# How do you solve the equation 6x^2-17x+12=0 by completing the square?

Sep 20, 2017

$x = \frac{3}{2} \text{ }$ and $\text{ } x = \frac{4}{3}$

#### Explanation:

To cut down on arithmetic involving fractions, multiply by $24 = 6 \cdot {2}^{2}$ in order to make the leading term a perfect square and provide a factor $2$ for the middle coefficient.

Once we reach a difference of squares, use the difference of squares identity:

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

with $A = \left(12 x - 17\right)$ and $B = 1$.

So:

$0 = 24 \left(6 {x}^{2} - 17 x + 12\right)$

$\textcolor{w h i t e}{0} = 144 {x}^{2} - 408 x + 288$

$\textcolor{w h i t e}{0} = {\left(12 x\right)}^{2} - 2 \left(12 x\right) \left(17\right) + {17}^{2} - 1$

$\textcolor{w h i t e}{0} = {\left(12 x - 17\right)}^{2} - {1}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(12 x - 17\right) - 1\right) \left(\left(12 x - 17\right) + 1\right)$

$\textcolor{w h i t e}{0} = \left(12 x - 18\right) \left(12 x - 16\right)$

$\textcolor{w h i t e}{0} = 6 \left(2 x - 3\right) \left(4\right) \left(3 x - 4\right)$

$\textcolor{w h i t e}{0} = 24 \left(2 x - 3\right) \left(3 x - 4\right)$

Hence roots:

$x = \frac{3}{2} \text{ }$ and $\text{ } x = \frac{4}{3}$