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

1 Answer
Feb 17, 2016

Answer:

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#