SOLUTION: 1. In solving the equation (x + 1)(x – 2) = 4, Eric stated that the solution would be
x + 1 = 4 => x = 3
or
(x – 2) = 4 => x = 6
However, at least one of these solutions fails
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Polynomials-and-rational-expressions
-> SOLUTION: 1. In solving the equation (x + 1)(x – 2) = 4, Eric stated that the solution would be
x + 1 = 4 => x = 3
or
(x – 2) = 4 => x = 6
However, at least one of these solutions fails
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Question 407321: 1. In solving the equation (x + 1)(x – 2) = 4, Eric stated that the solution would be
x + 1 = 4 => x = 3
or
(x – 2) = 4 => x = 6
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.
You can put this solution on YOUR website! It doesn't work because if we assume x+1 = 4, we fail to take into account the x-2 expression. However, if we had the equation (x+1)(x-2) = 0, then we can set x+1 = 0, x-2 = 0 because the other term is irrelevant.
The best way is to expand and get --> . This factors to --> x = 3, -2.
