Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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 ${b^2 - 4ca}\leq 0$.

share|cite|improve this answer
@Babak-sorouh:hello sir .thanks – Maisam Hedyelloo Feb 4 '13 at 17:08

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.

share|cite|improve this answer
Well put..........+1 – amWhy Feb 5 '13 at 0:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.