For a quadratic equation ax^2 + bx + c = 0 with real coefficients, if b^2 - 4ac > 0, what is the nature of the roots?

For a quadratic equation ax^2 + bx + c = 0 with real coefficients, if b^2 - 4ac > 0, what is the nature of the roots?

• A. Two distinct real roots
• B. One real root
• C. No real roots
• D. All real numbers

Answer: (A)

The key idea is the discriminant of a quadratic with real coefficients. If b^2 - 4ac is positive, the square root in the quadratic formula is real and nonzero, so you get two different real solutions: x = (-b ± sqrt(b^2 - 4ac)) / (2a). Since the discriminant is greater than zero, those two values are distinct, giving two distinct real roots. This assumes a ≠ 0, since it’s a quadratic. If the discriminant were zero you'd have one real (repeated) root, if it were negative there would be no real roots, and “all real numbers” isn’t possible for a single quadratic.

The key idea is the discriminant of a quadratic with real coefficients. If b^2 - 4ac is positive, the square root in the quadratic formula is real and nonzero, so you get two different real solutions: x = (-b ± sqrt(b^2 - 4ac)) / (2a). Since the discriminant is greater than zero, those two values are distinct, giving two distinct real roots. This assumes a ≠ 0, since it’s a quadratic. If the discriminant were zero you'd have one real (repeated) root, if it were negative there would be no real roots, and “all real numbers” isn’t possible for a single quadratic.