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

2 Answers
Apr 19, 2017

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)

Apr 19, 2017

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.