How do you find the roots, real and imaginary, of #y=12x^2 -7x +28-(x-2)^2 # using the quadratic formula?

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Answer 1 · 2018-04-13T15:46:05

#x=3/22+-sqrt1047/22i#

The quadratic formula states that, for a quadratic in the form

#y=color(red)(a)x^2+color(green)(b)x+color(blue)(c) => x=(-color(green)(b)+-sqrt(color(green)(b)^2-4color(red)(a)color(blue)(c)))/(2color(red)(a))#

We need to put our given quadratic in the form #y=color(red)(a)x^2+color(green)(b)x+color(blue)(c)#

#y=12x^2-7x+28-(x-2)^2#

#=12x^2-7x+28-(x^2-4x+4)#

#=color(red)(11)x^2color(green)(-3)x+color(blue)(24)#

From the quadratic formula:

# x=(-color(green)((-3))+-sqrt(color(green)((-3))^2-4xxcolor(red)(11)xxcolor(blue)(24)))/(2xxcolor(red)(11))#

#=(3+-sqrt(-1047))/22#

We can separate #sqrt(-1047)# into #sqrt1047xxsqrt(-1)#. Since #i=sqrt(-1)#:

#x=(3+-sqrt1047 i)/22#

#x=3/22+-sqrt1047/22i#