Question

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

Answer 1

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

Explanation:

`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.