Let's solve the given equation step by step.

Part (a): Show that $3p^2 - 10p - 8 > 0$

We are given the equation: $4(p - 2x) = \frac{12 + 15p}{x + p}$ where $p$ is a constant, and the equation has two distinct real roots. First, we will manipulate the equation to derive the necessary inequality for $p$.

Step 1: Multiply both sides by $x + p$ to eliminate the denominator:

$4(p - 2x)(x + p) = 12 + 15p$

Step 2: Expand the left-hand side:

$4(p - 2x)(x + p) = 4(p(x + p) - 2x(x + p))$ $= 4(px + p^2 - 2x^2 - 2px)$ $= 4(p^2 - 2x^2 + px - 2px)$ $= 4(p^2 - 2x^2 - px)$

So the equation becomes: $4(p^2 - 2x^2 - px) = 12 + 15p$

Step 3: Rearrange into standard quadratic form:

Now, we focus on the $x$-terms to write the equation in standard quadratic form $ax^2 + bx + c = 0$.

From the equation: $4(-2x^2 - px + p^2) = 12 + 15p$ We can rewrite this as: $-8x^2 - 4px + 4p^2 = 12 + 15p$ $-8x^2 - 4px + (4p^2 - 15p - 12) = 0$

Thus, the quadratic in $x$ is: $-8x^2 - 4px + (4p^2 - 15p - 12) = 0$ Now, the condition for two distinct real roots is that the discriminant of this quadratic must be positive. The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by: $\Delta = b^2 - 4ac$

Here, $a = -8$, $b = -4p$, and $c = 4p^2 - 15p - 12$.

Step 4: Calculate the discriminant:

$\Delta = (-4p)^2 - 4(-8)(4p^2 - 15p - 12)$ $= 16p^2 - 4(-8)(4p^2 - 15p - 12)$ $= 16p^2 + 32(4p^2 - 15p - 12)$ $= 16p^2 + 128p^2 - 480p - 384$ $= 144p^2 - 480p - 384$

For two distinct real roots, we require the discriminant $\Delta > 0$. Therefore, we need: $144p^2 - 480p - 384 > 0$ Dividing the entire inequality by 48 to simplify: $3p^2 - 10p - 8 > 0$ Thus, we have shown that $3p^2 - 10p - 8 > 0$, which completes part (a).

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Part (b): Find the range of possible values of $p$

Now, we will solve the inequality $3p^2 - 10p - 8 > 0$ to find the range of $p$.

Step 1: Solve the quadratic equation $3p^2 - 10p - 8 = 0$

We use the quadratic formula: $p = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(-8)}}{2(3)}$ $p = \frac{10 \pm \sqrt{100 + 96}}{6}$ $p = \frac{10 \pm \sqrt{196}}{6}$ $p = \frac{10 \pm 14}{6}$ So the two roots are: $p = \frac{10 + 14}{6} = 4 \quad \text{and} \quad p = \frac{10 - 14}{6} = -\frac{2}{3}$

Step 2: Analyze the inequality $3p^2 - 10p - 8 > 0$

The quadratic inequality $3p^2 - 10p - 8 > 0$ is satisfied when $p$ lies outside the roots $p = -\frac{2}{3}$ and $p = 4$.

Thus, the solution to the inequality is: $p < -\frac{2}{3} \quad \text{or} \quad p > 4$

Final Answer:

The range of possible values of $p$ is: $p \in (-\infty, -\frac{2}{3}) \cup (4, \infty)$

Let me know if you'd like further details or explanations on any part of this solution!

Related Questions:

1. How do you derive the quadratic discriminant formula?
2. What are the properties of quadratic inequalities?
3. What is the significance of having two distinct real roots for a quadratic equation?
4. Can the method used here be applied to non-quadratic equations?
5. How do different values of constants like $p$ affect the nature of the roots?

Tip: When solving quadratic inequalities, always check the signs of intervals around the roots to determine where the inequality holds.