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

1 Answer
David G.
Feb 17, 2016

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

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