Let `a ,b ,c` be real. If `a x^2+b x+c=0` has two real roots `alphaa n dbeta,w h e r ealpha<-1a n dbeta>1` , then show that `1+c/a+|b/a|<0`

Explanation

To solve the problem, we need to show that if the quadratic equation \( ax^2 + bx + c = 0 \) has two real roots \( \alpha \) and \( \beta \) such that \( \alpha < -1 \) and \( \beta > 1 \), then the inequality \( 1 + \frac{c}{a} + \left| \frac{b}{a} \right| < 0 \) holds.

### Step 1: Understanding the Roots
Given that the roots \( \alpha \) and \( \beta \) are real and \( \alpha < -1 \) and \( \beta > 1 \), we can conclude that the quadratic opens upwards (i.e., \( a > 0 \)) or downwards (i.e., \( a < 0 \)) based on the nature of the roots.

### Step 2: Evaluating the Function at Specific Points
We evaluate the quadratic function \( f(x) = ax^2 + bx + c \) at \( x = -1 \) and \( x = 1 \):
- For \( x = -1 \):
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