# What would be the best method to solve -3x^2+12x+1=0?

May 21, 2018

$x = 2 \pm \frac{\sqrt{39}}{3}$

#### Explanation:

Given:

$- 3 {x}^{2} + 12 x + 1 = 0$

This is in the form:

$a {x}^{2} + b x + c = 0$

with $a = - 3$, $b = 12$ and $c = 1$

In order to decide which method to use, we can first examine the discriminant:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{12}}^{2} - 4 \left(\textcolor{b l u e}{- 3}\right) \left(\textcolor{b l u e}{1}\right) = 144 + 12 = 156$

Since $\Delta > 0$ we can tell that the quadratic has $2$ distinct real solutions, but because it is not a perfect square we can also tell that those solutions are irrational.

So there is no rational factorisation and we can discount the use of an AC method or similar.

We can use the quadratic formula or we can complete the square.

If we choose to use the quadratic formula, then we find:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- \left(\textcolor{b l u e}{12}\right) \pm \sqrt{156}}{2 \left(\textcolor{b l u e}{- 3}\right)}$

$\textcolor{w h i t e}{x} = \frac{- 12 \pm 2 \sqrt{39}}{- 6}$

$\textcolor{w h i t e}{x} = 2 \pm \frac{\sqrt{39}}{3}$

When completing the square, note that $b = 12$ is already divisible by $a = - 3$ and by $2$. So to do most our arithmetic using integers, we can premultiply by $- 3$ to make the leading term square:

$0 = - 3 \left(- 3 {x}^{2} + 12 x + 1\right)$

$\textcolor{w h i t e}{0} = 9 {x}^{2} - 36 x - 3$

$\textcolor{w h i t e}{0} = {\left(3 x\right)}^{2} - 2 \left(3 x\right) \left(6\right) + {\left(6\right)}^{2} - 39$

$\textcolor{w h i t e}{0} = {\left(3 x - 6\right)}^{2} - {\left(\sqrt{39}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(3 x - 6\right) - \sqrt{39}\right) \left(\left(3 x - 6\right) + \sqrt{39}\right)$

$\textcolor{w h i t e}{0} = \left(3 x - 6 - \sqrt{39}\right) \left(3 x - 6 + \sqrt{39}\right)$

So:

$3 x = 6 \pm \sqrt{39}$

So:

$x = 2 \pm \frac{\sqrt{39}}{3}$