document.write( "Question 701571: For the equation x2 + 3x + j = 0, find all the values of j such that the equation has two real number solutions. Show your work. \n" ); document.write( "
Algebra.Com's Answer #432488 by josgarithmetic(39620)\"\" \"About 
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Assume you mean \"x%5E2%2B3%2Ax%2Bj=0\". Either know what is the discriminant, or try filling in solution to quadratic equation.\r
\n" ); document.write( "\n" ); document.write( "\"x=%28-3%2B-sqrt%289-4%2Aj%29%29%2F2\"\r
\n" ); document.write( "\n" ); document.write( "Notice if you want only Real solutions, then the square root's radicand must be positive. \"9-4%2Aj%3E=0\". Only a simple matter to find j. \r
\n" ); document.write( "\n" ); document.write( "-4j>=-9
\n" ); document.write( "j<=(9/4) [multiplied both sides by -(1/4)]
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