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

1 Answer
David G.
Feb 4, 2016

The two imaginary roots of this equation are =1/2 - (sqrt(207)i)/6 and 1/2 + (sqrt(207)i)/6.

Explanation:

For a quadratic equation in standard form, ax^2 + bx + c = 0, the quadratic formula is a means of finding the roots ('zeros') of the equation - the points where the quadratic (which will be parabola-shaped) crosses the x-axis, which is the line y=0.

We can take the equation we are given and set y=0 and it will be in standard form.

The quadratic formula has the form:

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

If the piece under the square root sign is positive, there will be two real roots. If it is negative there may be imaginary roots.

In this case:

x=(-3+-sqrt(3^2-4*(-3)*(-18)))/(2(-3)) = (-3+-sqrt(9-216))/-6

x=(-3+-sqrt(-207))/-6 = (-3+-sqrt(207)sqrt(-1))/-6 = (-3+-sqrt(207)i)/-6

x=1/2 - (sqrt(207)i)/6 and 1/2 + (sqrt(207)i)/6.

In this case both roots are imaginary.

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