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

2 Answers
Apr 19, 2017

Answer:

#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

Answer:

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.