How do you find the roots, real and imaginary, of #y=-(2x+2)^2+6x^2+3x-12 # using the quadratic formula?

1 Answer
Jan 5, 2018

Answer:

See a solution process below:

Explanation:

First, we need to expand the #(2x + 2)^2# term using this rule for quadratics:

#(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#

#(color(red)(2x) + color(blue)(2))^2 = (color(red)(2x) + color(blue)(2))(color(red)(2x) + color(blue)(2)) =#

#(color(red)(2x))^2 + (2 xx color(red)(2x) xx color(blue)(2)) + color(blue)(2)^2 = 4x^2 + 8x + 4#

Substituting and simplifying gives:

#y = -(2x + 2)^2 + 6x^2 + 3x - 12#

#y = -(4x^2 + 8x + 4) + 6x^2 + 3x - 12#

#y = -4x^2 - 8x - 4 + 6x^2 + 3x - 12#

#y = -4x^2 + 6x^2 - 8x + 3x - 4 - 12#

#y = (-4 + 6)x^2 + (-8 + 3)x + (-4 - 12)#

#y = 2x^2 + (-5)x + (-16)#

#y = 2x^2 - 5x - 16#

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)(2)# for #color(red)(a)#

#color(blue)(-5)# for #color(blue)(b)#

#color(green)(-16)# for #color(green)(c)# gives:

#x = (-color(blue)(-5) +- sqrt(color(blue)(-5)^2 - (4 * color(red)(2) * color(green)(-16))))/(2 * color(red)(2))#

#x = (5 +- sqrt(25 - (-128)))/4#

#x = (5 +- sqrt(25 + 128))/4#

#x = (5 +- sqrt(153))/4#

#x = (5 +- sqrt(9 xx 17))/4#

#x = (5 +- sqrt(9)sqrt(17))/4#

#x = (5 +- 3sqrt(17))/4#