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
(x-2)(x-1)=30,x^2-2x-x+2=30,x^2-3x+2-30=0since
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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
(x-2)(x-1)=30,x^2-2x-x+2=30,x^2-3x+2-30=0since
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Question 387723: 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
(x-2)(x-1)=30,x^2-2x-x+2=30,x^2-3x+2-30=0since x^2-3x-28=0 by factorization x^2+4x-7x-30=0,x(x+4)-7(x+4)=0 thrfr (x-7)(x+4)=0hence x=7
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. Found 2 solutions by Earlsdon, irshad_ahmed0@yahoo.com:Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Because "Eric" is applying the "zero product" principle to a product that is clearly not equal to zero.
As they say in the old country..."You can't do that there 'ere"
Here's the correct solution: Multiply the factors on the left side. Subtract 30 from both sides. Solve the quadratic equation by factoring. Here you can apply the zero product principle. or so... or
The zero product principle states:
If then either or or both.
So: therefore or is not valid.
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