SOLUTION: What are all the real values of x which satisfy the equation: {{{ (x-7)(x^2-6x+4)=-(x-7) }}}

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)
:
divide both sides of = by (x-7)
:
x^2 -6x +4 = -1
:
x^2 -6x +5 = 0
:
(x-5) * (x-1) = 0
:
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x = 5 or 1
:
check answer
:
x = 5
(5-7)(5^2 -6(5) +4) = -(5-7)
(-2)(25 -30 +4) = -(-2)
(-2)(-1) = 2
2 = 2
:
x = 1
(1-7)(1^2 -6(1) +4) = -(1-7)
(-6)(1 -6 +4) = -(-6)
(-6)(-1) = 6
6 = 6
:
our answers check
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:
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.
All real numbers that satisfy this equation are 7, 5 and 1.

Answer by greenestamps(13200)   (Show Source): 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]

The solutions are x = 1, 5, and 7.
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