Question

How do you find the roots, real and imaginary, of `y=-(x -4 )^2-12x+16` using the quadratic formula?

Answer 1

The two real roots are:

`x_1\approx 18.24621\quad,quad x_2\approx 1.75379`

Explanation:

First let’s simplify the first parenthetical term, `-(x-4)^2`, by distributing the negative sign and FOILing it.

`-(x-4)^2`

`color(magenta)(\implies) (-x+4)(-x+4)`

`color(magenta)(\implies) x^2-4x-4x+16`

`color(magenta)(\implies) x^2-8x+16`

Add that to the rest of the original expression:

`color(red)(x^2)-color(blue)(8x)+color(green)(16)-color(blue)(12x)+color(green)(16)`

`color(magenta)(\implies) color(red)(x^2)-color(blue)(20x)+color(green)(32)`

This is in standard form, `ax^2+bx+c`, so we can solve using the quadratic formula. Here,

• `a=1`

• `b=20`

• `c=32`

`x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}`

`color(magenta)(\implies)x=\frac{20\pm\sqrt{(-20)^2-4(1)(32)}}{2(10)}`

`color(magenta)(\implies)x=\frac{20\pm\sqrt{400-128}}{2}`

`color(magenta)(\implies)x=\frac{20\pm\sqrt{272}}{2}`

`color(magenta)(\implies)x=\frac{20\pm 4\sqrt{17}}{2}`

`color(magenta)(\implies) x=10\pm 2\sqrt{17}`

`color(magenta)(\therefore) x_1\approx 18.24621\quad,\quad x_2\approx 1.75379`