What is the necessary and sufficient conditions for a quadratic equation to have one positive root?

Question

the equation $ax^2 + bx + c = 0$ has positive roots if

• $b^2-4ac > 0$
• $-b/a > 0$
• $c/a > 0$

I would like to know is there any necessary and sufficient conditions for it to have exactly one positive root?

Answer 1

The roots are $\frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$. If $b^2 - 4ac=0$ then there is a single double root which is positive if $b$ and $a$ are negative signs.

Otherwise to have one positive root there must be second negative root (or zero depending on how you feel about zero). So $- b + \sqrt{b^2 - 4ac}$ and $-b - \sqrt{b^2 - 4ac}$ must be different signs.

That means $\sqrt{b^2 - 4ac} > |b|$ ($\ge$ if you accept a zero root as non-positive). But if $4ac$ is positive and $b^2 - 4ac \ge 0$ then $\sqrt{b^2 - 4ac} < \sqrt{b^2} = |b|$. But if $4ac$ is negative then $\sqrt{b^2 - 4ac} > \sqrt {b^2} = |b|$.

So if $a$ and $c$ are opposite signs (and there are any roots at all) then one root will be positive and one negative.

Answer 2

Assuming $a,b,c$ are reals, the following conditions guarantee that out of the two real roots one is negative and one is positive: $$b^2-4ac>0 \text{ and } c/a<0.$$ The first condition guarantees that the equation has two distinct real roots. Since the product of the roots is equal to $c/a$, and we want only one of them to be positive, the other one has to be negative, hence the second condition.

Note that if we treat the case of repeated roots as a single root (as opposed to thinking of it as two equal roots), then the answer should also include the second case: $$b^2-4ac=0 \text{ and } b/a<0.$$ The second inequality comes from the fact that the $x$-coordinate of the vertex is $x_0=-\frac{b}{2a}$, and we want it to be a positive number.

UPDATE. After sleeping on it, in the morning I realized that I had a number of inaccuracies in my own answer:

• We implicitly assumed that the equation is indeed a quadratic one, i.e. that $a\neq0$, even though it wasn't stated anywhere. So to be completely rigorous, we should account for the case of $a=0$ too.

• I was wrong when I said that when one root is positive, "the other one has to be negative". The correct statement is that the other one has to be non-positive, i.e. negative or zero.

• Not a mistake, but rather a comment: for nonzero numbers, $c/a<0$ is equivalent to $ac<0$. Same for $b/a$.

• As implicitly indicated in other great answers, if $ac<0$, then $b^2-4ac>0$ automatically, so we don't have to include this condition in the first case.

Taking all of that into account, I believe the following should be a complete answer, consisting of several possible cases:

The equation $ax^2+bx+c=0$, where $a,b,c$ are reals, has exactly one positive root iff:

• $ac<0$ (two real roots, one positive and one negative); or

• $c=0$ and $ab<0$ (two real roots, one positive and one zero); or

• $a=0$ and $bc<0$ (linear equation whose only root is positive); or

• $b^2-4ac=0$ and $ab<0$ (repeated root which is positive; only include this case if treating it as one root rather than two equal roots).

Answer 3

It depends on whether you count the special case of a double root in the positive reals. This happens iff $$b^2=4ac\land ab<0.$$ Apart from that, you have exactly one simple positive root iff the sign at $x=0$ differs from the sign as $x\to\pm\infty$, i.e., $$ac<0.$$

Answer 4

The roots of quadratic equation, $$ax^2+bx+c=0$$ are

$$x=\frac {-b\pm \sqrt {b^2-4ac}}{2a}$$

If you multiply the roots of a quadratic equation you get $ \frac {c}{a}$

Thus in order for a negative root, all you need is $$ \frac {c}{a} <0 $$