How do you find the discriminant of x^2-12x+4=0?

Dec 1, 2016

$\Delta = {b}^{2} - 4 a c$

Explanation:

${x}^{2} - 12 x + 4 = 0$

is in the form:

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

with $a = 1$, $b = - 12$ and $c = 4$

It has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 12\right)}^{2} - 4 \left(1\right) \left(4\right) = 144 - 16 = 128 = 2 \cdot {8}^{2}$

Since $\Delta > 0$ this quadratic equation has two distinct Real roots, but since $\Delta$ is not a perfect square, those roots are irrational.

We can find the roots using the quadratic formula:

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

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

$\textcolor{w h i t e}{x} = \frac{12 \pm \sqrt{2 \cdot {8}^{2}}}{2}$

$\textcolor{w h i t e}{x} = \frac{12 \pm 8 \sqrt{2}}{2}$

$\textcolor{w h i t e}{x} = 6 \pm 4 \sqrt{2}$