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I recently encountered a quadratic equations property that

$ax^2+bx+c>0$ $ \forall$ $x\in \Re \Rightarrow D<0$ $and$ $a>0$

and $ax^2+bx+c<0$ $ \forall$ $x\in \Re \Rightarrow D<0$ $and$ $a<0$.

Now, i tried to prove it algebraically and through graphs. Its clear that the equation seems much simple by just making a parabola and as $D<0$ the parabola never touches the $x$ axis and hence the two equations follow.

Can there be any algebraic proof the relation mentioned above.

Harsh Sharma
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    How is the discriminant related to the roots? Once you have that in mind, the first question is easily answered. For the second, use continuity and look at the limit as $x$ goes to infinity – b00n heT Jun 20 '16 at 21:00

2 Answers2

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We suppose $a \neq 0$, otherwise it is not a quadratic equation.

$$ax^2+bx+c>0,\forall x \in \mathbb{R} \Rightarrow ax^2+bx+c\neq0,\forall x \in \mathbb{R}\Rightarrow D<0$$

Also $\lim_{x\to \infty}ax^2+bx+c=\frac{a}{|a|}\infty$. So :

$$ax^2+bx+c>0,\forall x \in \mathbb{R} \Rightarrow \lim_{x\to \infty}ax^2+bx+c =+\infty \Rightarrow \frac{a}{|a|}=1 \Rightarrow a>0$$

Finally : $$ax^2+bx+c>0,\forall x \in \mathbb{R} \Rightarrow D<0 \space \text{and} \space a>0$$

You can do the same kind of reasonning for the second inequality.

Bérénice
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For $a>0$ we have: $$ ax^2+bx+c>0 \quad \iff \quad x^2+\frac{b}{a}+\frac{c}{a}>0 $$ so: $$ x^2+\frac{b}{a}>-\frac{c}{a} $$ $$ x^2+\frac{b}{a} +\left(\frac{b}{2a} \right)^2>-\frac{c}{a}+\left(\frac{b}{2a} \right)^2 $$ $$ \left(x+\frac{b}{2a} \right)^2>-\frac{c}{a}+\left(\frac{b}{2a} \right)^2 $$ that is always verified ( since The LHS is a square) if $$ -\frac{c}{a}+\left(\frac{b}{2a} \right)^2<0 $$ i.e. $$ \frac{b^2-4ac}{4a^2}=\frac{\Delta}{4a^2} <0 $$ and since $4a^2>0$ we have $\Delta<0$.

You can do the same for the other case.

Emilio Novati
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