document.write( "Question 29439: x^2+x<12 \n" ); document.write( "

Algebra.Com's Answer #716215 by ikleyn(52786) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( " < 12.    (1)\r\n" );
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document.write( "Subtract 12 from both sides. You will get an EQUIVALENT inequality\r\n" );
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document.write( " < 0.    (2)\r\n" );
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document.write( "    It is the same as to move the term 12 in the original equation from the right side\r\n" );
document.write( "    to its left side changing the sign.\r\n" );
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document.write( "In the inequality (2), factor left side. You will get an EQUIVALENT inequality\r\n" );
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document.write( "(x+4)*(x-3) < 0.     (3)\r\n" );
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document.write( "You have quadratic function in the left side.\r\n" );
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document.write( "This quadratic function is ZERO at x= -4 and at x=3.\r\n" );
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document.write( "It is POSITIVE at x < -4,  NEGATIVE  in the interval  -4 < x < 3,  and POSITIVE again in the interval x > 3.\r\n" );
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document.write( "So, the solution to your original inequality is the set of all real numbers \r\n" );
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document.write( "-4 < x < 3,  or, in interval notation (-4,3).\r\n" );
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document.write( "To check, take the number x = -3, for example, and calculate the quadratic function (1):\r\n" );
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document.write( "(-3)^2 + (-3) = 9 - 3 = 6.  It is less than 12, so it is correct.\r\n" );
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document.write( "Answer.  The solution to your original inequality is the set of all real numbers  -4 < x < 3,  or, in interval notation, the interval (-4,3).\r\n" );
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\n" ); document.write( "\n" ); document.write( "The solution by @amalm06 is TOTALLY WRONG.\r
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\n" ); document.write( "\n" ); document.write( "Below please see the plot to your inequality.\r
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\n" ); document.write( "\n" ); document.write( "Plot y = $\"x%5E2%2Bx\"$ (red) and y = 12 (green)\r
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\n" ); document.write( "\n" ); document.write( "You are searching for those x where the parabola is BELOW the horizontal green line.\r
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