Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$ 
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$

My Attempt:
Given
\begin{align}
  a(a+7b+49c)+c(a-b+c) &< 0 \\
  49a \left( \dfrac {a}{49} + \dfrac {b}{7} + c \right)+c(a-b+c) &< 0 \\
  49a \cdot f\left( \dfrac {1}{7} \right) + f(0)\cdot f(-1) &< 0
\end{align}
 A: $$a(a+7b+49c)+c(a-b+c)\lt 0$$
is equivalent to
$$ac\lt \frac{-a^2-7ab+bc-c^2}{50}$$
So, we get
$$\begin{align}b^2-4ac&\gt b^2-\frac{4}{50}(-a^2-7ab+bc-c^2)\\\\&=\frac{1}{50}(50b^2+4a^2+28ab-4bc+4c^2)\\\\&=\frac{1}{50}\left(4\left(a+\frac{7b}{2}\right)^2+(b-2c)^2\right)\ge 0\end{align}$$
A: a) Suppose that $a$ and $c$ have opposite sign. Then $b^2-4ac\geq 0$, and we are done.
b) Suppose that $f(1/7)$ and $f(-1)$ have opposite sign. Then by the intermediate value theorem, $ax^2+bx+c=0$ have a real root, hence two real roots, and we are done.
c) Suppose now that $a,c$ have the same sign, say $e_1$, and that $f(1/7)$ and $f(-1)$ have also the same  sign, say $e_2$.
By the condition, we cannot have $e_1=e_2$. Hence $e_2=-e_1$; for example, if $e_1=1$, then $f(x)\to +\infty$ if $x\to +\infty$, and $f(1/7)\leq 0$, hence $f(x)=0$ has one, hence two real roots. The same for the case $e_1=-1$.  
A: We can assume that $a=1$ and let $f(x) =x^2+bx+c$. 
If $c< 0$ then the graph of $f$ cuts the $y-$ axis under the $x-$ axis, so it must have real roots. 
If $c=0$ then 0ne root is $0$ and the second is $-b$. 
If $c>0$, since we have $$49f({1\over 7})+cf(-1)=0$$
we have 2 possibilities. 
a) If $f(-1)=f(1/7)=0$ we are done.
b) If $f(-1)>0>f(1/7)$ or $f(-1)<0<f(1/7)$ then we have one real root in $(-1,{1\over 7})$ and we are done.  
