How do you find the roots, real and imaginary, of #y=(x +4)^2+4x^2-3x + 2# using the quadratic formula?

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Answer 1 · 2017-09-03T17:28:02

See a solution process below:

First, we need to write the right side of the equation in standard form:
#y = (x + 4)^2 + 4x^2 - 3x + 2#

#y = x^2 + 8x + 16 + 4x^2 - 3x + 2#

#y = 1x^2 + 4x^2 + 8x - 3x + 16 + 2#

#y = (1 + 4)x^2 + (8 - 3)x + (16 + 2)#

#y = 5x^2 + 5x + 18#

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

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)(5)# for #color(red)(a)#

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

#color(green)(18)# for #color(green)(c)# gives:

#x = (-color(blue)(5) +- sqrt(color(blue)(5)^2 - (4 * color(red)(5) * color(green)(18))))/(2 * color(red)(5))#

#x = (-color(blue)(5) +- sqrt(25 - 360))/10#

#x = (-5 +- sqrt(-335))/10#