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Concerning the quadratic formula. What does it mean if $b^2-4ac>0$, $b^2-4ac<0$, and $b^2-4ac=0$?

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    $\begingroup$ What do you exactly mean? Those expresions are inequalities, and their content is the discriminant of a quadratic expresion $ax^{2}+bx+c=0$. Can you explain your question please? You could add another flag too, just like "homework" or "algebra-precalculus". $\endgroup$ – Daniel Jun 4 '14 at 23:29
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    $\begingroup$ Due to the nature of the quadratic formula I removed the square root signs in the three inequalities. $\endgroup$ – Américo Tavares Jun 4 '14 at 23:36
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$\Delta={ b^2-4ac}>0$ means that the equation has two real solutions.

$\Delta= { b^2-4ac}<0$ means that the equation has no real solutions, but two complex solutions.

$\Delta={ b^2-4ac}=0$ means that the equation has one solution.

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    $\begingroup$ And when it's $0$, the solution is real. I think it's worth pointing out that all of this can be determined by a careful look directly at the quadratic formula and thinking about what happens in each of these cases. $\endgroup$ – jpmc26 Jun 5 '14 at 0:37
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    $\begingroup$ Yes, you're right!!! $\endgroup$ – Mary Star Jun 5 '14 at 0:39
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Since one has the term $\sqrt{b^2-4ac}$ in solution of the quadratic formula, if $b^2-4ac>0$, then the equation has two real solutions (since $\sqrt{b^2-4ac}$ is real, and $b$ and $2a$ are also real). When $b^2-4ac<0$, then the quadratic equation has two complex solutions (since $\sqrt{b^2-4ac}$ is complex imaginary). If $b^2-4ac=0$, then $\sqrt{b^2-4ac}=0$, implying that the solution is $x=\dfrac{-b}{2a}$ (can you see why?).

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