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

Question

1486 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-15T12:48:08

$x =(-1+-sqrt(17))/2$

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

This has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(-1+-sqrt(1^2-(4*1*(-4))))/(2*1)$

$=(-1+-sqrt(1+16))/2$

$=(-1+-sqrt(17))/2$

Footnote

Notice the expression $+-sqrt(b^2-4ac)$ in the formula.

The radicand $b^2-4ac$ 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.

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