SOLUTION: Hi, I really need help please. In solving the equation (x + 1)(x � 2) = 54, Eric stated that the solution would be x + 1 = 54 => x = 53 or (x � 2) = 54 => x = 56 However,

Question 494576: Hi, I really need help please.

In solving the equation (x + 1)(x � 2) = 54, Eric stated that the solution would be
x + 1 = 54 => x = 53
or
(x � 2) = 54 => x = 56
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 yourself, and explain your reasoning.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
In solving the equation (x + 1)(x � 2) = 54, Eric stated that the solution would be
x + 1 = 54 => x = 53
or
(x � 2) = 54 => x = 56
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 yourself, and explain your reasoning.
===============================================================
Eric's reasoning is only valid if the RHS of the equation is equal to 0.
Then if either of the two factors are equal to 0, this makes the whole
expression equal to 0. If the RHS is not equal to 0, then the two factors when
multiplied together have to equal the RHS, not each of the factors individually.
To solve, first multiply the two factors and then subtract the RHS:
(x+1)(x-2) = 54
x^2 - x - 2 = 54
x^2 - x - 56 = 0
The above equation can be factored as:
(x-8)(x+7) = 0
We see from this that the solutions are x = 8 and x = -7
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