document.write( "Question 1014156: Find a and b such that the zeros of ax^2 + bx +24 are 3 and 4. \n" ); document.write( "
Algebra.Com's Answer #630507 by macston(5194)\"\" \"About 
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\n" ); document.write( "equation with roots 3 and 4:
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\n" ); document.write( "\"%28x-3%29%28x-4%29=0\"
\n" ); document.write( "\"x%5E2-7x%2B12=0\"
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\n" ); document.write( "To make c=24, we must multiply by 2.
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\n" ); document.write( "\"2x%5E2-14x%2B24=0\"
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\n" ); document.write( "ANSWER: a=2 and b=-14
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\n" ); document.write( "CHECK:
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"2x%5E2%2B-14x%2B24+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%28-14%29%5E2-4%2A2%2A24=4\".
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\n" ); document.write( " Discriminant d=4 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28--14%2B-sqrt%28+4+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%28-14%29%2Bsqrt%28+4+%29%29%2F2%5C2+=+4\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%28-14%29-sqrt%28+4+%29%29%2F2%5C2+=+3\"
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\n" ); document.write( " Quadratic expression \"2x%5E2%2B-14x%2B24\" can be factored:
\n" ); document.write( " \"2x%5E2%2B-14x%2B24+=+2%28x-4%29%2A%28x-3%29\"
\n" ); document.write( " Again, the answer is: 4, 3.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-14%2Ax%2B24+%29\"
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