What are the reasons for these conditions for quadratics?

Question

My book says that for the quadratic $ax^2+bx+c$ to be greater than zero, $a>0$ and $D<0$. Where did they get these conditions from?

Answer 1

If the discriminant is less than zero, there are no solutions to the equation $ax^2 + bx + c = 0$. That means if $D < 0$ the quadratic does not touch the $x$-axis anywhere. The second condition, $a>0$ ensures the quadratic is opening upwards (concave up) by ensuring the leading coefficient is positive.

That means the quadratic never touches the $x$-axis and is concave up meaning its minimum is its vertex with no absolute maximum. Thus it is never less than or equal to $0$.

Answer 2

Your function $$f(x)=ax^2+bx+c$$ is equaivelent to $$a\left(x^2+2\frac{b}{2a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}\right)=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$$

Answer 3

We need to proof that \begin{cases} a>0 \\ b^2-4ac<0 \end{cases}

Step one:

With $a\ne 0$:

\begin{equation}\begin{aligned} ax^2+bx+c>0&\Leftrightarrow x^2+\dfrac{b}{a}x+\dfrac{c}{a}>0 \\ &\Leftrightarrow x^2+2\times x\times \dfrac{b}{2a}+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}+\dfrac{4ac}{4a^2}>0 \\ &\Leftrightarrow \left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a^2}>0 \end{aligned}\end{equation}

If $b^2-4ac\ge 0$ then there always exist (real number) $x$ that satisfies $$ax^2+bx+c=0\Leftrightarrow \left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{\sqrt{b^2-4ac}}{2a}\right)^2=0\Leftrightarrow x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},$$

contradiction, so we conclude that $b^2-4ac<0$ is one of the conditions needed so that $ax^2+bx+c>0$ for all real numbers $x$.

Step two:

Now assume that $a<0$ ($a\ne 0$ because $b^2-4ac<0$).

We already proved that $b^2-4ac<0\Rightarrow 4ac>b^2\ge 0\Rightarrow ac>0\Rightarrow c<0\text{ because $a<0$}$.

If $a<0$ and $c<0$ then assume that $a_1=-a;c_1=-c$, then $a_1;c_1>0$ and we will have \begin{equation}\begin{aligned} ax^2+bx+c&=-a_1x^2+bx-c_1 \\ &= -(a_1x^2-bx+c_1)\\ &= -\left(a_1\left(x+\frac{-b}{2a_1}\right)^2-\frac{b^2-4a_1c_1}{4a_1}\right)\\ &=\frac{b^2-4a_1c_1}{4a_1}-a_1\left(x-\dfrac{b}{2a_1}\right)^2 \\ &\le \dfrac{b^2-4ac}{4a_1}\text{ because $\begin{cases} a_1c_1=(-a)(-c)=ac \\ a_1\left(x-\dfrac{b}{2a_1}\right)^2\ge 0\text{ $(a_1>0)$} \end{cases}$} \end{aligned}\end{equation}

We have already proved that $b^2-4ac<0$ and also because $a_1>0$,

$$ax^2+bx+c\le \dfrac{b^2-4ac}{4a_1}<0 \text{ for all $x$.}$$

This is not correct because $ax^2+bx+c>0$, this finishes the proof and I will sum up all information in this table (true for all real numbers $x$):

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline &a>0&a<0\ \\ \hline b^2-4ac<0 &ax^2+bx+c>0&ax^2+bx+c<0\\\hline b^2-4ac\ge 0&\text{The equation has two roots}&\text{The equation has two roots} \\\hline\end{array}