How do you find the roots, real and imaginary, of #y=-5(x-2)^2+3x^2-10x-5 # using the quadratic formula?

2 Answers
Apr 28, 2017

Answer:

#x=5/2 +- 5/2i#

Explanation:

#y=−5(x−2)^2+3x^2−10x−5#

by expanding and simplifing,
#y=−5(x^2−4x+4)+3x^2−10x−5#
#y=−5x^2+20x-20+3x^2−10x−5#
#y=−2x^2+10x-25#

Using the quadratic formula: #x=(-b+- sqrt(b^2-4ac))/(2a)#,
#x=(-10+- sqrt(10^2-4(-2)(-25)))/(2(-2))#

Simplifying,
#x=5/2 +- 5/2i#

Apr 28, 2017

Answer:

We first work this into a 'proper' quadratic equation:

Explanation:

#y=-5(x^2-4x+4)+3x^2-10x-5->#

#y=-5x^2+20x-20+3x^2-10x-5->#

#y=-2x^2+10x-25#

Now we may use the formula:

#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

#x_(1,2)=(-10+-sqrt(10^2-4*(-2) * (-25)))/(2*(-2))#

#x_(1,2)=(-10+-sqrt(100-200))/(-4)=(-10+-sqrt(-100))/(-4)=(-10+-10i)/(-4)#

#x_(1,2)=(10+-10i)/4=2 1/2+-2 1/2ior2 1/2(1+-i)#

You can see from the graph that there are no real roots.
graph{-2x^2+10x-25 [-85.1, 102.3, -78, 15.76]}