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

1 Answer
Dec 21, 2017

Answer:

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#