We are given the quadratic equation:

$p x^2 + q x + r = 0$

with two real roots $\alpha$ and $\beta$, where $\alpha < -1$ and $\beta > 1$. We are asked to find the value of $1 + \frac{r}{p} + \left| \frac{q}{p} \right|$.

Step 1: Use Vieta's formulas

For a quadratic equation $p x^2 + q x + r = 0$, Vieta's formulas tell us:

1. The sum of the roots: $\alpha + \beta = -\frac{q}{p}$
2. The product of the roots: $\alpha \beta = \frac{r}{p}$

Step 2: Use root information

Given that $\alpha < -1$ and $\beta > 1$, let's take specific values to simplify the calculations. Let’s choose $\alpha = -2$ and $\beta = 2$, which satisfy the condition $\alpha < -1$ and $\beta > 1$.

Using these values: $\alpha + \beta = -2 + 2 = 0 \quad \Rightarrow \quad -\frac{q}{p} = 0 \quad \Rightarrow q = 0$ $\alpha \beta = (-2)(2) = -4 \quad \Rightarrow \quad \frac{r}{p} = -4 \quad \Rightarrow r = -4p$

Step 3: Calculate the required expression

We need to find the value of: $1 + \frac{r}{p} + \left| \frac{q}{p} \right|$ Substitute $\frac{r}{p} = -4$ and $\frac{q}{p} = 0$: $1 + (-4) + \left| 0 \right| = 1 - 4 + 0 = -3$

Final Answer:

The value of $1 + \frac{r}{p} + \left| \frac{q}{p} \right|$ is $-3$.

Would you like a deeper explanation or have any questions?

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Here are 5 related questions to expand on this problem:

1. How does Vieta's formulas help solve quadratic equations with real roots?
2. What conditions guarantee that a quadratic equation has real roots?
3. How do transformations affect the graph of a quadratic function?
4. What happens when both roots of a quadratic equation are negative?
5. How can we factor a quadratic equation with given roots?

Tip: When solving quadratic equations, always check if the discriminant $\Delta = b^2 - 4ac$ is non-negative for real roots.