Question

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

Answer 1

`y=(x-2)(x+1)-3x`
`color(white)("XXX")`has two real roots at `x=2+sqrt(6) and x=2-sqrt(6)`

Explanation:

In order to use the quadratic formula it is first necessary to convert the given equation into standard form.

`y=(x-2)(x+1)-3x`

`rarr y=x^2-x-2-3x`

`rarr y=x^2-4x-2` (standard form)

Roots for an equation of the form:
`color(white)("XXX")y=ax^2+bx+c`
are given by the quadratic formula
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

In this case
`color(white)("XXX")a=1,`
`color(white)("XXX")b=-4`
`color(white)("XXX")c=-2`

So the roots are
`color(white)("XXX")x=(4+-sqrt((-4)^2-4(1)(-2)))/(2(1))`

`color(white)("XXX")x=(4+-sqrt(24))/2`

`color(white)("XXX")x=2+-sqrt(6)`

Answer 2

We have only real roots, which are `2-sqrt6` and `2+sqrt6`

Explanation:

`y=(x-2)(x+1)-3x=x^2-2x+x-2-3x=x^2-4x-2`

The roots of `x^2-4x-2`

given by quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)` are

`x=(-(-4)+-sqrt((-4)^2-4xx1xx(-2)))/(2xx1)=(4+-sqrt(16+8))/2` or

`x=(4+-sqrt24)/2=(4+-2sqrt6)/2=2+-sqrt6`

Hence, roots are `2-sqrt6` and `2+sqrt6`