Question

How do you find the roots, real and imaginary, of `y= x^2 - 3x - (2x+4)^2` using the quadratic formula?

Answer 1

real roots: `x=-16/3,-1`
imaginary roots: none

Explanation:

`1` Start by converting the equation into standard form.

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

`y=x^2-3x-(color(crimson)(2x)` `color(blue)(+4))(color(orange)(2x)` `color(teal)(+4))`

`y=x^2-3x-(color(crimson)(2x)(color(orange)(2x))` `color(crimson)(+2x)(color(teal)4)` `color(blue)(+4)(color(orange)(2x))` `color(blue)(+4)(color(teal)4))`

`y=x^2-3x-(4x^2+8x+8x+16)`

`y=x^2-3x-(4x^2+16x+16)`

`y=x^2-4x^2-3x-16x-16`

`y=color(brown)(-3)x^2` `color(turquoise)(-19)x` `color(violet)(-16)`

`2`. Identify the `color(brown)a,color(turquoise)b,` and `color(violet)c` values of the quadratic equation. Then plug the values into the quadratic formula to solve for the roots.

`color(brown)(a=-3)color(white)(XXXXX)color(turquoise)(b=-19)color(white)(XXXXX)color(violet)(c=-16)`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(color(turquoise)(-19))+-sqrt((color(turquoise)(-19))^2-4(color(brown)(-3))(color(violet)(-16))))/(2(color(brown)(-3)))`

`x=(19+-sqrt(361-192))/-6`

`x=(19+-sqrt(169))/-6`

`x=(19+-13)/-6`

`x=(19+13)/-6color(white)(X),color(white)(X)(19-13)/-6`

`x=32/-6color(white)(X),color(white)(X)6/-6`

`color(green)(|bar(ul(color(white)(a/a)x=-16/3color(white)(i),-1color(white)(a/a)|)))`

`:.`, the real roots are `x=-16/3` or `x=-1`.