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

1 Answer
Nov 9, 2017

Answer:

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)#