document.write( "Question 387723: In solving the equation (x 1)(x 2) = 30, Eric stated that the solution would be
\n" ); document.write( "x 1 = 30 => x = 31
\n" ); document.write( "or
\n" ); document.write( "x 2) = 30 => x = 32
\n" ); document.write( "(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
\n" ); document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #274136 by Earlsdon(6294) $\"\"$ $\"About$

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.
\n" ); document.write( "As they say in the old country...\"You can't do that there 'ere\"
\n" ); document.write( "Here's the correct solution:
\n" ); document.write( " $\"%28x-1%29%28x-2%29+=+30\"$ Multiply the factors on the left side.
\n" ); document.write( " $\"x%5E2-3x%2B2+=+30\"$ Subtract 30 from both sides.
\n" ); document.write( " $\"x%5E2-3x-28+=+0\"$ Solve the quadratic equation by factoring.
\n" ); document.write( " $\"%28x%2B4%29%28x-7%29+=+0\"$ Here you can apply the zero product principle.
\n" ); document.write( " $\"x%2B4+=+0\"$ or $\"x-7+=+0\"$ so...
\n" ); document.write( " $\"x+=+-4\"$ or $\"x+=+7\"$
\n" ); document.write( "The zero product principle states:
\n" ); document.write( "If $\"a%2Ab+=+0\"$ then either $\"a+=+0\"$ or $\"b+=+0\"$ or both.
\n" ); document.write( "So:
\n" ); document.write( " $\"%28x-1%29%28x-2%29+=+30\"$ therefore $\"x-1+=+30\"$ or $\"x-2+=+30\"$ is not valid.\r
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