Question

How do you find the roots, real and imaginary, of `y= (x-3)^2+2x-5` using the quadratic formula?

Answer 1

In order to use the quadratic equation, we must first get our equation to be in standard form. To do that, we need to simplify everything and then combine like-terms.

`y=(x-3)^2+2x-5` becomes `y=(x-3)(x-3)+2x-5`. This becomes `x^2-3x-3x+9+2x-5` or `x^2-6x+9+2x-5`. That simplifies to `x^2-4x+4`, which gives us our standard form .

Now we just use the quadratic formula: `(-b)/(2*a)+-(sqrt(b^2-4*a*c))/(2*a)`

`color(green)(x)^2-color(blue)(4)x+color(red)(4)`
`color(green)(1=a)`
`color(blue)(4=b)`
`color(red)(4=c)`

`x=(-(color(blue)(4))+-(sqrt((color(blue)(4))^2-4*(color(green)(1))*(color(red)(4)))))/(2*color(green)(1))`

`x=(-4+-sqrt(16-16))/2`

`x=(-4)/2`

`x=-2`

This seems to be the only root, but let's double check by graphing
graph{y=(x-3)^2+2x-5}

Yep, our only root is at `x=2`! Good job!