Question

How do you find the roots, real and imaginary, of `y=-(x-3)^2+4x^2-x-5` using the quadratic formula?

Answer 1

Roots are real (irrational) numbers `(-5+sqrt193)/6` and `(-5-sqrt193)/6`

Explanation:

Simplifying `y=-(x-3)^2+4x^2-x-5=-(x^2-6x+9)+4x^2-x-5`

or `y=3x^2+5x-14`

Its root will be given by the equation `3x^2+5x-14=0`.

Using quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)`

these are `(-5+-sqrt(5^2-4*3*(-14)))/(2*3)=(-5+-sqrt(25+168))/6=(-5+-sqrt193)/6`

Hence roots are irrational numbers `(-5+sqrt193)/6` and `(-5-sqrt193)/6`