How do you find the roots, real and imaginary, of #y= 5x^2-x-2x(x-1) # using the quadratic formula?

1 Answer
Dec 27, 2017

Answer:

#x_1=0#
#x_2=-1/3#

Explanation:

For the function in the form of:
#ax^2+bx+c=0#

Quadratic Formula:
#x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}#


#y=5x^2-x-2x(x-1)=5x^2-x-2x^2+2x=3x^2+x#

Let #y=0#:
#3x^2+1x+0=0#
where:
#a=3#
#b=1#
#c=0#
#=>#
#x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}={-1+-sqrt{1^2-4*3*0}}/{2*3}=#
#={-1+-sqrt{1}}/{6}={-1+-1}/{6}#
#=>#
#x_1={-1+1}/6=0/6=0#
#x_2={-1-1}/6={-2}/6=-1/3#