How do you find the zeros, real and imaginary, of y=2(x+4)^2+4 using the quadratic formula?

1 Answer
May 21, 2018

x = -4 +- isqrt(2)

Explanation:

2 Ways to Solve:

Solve by Square Roots:
2(x + 4)^2 + 4 = 0
2(x+4)^2 = -4
(x + 4)^2 = -2
(x + 4) = +- sqrt(-2)
x = -4 +- isqrt(2)

Using Quadratic Formula:
Need to multiply out first
y = 2(x +4)^2 + 4
y = 2(x+4)(x+4) + 4
y = 2(x^2 + 8x + 16)+4
y = 2x^2 + 16x + 32 + 4
y = 2x^2 + 16x + 36

x = (-b +- sqrt(b^2 -4ac))/(2a)
x = (-(16) +- sqrt((16)^2 - 4(2)(36)))/(2(2))
x = (-16 +- sqrt(256-288))/4
x = (-16 +- sqrt(-32))/4
x = (-16 +- 4isqrt(2))/4
x = -4 +- isqrt(2)