13 \pm 5}{12}$

$\textcolor{w h i t e}{\text{XXX")=18/12= 3/2color(black)(" or }} = \frac{8}{12} = \frac{2}{3}$

Oct 6, 2016

$y = \frac{2}{3}$ or $y = \frac{3}{2}$

Explanation:

Use an AC method:

Look for a pair of factors of $A C = 6 \cdot 6 = 36$ with sum $B = 13$

The pair $9 , 4$ works.

Use this pair to split the middle term, then factor by grouping as follows:

$0 = 6 {y}^{2} - 13 y + 6$

$\textcolor{w h i t e}{0} = 6 {y}^{2} - 9 y - 4 y + 6$

$\textcolor{w h i t e}{0} = \left(6 {y}^{2} - 9 y\right) - \left(4 y - 6\right)$

$\textcolor{w h i t e}{0} = 3 y \left(2 y - 3\right) - 2 \left(2 y - 3\right)$

$\textcolor{w h i t e}{0} = \left(3 y - 2\right) \left(2 y - 3\right)$

Hence $y = \frac{2}{3}$ or $y = \frac{3}{2}$