document.write( "Question 995840: Explain why x^2+bx-c=0 must have one positive real and one negative real solution when b and c are real numbers and c>0 .\r
\n" ); document.write( "\n" ); document.write( "For what values of b will x^2+bx+1=0have real solutions? Clearly explain the reasoning you followed to reach your conclusion
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Algebra.Com's Answer #614506 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
The first part is wrong. That cannot be justified. An example was already given a few minutes ago.\r
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\n" ); document.write( "\n" ); document.write( "Use the discriminant for the second question.
\n" ); document.write( "\"x%5E2%2Bbx%2B1=0\" --- given equation;
\n" ); document.write( "\"b%5E2-4%2A1%2A1\" --- the discriminant, using a=1 and c=1;
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\n" ); document.write( "\"b%5E2-4%3E=0\" --- discriminant must be non-zero for the original quadratic equation to have real solutions.
\n" ); document.write( "\"b%5E2=4\"
\n" ); document.write( "\"highlight%28b=0%2B-+2%29\"
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