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

1 Answer
Apr 4, 2016

Answer:

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#