document.write( "Question 610240: 3x2 + 7x > -12 \n" ); document.write( "

Algebra.Com's Answer #384402 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"3x%5E2+%2B+7x+%3E+-12\"$
\n" ); document.write( "Like quadratic equations, you want one side to be zero and then factor the other side. Adding 12 to each side we get:
\n" ); document.write( " $\"3x%5E2+%2B+7x+%2B+12+%3E+0\"$
\n" ); document.write( "Then we factor. But this won't factor!

\n" ); document.write( "We can use the Quadratic Formula to tell us something helpful:
\n" ); document.write( " $\"x+=+%28-%287%29+%2B-+sqrt%28%287%29%5E2+-+4%283%29%2812%29%29%29%2F2%283%29\"$
\n" ); document.write( "Simplifying:
\n" ); document.write( " $\"x+=+%28-%287%29+%2B-+sqrt%2849+-+4%283%29%2812%29%29%29%2F2%283%29\"$
\n" ); document.write( " $\"x+=+%28-%287%29+%2B-+sqrt%2849+-+144%29%29%2F2%283%29\"$
\n" ); document.write( " $\"x+=+%28-%287%29+%2B-+sqrt%28-95%29%29%2F2%283%29\"$
\n" ); document.write( "At this point, with the negative in the square root, we can stop. The formula is trying to tell us x value(s) that make $\"3x%5E2%2B7x%2B12\"$ equal to zero. Since we cannot have a negative inside a square root, this tells us that $\"3x%5E2%2B7x%2B12\"$ is never equal to zero.

\n" ); document.write( "If $\"3x%5E2%2B7x%2B12\"$ is never equal to zero, it must always be positive or always be negative. We can quickly see that when x = 0, $\"3x%5E2%2B7x%2B12\"$ will be a 12. 12 is positive. Since $\"3x%5E2%2B7x%2B12\"$ is always positive or negative and since we have found a case where it is positive, $\"3x%5E2%2B7x%2B12\"$ must always be positive. So the solution to
\n" ); document.write( " $\"3x%5E2+%2B+7x+%2B+12+%3E+0\"$
\n" ); document.write( "which says that $\"3x%5E2+%2B+7x+%2B+12\"$ is positive, is:
\n" ); document.write( "All real numbers. In other words, no matter what number you use for x, $\"3x%5E2+%2B+7x+%2B+12\"$ will work out to be positive. \n" ); document.write( "

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