How do you solve 6x^2 + 13x = 5 by completing the square?

Aug 1, 2016

$x = \frac{1}{3} , \mathmr{and} x = - \frac{5}{2}$

Explanation:

For ease of explanation - remember that the general form of a quadratic is $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{m a \ge n t a}{c}$

$\textcolor{red}{6} {x}^{2} + 13 x = 5$

The constant (c) has already been moved to the RHS.

We need to make the LHS into the square of a binomial (ie a perfect square)

Step 1. $\textcolor{red}{a}$ must be equal to 1. Divide through by 6.

$\frac{\cancel{6} {x}^{2}}{\cancel{6}} + \frac{\textcolor{b l u e}{13} x}{\textcolor{b l u e}{6}} = \frac{5}{6}$

Step 2: complete the square by adding the missing third term (to both sides)

${\left(\frac{\textcolor{b l u e}{b}}{2}\right)}^{2} \text{ "rArr ((color(blue)13)/(color(blue)6 xx2))^2 " } \Rightarrow {\left(\frac{13}{12}\right)}^{2}$

${x}^{2} + \frac{\textcolor{b l u e}{13} x}{\textcolor{b l u e}{6}} + {\left(\frac{13}{12}\right)}^{2} = \frac{5}{6} + {\left(\frac{13}{12}\right)}^{2}$

Step 3. Write as ${\left(x + \ldots .\right)}^{2}$

${\left(x + \frac{13}{12}\right)}^{2} = \frac{289}{144}$

Step 4: square root both sides.

$x + \frac{13}{12} = \pm \sqrt{\frac{289}{144}}$

$x = \frac{17}{12} - \frac{13}{12} \text{ } \mathmr{and} x = \left(- \frac{17}{12}\right) - \frac{13}{12}$

$x = \frac{4}{12} = \frac{1}{3} \text{ or } x = \frac{- 30}{12} = - \frac{5}{2}$