If the quadratic equation x^2 + 2px + p^2 - 4 = 0 has real roots, what is the condition on p?
Correct Answer: p can be any real number.
- Step 1: Identify the quadratic equation given: x^2 + 2px + (p^2 - 4) = 0.
- Step 2: Recall that for a quadratic equation ax^2 + bx + c = 0, the discriminant (D) is given by D = b^2 - 4ac.
- Step 3: In our equation, a = 1, b = 2p, and c = p^2 - 4.
- Step 4: Calculate the discriminant: D = (2p)^2 - 4(1)(p^2 - 4).
- Step 5: Simplify the discriminant: D = 4p^2 - 4(p^2 - 4).
- Step 6: Distribute the -4: D = 4p^2 - 4p^2 + 16.
- Step 7: Combine like terms: D = 0 + 16 = 16.
- Step 8: Since the discriminant D = 16 is always greater than or equal to 0, the quadratic equation has real roots for any value of p.
- Step 9: Conclude that there is no restriction on p; it can be any real number.
No concepts available.
