How do you find the roots, real and imaginary, of #y= x^2-39x-2(3x-1)^2 # using the quadratic formula?

1 Answer
Jul 17, 2017

Answer:

#0.04 " or " 2.644#

Explanation:

Right now, you have #y=x^2-39x-2(3x-1)^2#

First, we need to expand #(3x-1)^2#.

To expand the brackets in the form of #(color(red)(a)color(green)(x)+color(blue)(b))^2#, we do #(color(red)(a)color(green)(x))^2+2(color(red)(a)color(green)(x)color(blue)(b))+color(blue)(b)^2#

In this case we have #(color(red)(3)color(green)(x)+color(blue)(-1))^2#

#(3x)^2=9x^2#
#2*3x*-1=-6x#
#-1^2=1#
#9x^2-6x+1#

But we need 2 lots, so #2(9x^2-6x+1)=18^2-12x+2#

#y=x^2-39x+18x^2-12x+2=19x^2-51x+2#

The quadratic formula is #x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(orange)(c)))/(2color(red)(a))-=(-color(blue)(-51)+-sqrt(color(blue)(-51)^2-4(color(red)(19)*color(orange)(4))))/(2(color(red)(19)))=(color(blue)(51)+-sqrt(color(blue)2601-color(yellow)(152)))/(38)=0.04 " or " 2.644#