How do you solve #3x^2-8x-19=(x-1)^2#?

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Answer 1 · 2017-01-26T06:06:50

#x=5, -2#

#3x^2-8x-19=(x-1)^2#

#3x^2-8x-19=x^2-2x+1#

#3x^2color(red)(-x^2)-8xcolor(red)(+2x)-19color(red)(-1)=x^2color(red)(-x^2)-2xcolor(red)(+2x)+1color(red)(-1)#

#2x^2-6x-20=0#

Let's use the quadratic formula to solve for #x#:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(6+-sqrt((-6)^2-4(2)(-20)))/(2(2))#

#x=(6+-sqrt(36+160))/4#

#x=(6+-sqrt(196))/4#

#x=(6+-14)/4#

#x=20/4=5, =-8/4=-2#

And we can check this by graphing both sides of the original equation (note the points of intersection at #x=5, -2#:

graph{(y-3x^2+8x+19)(y-(x-1)^2)=0 [-10, 10, -2.11, 26.78]}