How do you find the roots, real and imaginary, of y= 5x^2-x-2x(x-1) y=5x2−x−2x(x−1) using the quadratic formula?

1 Answer
Tamir E.
Dec 27, 2017

x_1=0x1=0
x_2=-1/3x2=−13

Explanation:

For the function in the form of:
ax^2+bx+c=0ax2+bx+c=0

Quadratic Formula:
x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}x1,2=−b±√b2−4ac2a

y=5x^2-x-2x(x-1)=5x^2-x-2x^2+2x=3x^2+xy=5x2−x−2x(x−1)=5x2−x−2x2+2x=3x2+x

Let y=0y=0:
3x^2+1x+0=03x2+1x+0=0
where:
a=3a=3
b=1b=1
c=0c=0
=>⇒
x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}={-1+-sqrt{1^2-4*3*0}}/{2*3}=x1,2=−b±√b2−4ac2a=−1±√12−4⋅3⋅02⋅3=
={-1+-sqrt{1}}/{6}={-1+-1}/{6}=−1±√16=−1±16
=>⇒
x_1={-1+1}/6=0/6=0x1=−1+16=06=0
x_2={-1-1}/6={-2}/6=-1/3x2=−1−16=−26=−13

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