Question

How do you find the roots, real and imaginary, of `y= -2x^2 - 8x +16 - (x-5)^2` using the quadratic formula?

Answer 1

See a solution process below;

Explanation:

First, we need to expand the term in parenthesis on the right side of the equation using this rule:

`(color(red)(x) - color(blue)(y))^2 = (color(red)(x) - color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - 2color(red)(x)color(blue)(y) + color(blue)(y)^2`

Substituting `x` for `x` and `5` for `y` gives:

`y = -2x^2 - 8x + 16 - (x^2 - 10x + 25)`

`y = -2x^2 - 8x + 16 - x^2 + 10x - 25`

We can next group and combine like terms:

`y = -2x^2 - x^2 - 8x + 10x + 16 - 25`

`y = -2x^2 - 1x^2 - 8x + 10x + 16 - 25`

`y = (-2 - 1)x^2 + (-8 + 10)x + (16 - 25)`

`y = -3x^2 + 2x + (-9)`

`y = -3x^2 + 2x - 9`

We can now use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

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

Substituting:

`color(red)(-3)` for `color(red)(a)`

`color(blue)(2)` for `color(blue)(b)`

`color(green)(-9)` for `color(green)(c)` gives:

`x = (-color(blue)(2) +- sqrt(color(blue)(2)^2 - (4 * color(red)(-3) * color(green)(-9))))/(2 * color(red)(-3))`

`x = (-color(blue)(2) +- sqrt(4 - 108))/(-6)`

`x = (-color(blue)(2) +- sqrt(-104))/(-6)`

`x = (-color(blue)(2) +- sqrt(4 * -26))/(-6)`

`x = (-color(blue)(2) +- sqrt(4)sqrt(-26))/(-6)`

`x = (-color(blue)(2) +- 2sqrt(-26))/(-6)`

`x = (-1 +- sqrt(-26))/(-3)`

Or

`x = (1 +- sqrt(-26))/(3)`