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

May 15, 2016

$x = \frac{- 1 \pm \sqrt{17}}{2}$

#### Explanation:

${x}^{2} + x - 4$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 1$ and $c = - 4$

This has zeros given by the quadratic formula:

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

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

$= \frac{- 1 \pm \sqrt{1 + 16}}{2}$

$= \frac{- 1 \pm \sqrt{17}}{2}$

Footnote

Notice the expression $\pm \sqrt{{b}^{2} - 4 a c}$ in the formula.

The radicand ${b}^{2} - 4 a c$ is called the discriminant and often represented by the capital Greek letter delta $\Delta$.

By examining whether $\Delta > 0$, $\Delta = 0$ or $\Delta < 0$ we can tell what kind of zeros a quadratic (with Real coefficients) has:

• If $\Delta > 0$ then the quadratic has two distinct Real zeros. In addition, if $\Delta$ is a perfect square, then the zeros of the quadratic are rational (assuming the coefficients of the quadaratic are).

• If $\Delta = 0$ then the quadratic has one repeated Real zero.

• If $\Delta < 0$ then the quadratic has two non-Real Complex zeros which are Complex conjugates of one another.