document.write( "Question 693410: Finding finding the roots of a quadratic equation with leading coefficient greater than 1\r
\n" ); document.write( "\n" ); document.write( "Solve for u
\n" ); document.write( "3u^2-5u=2
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Algebra.Com's Answer #427383 by pmatei(79)\"\" \"About 
You can put this solution on YOUR website!
For quadratics you need everything on the left side of the equal sign:\r
\n" ); document.write( "\n" ); document.write( "\"3u%5E2-5u-2=0\"\r
\n" ); document.write( "\n" ); document.write( "There are 2 ways of solving quadratics. \r
\n" ); document.write( "\n" ); document.write( "A. You try to find to numbers that their product is equal to the product of the u^2 coefficient and the number without letter (in this case \"3%28-2%29=-6\"), and their sum is equal to the u coefficient (in this case \"-5\"). The numbers that multiplied give \"-6\" and added give \"-5\" are : \"-6\" and \"%2B1\".\r
\n" ); document.write( "\n" ); document.write( "Instead of \"-5u\" in the original equation write the two numbers multiplied by \"u\":\r
\n" ); document.write( "\n" ); document.write( "\"3u%5E2+-6u%2Bu-2=0\"\r
\n" ); document.write( "\n" ); document.write( "Group the four terms in 2 pairs: first two terms and last two terms.\r
\n" ); document.write( "\n" ); document.write( "\"%283u%5E2-6u%29%2B%28u-2%29=0\"\r
\n" ); document.write( "\n" ); document.write( "See if you can factor out anything from those parentheses (you always have at least the letter from the first parenthesis):\r
\n" ); document.write( "\n" ); document.write( "\"3u%28u-2%29%2B1%28u-2%29=0\"\r
\n" ); document.write( "\n" ); document.write( "If you done everything right the two parentheses are the same at this point. This repeated parenthesis is one of the 2 parentheses you want to solve the quadratic. The second one is put together by what you factored out:\r
\n" ); document.write( "\n" ); document.write( "\"%28u-2%29%283u%2B1%29=0\"\r
\n" ); document.write( "\n" ); document.write( "For the quadratic solutions take each parenthesis and set it equal to zero.\r
\n" ); document.write( "\n" ); document.write( "First solution: \r
\n" ); document.write( "\n" ); document.write( "\"u-2=0\"
\n" ); document.write( "\"u=2\"\r
\n" ); document.write( "\n" ); document.write( "Second solution:\r
\n" ); document.write( "\n" ); document.write( "\"3u%2B1=0\"
\n" ); document.write( "\"u=-1%2F3\"\r
\n" ); document.write( "\n" ); document.write( "This method does not always work. \r
\n" ); document.write( "\n" ); document.write( "B. The second method is to know the quadratic formula, and works every time:\r
\n" ); document.write( "\n" ); document.write( "If \"ax%5E2%2Bbx%2Bc=0\" then the solutions are given by:\r
\n" ); document.write( "\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" \r
\n" ); document.write( "\n" ); document.write( "In your case \"3u%5E2-5u-2=0\", \"a=3\", \"b=-5\", \"c=-2\".\r
\n" ); document.write( "\n" ); document.write( "\"u=%285+%2B-+sqrt%28+25-4%2A%283%29%2A%28-2%29+%29%29%2F6+\"\r
\n" ); document.write( "\n" ); document.write( "\"u=%285+%2B-+sqrt%28+25%2B24%29%29%2F6+\"\r
\n" ); document.write( "\n" ); document.write( "\"u=%285+%2B-+sqrt%28+49%29%29%2F6+\"\r
\n" ); document.write( "\n" ); document.write( "\"u=%285+%2B-+7%29%2F6+\"\r
\n" ); document.write( "\n" ); document.write( "First solution:\r
\n" ); document.write( "\n" ); document.write( "\"u+=+%285%2B7%29%2F6+=+12%2F6=2\"\r
\n" ); document.write( "\n" ); document.write( "Second solution:\r
\n" ); document.write( "\n" ); document.write( "\"u=%285-7%29%2F6=-2%2F6=-1%2F3\"
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