Question

How do you find the zeros, real and imaginary, of `=5(3x-3)^2+13` using the quadratic formula?

Answer 1

`x=1 pm isqrt(13/45)`

Explanation:

`5(3x-3)^2 = 13 i^2` implies
`45(x-1)^2 = 13 i^2`
`x-1 = pm i sqrt(13/45)`
`x=1 pm isqrt(13/45)`

Answer 2

Zeros are `1+0.5375i` and `1-0.5375i`, two complex conjugate numbers.

Explanation:

According to quadratic formula, zeros of `ax^2+bx+c` are given by `(-b+-sqrt(b^2-4ac))/(2a)`

As function is `5(3x-3)^2+13`

=`5(9x^2-18x+9)+13`

= `45x^2-90x+45+13`

= `45x^2-90x+58`

And zeros are `(-(-90)+-sqrt((-90)^2-4×45×58))/(2×45)`

= `(90+-sqrt(8100-10440))/90`

= `1+-sqrt(-2340)/90`

= `1+-i48.37/90`

= `1+-i×0.5375`

i.e. Zeros are `1+0.5375i` and `1-0.5375i`, two complex conjugate numbers.