One of the roots of the quadratic equation is positive and the other is negative [closed]

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.

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

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• The two solutions are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Can you use this to produce two inequalities that help you? – Ruben Jun 23 at 11:56
• @RubenduBurck i did the same but i am not able to solve inequality. . – darshh Jun 23 at 11:57
• You can use Descartes's sign rule. In the coefficients $a,b,c$ , there must be exactly one sign change. Moreover, we must have $c\ne 0$ – Peter Jun 23 at 11:58

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 $$ac<0$$.

Hint:

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

• I know but how can i find the sign of a,b,c to satisfy the given condition – darshh Jun 23 at 11:58
• Hint: You want the product of roots, which is $\dfrac c a$, to be negative – J. W. Tanner Jun 23 at 12:14
• Or $\;a>0,\,c<0\;$ or the other way around, the sign of $\;-\frac ba\;$ depends on what root's absolute value is higher... – DonAntonio Jun 23 at 13:04

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.

• What is vieta inequality nd how c must be negative – darshh Jun 23 at 12:12
• See DonAntonios answer, please. – Michael Hoppe Jun 23 at 12:15

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.

• Discriminant become greater than b – darshh Jun 23 at 12:11