Vectors and euclidean spaces

Question

I tried doing this problem and this is the problem

if $a, b,$ and $c$ are real numbers and $a+bx+cx^2 \geq 0$ for any real number $x$, explain why $b^2-4ac \leq 0$

this is how i am trying to do if we know $a +bx+cx^2 \geq 0$ then our disriminant is $\sqrt{b^2 - 4ca}$ but this cant be $\leq o$ because it would make our square root negative. and that woulnd exist so the solution wouldnt exist.

I am confused on this please help out, if you could give me some hints or a solution i would be happy, these are just practice questions i need help on

Answer 1

If $y=a+bx+cx^2 \geq 0$ then $y|_{\min}\ge 0$. $$x_{min}=\frac{-b}{2c}$$ therefore $ y(x_{min})\ge0$. If you calculate $$y\left(\frac{-b}{2c}\right)$$ you will obtain ${b^2 - 4ca}\leq 0$.

Answer 2

Let $ax^2+bx+c$ wherein $a,~b,~c\in\mathbb R$. This expression can be written as: $$I:ax^2+bx+c=\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2+c-\frac{b^2}{4a}$$ I assume $a>0$ since you assumed $I\geq0$. If $I$ wants to be greater than $0$ so we need $c-\frac{b^2}{4a}$ to be positive also. It means that $\frac{4ac-b^2}{4a}$ is needed to be positive. But $a>0$, so $b^2-4ac<0$. If $a=0$, so $I$ will reduce to $bx+c$ and again the discriminate is positive.