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

1 Answer
May 21, 2018

Answer:

#x = (5 +- i*sqrt(2))/3#

Explanation:

First, square #(x-2)# and combine like terms.
#y = 3(x-2)^2 + 2x - 3#
#y = 3(x-2)(x-2) + 2x - 3#
#y = 3(x^2 - 4x + 4) + 2x - 3#
#y = 3x^2 - 12x + 12 + 2x - 3#
#y = 3x^2 - 10x + 9#

Quadratic Formula:
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#
#x = (-(-10) +- sqrt((-10)^2 - 4(3)(9)))/(2(3))#
#x = (10 +- sqrt(100 - 108))/6#
#x = (10 +- sqrt(-8))/6#
#x = (10 +- 2i*sqrt(2))/6#

Simplify:
#x = (5 +- i*sqrt(2))/3#