Question

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

Answer 1

`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`