Show that if $a > 0$, then $ax^2 + bx + c ≥ 0$ for all values of $x$ if and only if $b^2 − 4ac ≤ 0$.

Question

So I'm fairly new to proofs and was wondering how to solve this problem. This is my train of thought.

I listed the terms as so: $ax^2 + bx + c ≥ 0 ≥ b^2 − 4ac$.

From there I tried to complete the square and ended up with this:

$$x ≥ \sqrt{-c/a} ≥ \sqrt{(b^2/a) - 4ac - c/a}.$$

I'm not quite exactly sure how to show that if $a > 0$, then $ax^2 + bx + c ≥ 0$ for all values of $x$ if and only if $b^2 − 4ac ≤ 0$ and would greatly enjoy any help.

Thanks. Forgive my formatting.

Answer 1

Note that $$ax^2+bx+c=a\cdot\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}=a\cdot\left(x+\frac b{2a}\right)^2-\frac{b^2-4ac}{4a}.$$

Answer 2

The roots of $ax^2 + bx + c =0$ are $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} $.

If the roots are complex, then $ax^2 + bx + c \ne 0$ for all real $x$.

If $b^2-4ac < 0$, the roots are complex. If $a > 0$, then $ax^2+bx+c > 0$ for large enough $x$, so $ax^2+bx+c > 0$ for all $x$ since it has no real roots.

If $b^2-4ac = 0$, the roots are real and equal. The double root is $\dfrac{-b}{2a} $ so $ax^2+bx+c =a(x+\frac{b}{2a})^2 \ge 0$ for all $x$ and is zero only for $x=\dfrac{-b}{2a} $.

Answer 3

1.If the $b^2-4ac < 0$ then equation has no real root so graph of the $ax^2+bx+c$ will always be above $X-axis$.

2.If $b^2-4ac=0$ then equation will have only one real root.

3.If $ax^2+bx+c$ is Non negative then it has at most one real root so $b^2-4ac$ is nonpositive

so $ax^2+bx+c$ is always non negative iff $b^2-4ac$ is non positive