Please help with this algebra problem

it says
In solving the equation (x + 4)(x – 7) = -18, Eric stated that the solution would be
x + 4 = -18 => x = -22
or
(x – 7) = -18 => x = -11
However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem, and explain your reasoning.

x= 5 and x = -2

(x+4)(x-7)= -18

You take x times x and you get x squared. Then you take x times -7 and you get -7x
then you take x times 4 = 4x and 4 times -7 = -28
combine like terms
x squared -7x + 4x - 28 = -18
x squared -3x = 10
then you take this and make it all equal to zero
x squared - 3x + 10 = o
now it is kind of a trial and error. You have to have two numbers when multiplied together equals 10 and the same numbers when added together equals a -3
(x-5)(x+2)=0
(x-5)=0
so x = 5
(x+2)= 0 so x is -2
so the answers are 5 and -2

Firstly, expand the brackets. You get:
x^2 - 3x - 28 = -18 (x^2 means x squared)
Add 18 to both sides
x^2 - 3x - 10 = 0
Factorise
(x-5)(x+2) = 0

So there are two solutions as if each bracket was 0, the whole thing would be 0. To make the first bracket 0, x would have to be 5. To make the second bracket 0, x would have to be -2.
So you have two solutions: x=5 and x=-2

And you can check these are correct by putting them into the original equation and makeing sure you get 18.

(5+4)(5-7)=9 x -2 = -18 so that is correct
(-2+4)(-2-7)=2 x -9 = -18 so that is also correct

Setting the factors to equal the numeric side (-18 in this case) is not correct.
The equation has to be rewritten so as to equal 0.

Here
(x + 4)(x – 7) = -18
x^2 - 3x - 28 = -18

add 18 to both sides
x^2 - 3x - 10 = 0

Now that the equation equals zero it can be factored and restated
(x - 5)(x +2) = 0

x - 5 = 0 => x = 5
or
x + 2 =0 => x = -2



m06302012