Question on quadratic inequations

Question

Given an equation $ax^2+bx+c$ $>$ or $<$ 0. Note that $a$ is kept positive.

If the discriminant $(\sqrt{b^2-4ac})$ of the quadratic equation $< 0$, then the solution either applies for all real $x$ or no real $x$ depending on the inequality sign of: (This one: $ax^2+bx+c$ $>$ or $<$ $0$)

Why so? Regards

Answer 1

If you look at the quadratic EQUATION $ax^2+bx+c=0$ then the discriminant will tell you whether it will have $0,1$ or $2$ real roots. If the sign of the discriminant is negative we have no roots, which means that the graph of the function $y=ax^2+bx+c$ is either in the upper half-plane or lower-half plane (Otherwise it will intersect the $x-$axis and we'll get a root.) But $y=ax^2+bx+c$ being in either one of the half-planes means that $y>0$ for no $x$ all for all $x$.

Answer 2

you can write as $$x^2+\frac{b}{a}x+\frac{c}{a}>0$$ then $$x^2+2\frac{b}{2a}x+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}>0$$ this ncan be written as $$\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}>0$$ and this is positive if $$b^2-4ac<0$$

Answer 3

The discriminant indicates the presence of roots. When it is negative there are no roots. In other words, the function keeps a constant sign, which is positive as $a$ is positive.

Then $+>0$ is true and $+<0$ is false.