If the quadratic equation x^2 + 2px + p^2 - 4 = 0 has real roots, what is the condition on p?

Practice Questions

1 question

p > 2
p < 2
p = 2
p >= 2

Questions & Step-by-step Solutions

1 item

Q: If the quadratic equation x^2 + 2px + p^2 - 4 = 0 has real roots, what is the condition on p?

Solution: The discriminant must be non-negative: (2p)^2 - 4(1)(p^2 - 4) >= 0 => 4p^2 - 4p^2 + 16 >= 0, which is always true. Thus, p can be any real number.

Steps: 9

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.

If the quadratic equation x^2 + 2px + p^2 - 4 = 0 has real roots, what is the condition on p?

Practice Questions

Questions & Step-by-step Solutions

Related Questions

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?