SOLUTION: In solving the equation (x – 1)(x – 2) = 30, Eric stated that the solution would be x – 1 = 30 => x = 31 or (x – 2) = 30 => x = 32 However, at least one of these solutions fail

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: In solving the equation (x – 1)(x – 2) = 30, Eric stated that the solution would be x – 1 = 30 => x = 31 or (x – 2) = 30 => x = 32 However, at least one of these solutions fail      Log On


   



Question 548394: In solving the equation (x – 1)(x – 2) = 30, Eric stated that the solution would be
x – 1 = 30 => x = 31
or
(x – 2) = 30 => x = 32
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 Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Eric's assumption was incorrect!
He should have done the following:
%28x-1%29%28x-2%29+=+30 Multiply the two factors on the left.
x%5E2-3x%2B2+=+30 Now subtract 30 from both sides.
x%5E2-3x-28+=+0 Solve the quadratic equation by factoring.
%28x%2B4%29%28x-7%29+=+0 Apply the zero product rule.
x%2B4+=+0 or x-7+=+0 so that...
x+=+-4 or x+=+7