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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 :) .

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    $\begingroup$ What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question) $\endgroup$ – Mark Bennet Jan 17 at 22:08
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    $\begingroup$ 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... $\endgroup$ – diracdeltafunk Jan 17 at 22:10
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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 $c>0$.

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

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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?

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  • $\begingroup$ Yes! With your and other answers, I understand even better :). Thank you! $\endgroup$ – Wolf M. Jan 17 at 22:25
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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 $c>0$.

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Short answer:

Perforce $$\text{sgn}(y(0))=\text{sgn}(y(1)).$$

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

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