Question

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

Answer 1

`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)`