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

Answer:

#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)#