How do you find the roots, real and imaginary, of #y= -13x^2 -5x +11(x-3)^2 # using the quadratic formula?

1 Answer
Aug 4, 2016

Answer:

We have two irrational zeros #-71/4-1/4sqrt5833# and #-71/4+1/4sqrt5833#

Explanation:

#y=-13x^2-5x+11(x-3)^2#

= #-13x^2-5x+11(x^2-6x+9)#

= #-13x^2-5x+11x^2-66x+99#

= #-2x^2-71x+99#

As discriminant #b^2-4ac=(-71)^2-4xx(-2)xx99#

= #5041+792=5833# which is positive but not a square of a rational number and hence we have irrational numbers as zeros of te function #y=-13x^2-5x+11(x-3)^2# and these are given by quadratic formula

#(-(-71)+-sqrt5833)/(-4)=-71/4+-1/4sqrt5833# or

#-71/4-1/4sqrt5833# and #-71/4+1/4sqrt5833#

Note: Only factors of #5833# are #19# and #307#.