Question

If x² + 2px - 2p + 8 > 0 for all real values of x, then the set of all possible values of p is

• A. (2,4)
• B. (-∞, -4)
• C. (2,∞)
• D. (-4,2)

Answer 1

The Correct Option is D

1. Understanding the Problem:
The given inequality is \(x^2 + 2px - 2p + 8 > 0\). We need to find the values of \(p\) such that the inequality holds for all real values of \(x\).

2. Analyzing the Quadratic Inequality:
We recognize that this is a quadratic inequality in terms of \(x\). The general form of a quadratic equation is \(Ax^2 + Bx + C\). For this inequality, we have:

\(A = 1, B = 2p, C = -2p + 8\).

3. Condition for the Inequality to Hold for All Real Values of \(x\):
For the quadratic inequality \(Ax^2 + Bx + C > 0\) to hold for all real values of \(x\), the discriminant must be negative. The discriminant \(\Delta\) for a quadratic equation \(Ax^2 + Bx + C = 0\) is given by:

\(\Delta = B^2 - 4AC\).

4. Calculating the Discriminant:
For our quadratic equation \(x^2 + 2px - 2p + 8\), the discriminant is:

\(\Delta = (2p)^2 - 4(1)(-2p + 8)\)

\(\Delta = 4p^2 - 4(-2p + 8)\)

\(\Delta = 4p^2 + 8p - 32\).

5. Setting the Discriminant Less Than Zero:
For the inequality to hold for all \(x\), we require \(\Delta < 0\), so:

\(4p^2 + 8p - 32 < 0\).

6. Solving the Quadratic Inequality:
Divide the entire inequality by 4 to simplify:

\(p^2 + 2p - 8 < 0\).

Now solve the quadratic equation \(p^2 + 2p - 8 = 0\) using the quadratic formula:

\(p = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}\).

So, \(p = \frac{-2 + 6}{2} = 2\) or \(p = \frac{-2 - 6}{2} = -4\).

7. Analyzing the Solution:
Since the quadratic inequality is \(p^2 + 2p - 8 < 0\), the solution lies between the roots \(p = -4\) and \(p = 2\). Therefore, the set of all possible values of \(p\) is \((-4, 2)\).

Final Answer:
The set of all possible values of \(p\) is \((-4, 2)\).