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......

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.
