# in solving (x+4)(x-7)=-18 eric stated the solution as x+4=-18=>=-22 or (x-7)=-18>x=-11 one of these fails to work when substitutedback to the original equation. why is that? please help eric...

in solving (x+4)(x-7)=-18 eric stated the solution as x+4=-18=>=-22 or (x-7)=-18>x=-11 one of these fails to work when substituted

back to the original equation. why is that? please help eric understand, solve the problem and explain

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### 1 Answer

To solve the equation, you will need to expand the brackets and then move -18 to the left sides so the right side is 0.

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

==> x^2 -3x - 28 = -18

==> x^2 -3x -10 = 0

==> Now that we have quadratic equation, we will factor.

==> (x-5)(x+3)= 0

==> Now, since the product of both terms is 0, then, either one of the terms should be zero.

==> (x-5)= 0 ==> x = 5

==> x+ 3= 0 ==> x = -3

Then, the solution is x = { 5,-3}.

The error Eric made is that we have the product of both terms is -18.

We can not use the same method when the right side is not zero because we do not know what each term represent.

For exmple, the terms could be 2 * -9 , 1*-18, -1*18, 3*-6 , ...etc.

Then, we have unlimited options for each terms and it is almost not possible to know the answer.

However, when the product is zero, then, we know for sure that either terms is 0.