<s>[INST] 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}
 [/INST] 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 $e_1=-1$.  
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