Question

find all values of p for which one root of equation x^2 - (p+1)x + p^2 + p - 8 = 0 is greater than 2 and the other root is smaller than 2

Answer 1

-2 < p < 3

Answer 2

To find the values of $p$ for which one root of the quadratic equation $x^2 - (p+1)x + p^2 + p - 8 = 0$ is greater than 2 and the other root is smaller than 2, we can follow these steps:

1. Step 1: Determine f(2).
   Let $f(x) = x^2 - (p+1)x + p^2 + p - 8$.
   Then,

   $$f(2) = 2^2 - (p+1)(2) + p^2 + p - 8 = 4 - 2p - 2 + p^2 + p - 8 = p^2 - p - 6.$$

   Factorize:

   $$p^2 - p - 6 = (p-3)(p+2).$$

2. Step 2: Use the condition for roots.
   If one root is greater than 2 and the other is less than 2, then 2 lies strictly between the roots. Since the coefficient of $x^2$ is positive, this happens if

   $$f(2) < 0 \quad \text{i.e.} \quad (p-3)(p+2) < 0.$$

   The inequality $(p-3)(p+2) < 0$ holds when

   $$-2 < p < 3.$$

3. Step 3: Verify real roots (optional but needed for existence).
   The discriminant $\Delta$ is given by:

   $$\Delta = (p+1)^2 - 4(p^2+p-8) = p^2 + 2p + 1 - 4p^2 - 4p + 32 = -3p^2 - 2p + 33.$$

   The discriminant must be nonnegative, i.e.

   $$-3p^2 - 2p + 33 \ge 0 \quad \Longrightarrow \quad 3p^2 + 2p - 33 \le 0.$$

   Solving $3p^2 + 2p - 33 = 0$ gives:

   $$p = \frac{-2 \pm \sqrt{4 + 396}}{6} = \frac{-2 \pm 20}{6}.$$

   Therefore,

   $$p = 3 \quad \text{or} \quad p = -\frac{11}{3}.$$

   Since the quadratic in $p$ opens upwards, the condition is satisfied for

   $$-\frac{11}{3} \le p \le 3.$$

   Intersecting this with $-2 < p < 3$ (from Step 2) results in:

   $$-2 < p < 3.$$

Explanation (Minimal Core):

• Compute $f(2) = (p-3)(p+2)$.
• For 2 to lie between the roots, require $(p-3)(p+2) < 0$ $\Longrightarrow -2 < p < 3$.
• Real roots exist in this interval.