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One of the roots of the quadratic $ax^2 + bx + c = 0$ is positive and the other is negative. Tell me the sign of a,b,c so that this happens.

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closed as off-topic by Peter, José Carlos Santos, Lee David Chung Lin, cmk, Davide Giraudo Jun 23 at 15:48

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  • $\begingroup$ The two solutions are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Can you use this to produce two inequalities that help you? $\endgroup$ – Ruben Jun 23 at 11:56
  • $\begingroup$ @RubenduBurck i did the same but i am not able to solve inequality. . $\endgroup$ – darshh Jun 23 at 11:57
  • $\begingroup$ You can use Descartes's sign rule. In the coefficients $a,b,c$ , there must be exactly one sign change. Moreover, we must have $c\ne 0$ $\endgroup$ – Peter Jun 23 at 11:58
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We need distinct real roots, so necessarily $\Delta=b^2-4ac>0$. Furthermore, the product of the roots, $\frac ca$ must be negative. As the sign of $\frac ca$ is the same as the sign of $ac$, we have $ac<0$.

We may note that $ac<0$ implies $\Delta>0$, so the first condition is redundant, and finally, the equation $ax^2+bx+c=0$ has a positive root and a negative root if and only if $ac<0$.

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Hint:

Viete equalitites: the sum of the roots is $\;-\cfrac ba\;$ and their product is $\;\cfrac ca\;$ .

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  • $\begingroup$ I know but how can i find the sign of a,b,c to satisfy the given condition $\endgroup$ – darshh Jun 23 at 11:58
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    $\begingroup$ Hint: You want the product of roots, which is $\dfrac c a$, to be negative $\endgroup$ – J. W. Tanner Jun 23 at 12:14
  • $\begingroup$ Or $\;a>0,\,c<0\;$ or the other way around, the sign of $\;-\frac ba\;$ depends on what root's absolute value is higher... $\endgroup$ – DonAntonio Jun 23 at 13:04
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Let $a$ be positive. Then by Vieta $c$ must be negative; here the sign of $b$ doesn't matter. If conversely $c$ is negative in that case we know that there must be two different solution one of different signs; again the sign of $b$ doesn't matter. Similarly if $a$ is negative. Hence $ac$ must be negative and the sign of $b$ doesn't matter.

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  • $\begingroup$ What is vieta inequality nd how c must be negative $\endgroup$ – darshh Jun 23 at 12:12
  • $\begingroup$ See DonAntonios answer, please. $\endgroup$ – Michael Hoppe Jun 23 at 12:15
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We need $\sqrt{b^2-4ac}\gt b$. This happens iff $b^2-4ac\gt b^2$. Which happens whenever $ac\lt0$. So $a$ and $c$ must have opposite signs.

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    $\begingroup$ Discriminant become greater than b $\endgroup$ – darshh Jun 23 at 12:11

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