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

1 Answer
May 21, 2018

x = -4 +- isqrt(2)x=4±i2

Explanation:

2 Ways to Solve:

Solve by Square Roots:
2(x + 4)^2 + 4 = 02(x+4)2+4=0
2(x+4)^2 = -42(x+4)2=4
(x + 4)^2 = -2(x+4)2=2
(x + 4) = +- sqrt(-2)(x+4)=±2
x = -4 +- isqrt(2)x=4±i2

Using Quadratic Formula:
Need to multiply out first
y = 2(x +4)^2 + 4y=2(x+4)2+4
y = 2(x+4)(x+4) + 4y=2(x+4)(x+4)+4
y = 2(x^2 + 8x + 16)+4y=2(x2+8x+16)+4
y = 2x^2 + 16x + 32 + 4y=2x2+16x+32+4
y = 2x^2 + 16x + 36y=2x2+16x+36

x = (-b +- sqrt(b^2 -4ac))/(2a)x=b±b24ac2a
x = (-(16) +- sqrt((16)^2 - 4(2)(36)))/(2(2))x=(16)±(16)24(2)(36)2(2)
x = (-16 +- sqrt(256-288))/4x=16±2562884
x = (-16 +- sqrt(-32))/4x=16±324
x = (-16 +- 4isqrt(2))/4x=16±4i24
x = -4 +- isqrt(2)x=4±i2