2

$$(x-1)^2 + 2x - 2 = (x+1)(x+2)$$ $$(x-1)^2 + 2(x-1) = (x+1) (x+2)$$ $$(x-1) (x-1+2) = (x+1) (x+2)$$ $$(x-1)\require{cancel}\cancel{(x+1)} = \require{cancel}\cancel{(x+1)}(x+2)$$ $$(x-1) = (x+2)$$ $$x-x = 2+1$$ $$0 = 3$$

But when i do it like this $$(x-1)^2 + 2x - 2 = (x+1) (x+2)$$ $$x^2 - 2x + 1 + 2x -2 = x^2 + 2x + x +2$$ $$x^2 - 1 = x^2 + 3x +2$$ $$-1 -2 = 3x$$ $$-3 = 3x$$ $$-1 = x$$

So in the first solution is my simplifying is wrong?

Guy
  • 8,857
  • 1
  • 28
  • 57
Shaadi Alfred
  • 71
  • 1
    You might like to format this rather better - at least put (double) line breaks in between different statements so we can see what the equations are, and which terms belong to which equation. At the moment it is a complete guess what this all means. – Mark Bennet Mar 14 '14 at 21:27
  • 4
    You divided by $x-1$, which is a division by a zero in case $x=-1$. You need to check that case before dividing. – Vedran Šego Mar 14 '14 at 21:31

1 Answers1

3

In the first set of equations, where you reached a contradiction, you divided both sides of the equation by $x+1$, but you can't divide by zero.

So you should have said at that point, either $x=-1$ or I can divide ... and since dividing gives a contradiction, I must have $x=-1$.

Then the two methods become consistent.

Mark Bennet
  • 100,194
  • 1
    @VedranŠego I agree it is necessary to take care, but if $x=-1$, both sides of the equation before dividing through are zero, so there is nothing to check. Dividing by zero is the only hazard here. – Mark Bennet Mar 14 '14 at 21:35
  • You're right, I wasn't paying attention. – Vedran Šego Mar 14 '14 at 21:37