Question

How do you find the roots, real and imaginary, of `y=x(x-1)-(3x-1)^2` using the quadratic formula?

Answer 1

Expand and simplify to find the equation in standard form, then apply the quadratic formula to obtain the roots: `x=(5+-isqrt(7))/16`.

Explanation:

The quadratic formula can be used with quadratic equations in standard form, `y=ax^2+bx+c`.

First expand and simplify the given equation to rearrange it into standard form.

`y=x(x-1)-(3x-1)^2`
`y=x^2-x-(9x^2-6x+1)`
`y=x^2-x-9x^2+6x-1`
`y=-8x^2+5x-1`

This equation is now in standard form, where `a=-8`, `b=5`, and `c=-1`.

To solve for the roots of the equation, set `y=0`:
`0 = -8x^2+5x-1`,

Apply the quadratic formula:
`x = (-b+-sqrt(b^2-4ac))/(2a)`
`x = (-(5)+-sqrt((5)^2-4(-8)(-1)))/(2(-8))`
`x = (-5+-sqrt(-7))/-16`
`x=(5+-sqrt(-7))/16`
`x=(5+-sqrt(-1)sqrt(7))/16`
`x=(5+-isqrt(7))/16`