Alternative proof that if $a,b,c \in \mathbb{R}$ and $(a+b+c)c<0$ then $b^2-4ac>0$?

Question

A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?

The question emerged from a reddit post https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/

Answer 1

We have $$ (b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 \ge 0 \\ \implies b^2 - 4ac \ge - 4(a+b+c)c > 0 $$

To give proper credit: The above approach was found after reading guest's answer and is just a simplification of that solution.

Answer 2

Here is a proof that works for any ordered commutative ring:

Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.

Solution:

First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $c\neq 0$. Then form the matrix $$M = \left( \begin{array}{cc} 2c & b \\ b & 2a \\ \end{array}\right)$$ whose determinant is precicely $4ac-b^2$, and the matrix $$ S = \left( \begin{array}{cc} 1 & 0 \\ 2 & 1 \\ \end{array}\right)$$ whose determinant is $1$. Now from the multiplicativity of the determinant $\det(SM) =\det(S)\det(M)=\det(M)$. But $$\det(SM) = \det \left( \begin{array}{cc} 2c & b \\ 4c+b & 2(a+b) \\ \end{array}\right) = 4c(a+b) - b^2-4bc.$$

But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$ The last element is always non-positive and the inequality is strict, so $\det(M)=4ac-b^2<0$.

Answer 3

If all you know is completing the square:

Suppose $b^2-4ac\leq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.

By completing the square we can rewrite $ax^2+bx+c=a\left(\left(x+\frac{b}{2a}\right)^2+d^2\right)$ for some real $d$.

We have $c=\frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:

$$\begin{align}(a+b+c)c&=\left(a+b+\frac{b^2}{4a}+ad^2\right)\left(\frac{b^2}{4a}+ad^2\right)\\ &=\left(a+b+\frac{b^2}{4a}\right)\frac{b^2}{4a}+\left(ad^2\frac{b^2}{4a}+\left(a+b+\frac{b^2}{4a}\right)ad^2\right)+\left(ad^2\right)^2\\ &=\frac{b^2}{4}\left(1+\frac{b}{a}+\frac{b^2}{4a^2}\right)+d^2\left(a^2+ab+\frac{b^2}{2}\right)+\left(ad^2\right)^2\\ &=\frac{b^2}{4}\left(1+\frac{b}{2a}\right)^2+d^2\left(a+\frac{b}{2}\right)^2+\left(ad^2\right)^2\\ &\geq0\text{.} \end{align}$$ Thus by contrapositive, $(a+b+c)c<0\,\Rightarrow\, b^2-4ac>0$.

Answer 4

Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c \lt 0 $ and $b^2-4ac \le 0$ are simultaneously true.

1) By assumption

$$ (a + b + c) c \lt 0 $$

2) Negation of goal: $ b^2 - 4ac \gt 0$

$$ b^2 - 4ac \le 0 $$

3) From 2 by addition

$$ b^2 \le 4ac $$

4) expand 1

$$ ac + bc + c^2 < 0 $$

5) From 4 by addition

$$ ac < -bc - c^2 $$

6) From 5 by multiplication

$$ 4ac < -4bc -4c^2 $$

7) transitivity $x \le y \lt z$ implies $x \lt z$

$$ b^2 \lt -4bc - 4c^2 $$

8) from 7 by addition.

$$ b^2 + 4bc + 4c^2 \lt 0 $$

9) (7) is a perfect square

$$ (b + 2c)^2 < 0 $$

10) By assumption, $a, b, c \in \mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.

$$ \bot $$

Therefore, $b^2 - 4ac \gt 0$ .