How do you find the roots, real and imaginary, of #y=-5(x-3)^2-45 # using the quadratic formula?

1 Answer
Jul 4, 2017

Answer:

#x=-3/2+-9/2i#

Explanation:

To make this more strait forward lets convert the equation back to the form of #y=ax^2+bx+c#

Square the brackets.

#y=-5(x^2-3x+9)-45#

Multiply out the -5

#y=-5x^2+15x-45-45#

#y=-5x^2+15x-90#

Thus the quadratic formula states that

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

where #a=-5"; "b=+15"; "c=-90#

#x=(-15+-sqrt((-15)^2-4(-5)(-90)))/(2(-5))#

#x=-3/2+-(sqrt(-2025))/(-10)#

If you are ever not sure about factors do a quick sketch of a prime factor tree on the side of the page.
Tony B

#x=-3/2+-(sqrt(5^2xx9^2xx(-1)))/(-10)#

#x=-3/2+-( cancel(5)^1xx9xxi)/(cancel(10)^2)#

As there is a #+-# sign in front of #(9xxi)/2# you can ignore any negative within it.

#x=-3/2+-9/2i#