How do you find the roots, real and imaginary, of y=4x^2-x+12 using the quadratic formula?

Dec 31, 2015

$x = \frac{1 \pm i \sqrt{191}}{8}$

Explanation:

For any quadratic equation $y = a {x}^{2} + b x + c$, the roots are found through the formula

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

In this case, when $y = 4 {x}^{2} - x + 12$,

$\left\{\begin{matrix}a = 4 \\ b = - 1 \\ c = 12\end{matrix}\right.$

Plug these into the quadratic equation.

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - \left(4 \times 4 \times 12\right)}}{2 \times 4}$

$x = \frac{1 \pm \sqrt{1 - 192}}{8}$

$x = \frac{1 \pm \sqrt{- 191}}{8}$

$x = \frac{1 \pm i \sqrt{191}}{8}$

The two solutions of this function are imaginary. The solution has no real solutions and will never cross the $x$-axis.

graph{4x^2-x+12 [-20, 20, -10, 42.6]}