SOLUTION: Solve the following equation by using an appropriate substitution (x^2-x)^2-18(x^2-x)+72=0

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Question 624872: Solve the following equation by using an appropriate substitution
(x^2-x)^2-18(x^2-x)+72=0
Answer by Maths68(1474) About Me  (Show Source):
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(x^2-x)^2-18(x^2-x)+72=0
Let y=(x^2-x)
y^2-18y+72=0
Now solve for y
y^2-12y-6y+72=0
y(y-12)-6(y-12)=0
(y-12)(y-6)=0
y-12=0 or y-6=0
y=12 or y=6
Put the value of y in to above equations.
y=12
x^2-x=12 or x^2-x=6
x^2-x-12=0 or x^2-x-6=0
x^2-4x+3x-12=0 or x^2-3x+2x-6=0
x(x-4)+3(x-4)=0 or x(x-3)+2(x-3)=0
(x-4)(x+3)=0 or (x-3)(x+2)=0
x-4=0 or x+3=0 or x-3=0 or x+2=0
x=4 or x=-3 or x=3 or x=-2

Check
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(x^2-x)^2-18(x^2-x)+72=0
Put x=4
((4)^2-4)^2-18((4)^2-4)+72=0
(16-4)^2-18(16-4)+72=0
(12)^2-18(12)+72=0
144-216+72=0
-72+72=0
0=0


(x^2-x)^2-18(x^2-x)+72=0
Put x=3
((3)^2-3)^2-18((3)^2-3)+72=0
(9-3)^2-18(9-3)+72=0
(6)^2-18(6)+72=0
36-108+72=0
-72+72=0
0=0



(x^2-x)^2-18(x^2-x)+72=0
Put x=-3
((-3)^2-(-3))^2-18((-3)^2-(-3))+72=0
(9+3)^2-18(9+3)+72=0
(12)^2-18(12)+72=0
144-216+72=0
-72+72=0
0=0



(x^2-x)^2-18(x^2-x)+72=0
Put x=-2
((-2)^2-(-2))^2-18((-2)^2-(-2))+72=0
(4+2)^2-18(4+2)+72=0
(6)^2-18(6)+72=0
(6)^2-18(6)+72=0
36-108+72=0
-72+72=0
0=0