Proving that $f(x)\ge 0$

Question

The equation $ax^2+bx+c=0$ does not have two distinct real roots, $a\neq0$ and $a+c>b$. Prove that $f(x)\ge0, \forall x \in R$.

Now, from the question I can interpret two things:

1. $\Delta (= b^2-4ac)\le0$
2. $a>0$

Proving the second point will be a solution to the question (graphically). How do I prove that $a>0$?

Answer 1

I think your statement is wrong. Try $b=1$, $c=2$ and $a=0$.

If we are talking about quadratic equation then $a\neq0$ and

since $a(-1)^2+b(-1)+c>0$, we obtain $a>0$ and $\Delta\leq0$, which says $ax^2+bx+c\geq0$ for all value of $x$.

Answer 2

This is a continuous function with $f(-1)>0$.

Answer 3

You can show that if a is negative, than the limit of f(x) as x tends to positive or negative infinity is always negative infinity, and that thus a must be positive to insure condition one. I don't know if you wanted limits in the proof but you could use them here.

Answer 4

Let $f(x)=ax^2+bx+c$ with $a \ne 0$.

Case 1: $f$ has only one real zero $x_0$ . Then $f(x)=a(x-x_0)^2 $

Now show that $a+b>c$ implies $a >0$. Hence $f(x) \ge 0$ for all $x$.

Case 2: $f(x) \ne 0$ for all $x$. From $f(-1) >0$, we get $f(x) > 0$ for all $x$.

Answer 5

Note that we have $(a-c)^2\geq 0$. This gives $$ a^2 - 2ac + c^2 \geq 0\\ a^2 + c^2 \geq 2ac$$ And since $2ac\geq b^2 - 2ac$, we get $$ a^2 + c^2 \geq b^2 - 2ac\\ a^2 + 2ac + c^2 \geq b^2\\ (a+c)^2 \geq b^2 $$ Now for the contradiction. Assume $a$ is negative. Then $c$ cannot be positive, because otherwise $b^2 - 4ac$ becomes positive. Therefore $a+c>b$ has negative numbers on either side of the inequality sign, which means that if we square it, we change the direction of the inequality. This yields $(a+c)^2 < b^2$, which contradicts the above inequality.

Answer 6

assuming $a\neq 0$

Observation 1. "does not have two distinct real roots", which means:

Either have only one real root. In that case $\Delta=0$.

Or have no real roots. In that case $\Delta<0$

Both cases implies $f$ is either always non-negative or always non-positive, i.e. $f(x)\geq 0$ for all $x\in\mathbb{R}$ or $f(x)\leq 0$ for all $x\in\mathbb{R}$

Observation 2. $f(-1)=a-b+c>0$.

Hence combining observation 1 observation 2 makes sure that $f(x)\geq 0$ for all $x\in\mathbb{R}$.