# The square trinomial $y=ax^2+ bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$ .

The square trinomial $$y=ax^2 + bx + c$$ has no roots and $$a + b + c > 0$$. Find the sign of the coefficient $$c$$. I'm having difficulties with this problem.

What I've tried: I realized that a quadratic equation doesn't have roots if the discriminant $$b^2 - 4ac < 0$$, so I've tried to combine that with the condition $$a + b + c > 0 <=> a > -b -c$$, but that didn't help me that much.

I would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .

• What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question) – Mark Bennet Jan 17 at 22:08
• You have the right idea. You know $0 ≤ b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign... – diracdeltafunk Jan 17 at 22:10

Call $$p(x)=ax^2+bx+c$$. Then $$p(1)=a+b+c >0$$ $$p(0)=c$$ If $$c$$ were negative, then there would be a root between $$0$$ and $$1$$. This contradicts our hypothesis, hence necessarily $$c>0$$.

• +1, beautiful argument – gt6989b Jan 17 at 22:10
• Thank you so much! I understand now and I really like this way of solving the problem. – Wolf M. Jan 17 at 22:25

That $$a+b+c > 0$$ gives $$ax^2+bx+c$$ evaluated at $$x=1$$ is $$a+b+c$$ which is positive.

Suppose that $$c$$ is negative. Then $$ax^2+bx+c$$ evaluated at $$x=0$$ is $$c$$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $$x \in (0,1)$$. So $$c$$ cannot be negative.

If $$c$$ is 0 then 0 is a root of $$ax^2 +bx+c$$.

So $$c$$ must be positive for there to be no real root.

It's easy to see that since $$b^2<4ac$$, you have $$c > b^2/(4a) > 0$$ if $$a>0$$ and $$c < b^2/(4a) < 0$$ if $$a < 0$$.

But you have no roots, so it is either a parabola opening down below $$x$$-axis or opening up above $$x$$-axis, and since $$a+b+c=1$$ it must be all above.

Can you conclude?

• Yes! With your and other answers, I understand even better :). Thank you! – Wolf M. Jan 17 at 22:25

Since the polynomial has no roots, its graph is either strictly above or below the $$x$$-axis. But $$f(1)=a+b+c>0$$, so the graph is above the $$x$$-axis. The parabola then intersects the positive part $$y$$-axis, but this intersection point is $$(0,c)$$, so $$c>0$$.

Perforce $$\text{sgn}(y(0))=\text{sgn}(y(1)).$$
• If you prefer, $\text{sgn}(c)=\text{sgn}(a+b+c)$. – Yves Daoust Jan 17 at 22:22