Question 1096723: What are all the real values of x which satisfy the equation: Found 3 solutions by rothauserc, ikleyn, greenestamps:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! (x-7)(x^2-6x+4)=-(x-7)
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divide both sides of = by (x-7)
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x^2 -6x +4 = -1
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x^2 -6x +5 = 0
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(x-5) * (x-1) = 0
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x = 5 or 1
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check answer
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x = 5
(5-7)(5^2 -6(5) +4) = -(5-7)
(-2)(25 -30 +4) = -(-2)
(-2)(-1) = 2
2 = 2
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x = 1
(1-7)(1^2 -6(1) +4) = -(1-7)
(-6)(1 -6 +4) = -(-6)
(-6)(-1) = 6
6 = 6
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our answers check
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You can put this solution on YOUR website! The first response missed one solution because he divided both sides of the equation by x-7. That turned out to be invalid, because x=7 is one of the solutions.
Never divide both sides of an equation by an expression containing a variable. Use factoring instead.
And the second solution gave the answers without showing the process....
So let me fill in the details, in case you are still not quite sure.
[except for linear polynomials, you always want to get everything on one side set equal to 0 and then factor]
