Concept of roots in Quadratic Equation $a$ , $b$, $c$ are real numbers where a is not equal to zero and the quadratic equation 
\begin{align} 
ax^2 + bx +c =0   
\end{align} 
has no real roots then prove that $c(a+ b+ c)>0$  and  $a(a+ b + c) >0$ 
My Approach : As the equation has no real roots then its discriminant is less than zero.  So the graph of the equation will be above  $x$-axis or below $x$-axis . 
I am able to conclude signs of $a$ , $b$, $c$ but still not getting appropriate answer. 
Please explain the concept......
 A: Continuing from where you left.
Let $p(x)=ax^2+bx+c$
Observe that 


*

*$p(1)=(a+b+c)$

*$p(0)=c$

*$\lim_{x \rightarrow \infty}\frac{p(x)}{x^2} =a$


So, $$p(1) \cdot p(0)=c(a+b+c) >0$$ $[p(0)$ and $p(1)$ have same sign]
Also, $$p(1) \cdot \lim_{x\rightarrow \infty} \frac{p(x)}{x^2} =a(a+b+c) >0$$ [$p(1)$ and $\lim_{x \rightarrow \infty} \frac{p(x)}{x^2}$ have the same sign because $x^2 > 0$, for all $x \in \mathbb{N}$]
Hence, the claim is proved.
A: For the second part, another justification is that since the discriminant is less than zero, then it must hold that $a,c$ have the same sign. Why? If $a,c$ had opposite signs then the discriminant $\Delta = b^2-4ac$ would have been positive.
Since $c,(a+b+c)$ have the same sign and $c,a$ have the same sign then $a,(a+b+c)$ have the same sign as well, thus:
$$a(a+b+c) >0.$$
A: If the quadratic equation has no real roots, then the graph of the corresponding quadratic function ($y = ax^2 + bx + c$) will not cross the x-axis. (Case-1 I will just discuss the case that the function states in the upper part of the x-axis.) Then, 
(1) the parabola must be opening upward. This means $a > 0$; 
(2) it will somehow cut the y-axis at $(0, c)$ where $c > 0$; and
(3) for all real t, $y(t) = at^2 + bt + c >0$. This is especially true for $t = 1$. That is $y(1) = … = a + b + c > 0$.
Required results for this case follow.
