# How do you complete the square when a quadratic equation has a coefficient?

Feb 20, 2015

It is easier to explain with an example:

$3 {x}^{2} + 7 x - 5 = 0 \Rightarrow 3 \left({x}^{2} + \frac{7}{3} x\right) - 5 = 0$

Now, since $\frac{7}{3} x$ will have to be the double product of the first and the second terms of the square, if we divide it for the double of the first ($2 x$) we can obtain the second:

$\frac{\frac{7}{3} x}{2 x} = \frac{7}{6}$, so:

$3 \left({x}^{2} + \frac{7}{3} x + {\left(\frac{7}{6}\right)}^{2} - {\left(\frac{7}{6}\right)}^{2}\right) - 5 = 0 \Rightarrow$

$3 \left({x}^{2} + \frac{7}{3} x + \frac{49}{36} - \frac{49}{36}\right) - 5 = 0 \Rightarrow$

$3 \left({x}^{2} + \frac{7}{3} x + \frac{49}{36}\right) - 3 \cdot \frac{49}{36} - 5 = 0 \Rightarrow$

$3 {\left(x + \frac{7}{6}\right)}^{2} - \frac{49}{12} - 5 = 0 \Rightarrow$

$3 {\left(x + \frac{7}{6}\right)}^{2} = \frac{49 + 60}{12} \Rightarrow$

$3 {\left(x + \frac{7}{6}\right)}^{2} = \frac{109}{12} \Rightarrow$ and, if you want to solve:

${\left(x + \frac{7}{6}\right)}^{2} = \frac{109}{36} \Rightarrow \left(x + \frac{7}{6}\right) = \pm \sqrt{\frac{109}{36}} \Rightarrow$

$x = - \frac{7}{6} \pm \frac{\sqrt{109}}{6} = \frac{- 7 \pm \sqrt{109}}{6}$.