document.write( "Question 1163693: Decide what values of the variable cannot possibly be solutions for the equation. Do not solve.
\n" ); document.write( "1x−4+1x+3=1x2−x−12
\n" ); document.write( "What values of x cannot be solutions of the equation?
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Algebra.Com's Answer #787874 by MathTherapy(10556) $\"\"$ $\"About$

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Decide what values of the variable cannot possibly be solutions for the equation. Do not solve.
\n" ); document.write( "1x−4+1x+3=1x2−x−12
\n" ); document.write( "What values of x cannot be solutions of the equation?
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document.write( "Using the discriminant, , we get: 
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document.write( "Since 53 is > 0, and NOT a perfect square, this means, as you might know, that the ROOTS/SOLUTIONS/ZEROES of the above quadratic will be REAL, IRRATIONAL, and UNEQUAL.
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document.write( "While they will STILL be UNEQUAL (), they can NEVER be IMAGINARY or RATIONAL. \n" );
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