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

1 Answer
Feb 4, 2016

Answer:

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.