Question

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

Answer 1

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`.