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

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

Options:

Correct Answer: p >= 2

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.

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

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.

Discriminant of a Quadratic Equation – The discriminant (D) of a quadratic equation ax^2 + bx + c = 0 is given by D = b^2 - 4ac. It determines the nature of the roots: if D > 0, there are two distinct real roots; if D = 0, there is one real root; if D < 0, there are no real roots.
Conditions for Real Roots – For a quadratic equation to have real roots, the discriminant must be non-negative (D >= 0).