Question

How do you find the zeros, real and imaginary, of `y= 3x^2+18x-24` using the quadratic formula?

Answer 1

In this case both roots are real:

`x=(-18-24.7)/6~~-7.1` and `x=(-18+24.7)/6~~1.1`

Explanation:

The zeroes, or solutions, are the points where the line crosses the `x` axis, the line `y=0`.

Set the equation equal to `0`:

`3x^2+18x-24=0`

This is the standard form of a quadratic equation:

`ax^2+bx+c=0`

Solve using the quadratic formula:

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

`=(-18+-sqrt(18^2-4*3*(-24)))/(2*3)`

Notice the piece under the square root sign (radical). If it was `0` there would only be one root, if it was negative the roots would be imaginary, but since it is positive both the roots are real.

`x=(-18+-sqrt(612))/6`

`x=(-18-24.7)/6~~-7.1` and `x=(-18+24.7)/6~~1.1`