How do you find the roots, real and imaginary, of y=3(x -2)^2+2x+3 y=3(x2)2+2x+3 using the quadratic formula?

1 Answer
May 21, 2018

x = (5 +- i*sqrt(2))/3x=5±i23

Explanation:

First, square (x-2)(x2) and combine like terms.
y = 3(x-2)^2 + 2x - 3y=3(x2)2+2x3
y = 3(x-2)(x-2) + 2x - 3y=3(x2)(x2)+2x3
y = 3(x^2 - 4x + 4) + 2x - 3y=3(x24x+4)+2x3
y = 3x^2 - 12x + 12 + 2x - 3y=3x212x+12+2x3
y = 3x^2 - 10x + 9y=3x210x+9

Quadratic Formula:
x = (-b +- sqrt(b^2 - 4ac))/(2a)x=b±b24ac2a
x = (-(-10) +- sqrt((-10)^2 - 4(3)(9)))/(2(3))x=(10)±(10)24(3)(9)2(3)
x = (10 +- sqrt(100 - 108))/6x=10±1001086
x = (10 +- sqrt(-8))/6x=10±86
x = (10 +- 2i*sqrt(2))/6x=10±2i26

Simplify:
x = (5 +- i*sqrt(2))/3x=5±i23