Consider the cubic equation $x^3+px^2+qx+r=0$, where p, q, r are real numbers. Which of the following statements is correct?
- If $p^2-2q < 0$, then the equation has one real and two imaginary root.
- If $p^2-2q\geq 0$, then the equation has all real roots.
- If $p^2-2q > 0$, then the equation has all real and distinct roots.
- If $4p^3-27q^2 > 0$, then the equation has real and distinct roots.
For a cubic equation x^3 + px^2 + qx + r = 0, the nature of roots is determined by the discriminant. The condition p^2 - 2q < 0 relates to the derivative and the existence of local extrema, which implies one real root and two complex roots.
For a cubic equation x^3 + px^2 + qx + r = 0, if we square the sum of the roots, we get the identity (sum of roots)^2 = sum of squares of roots + 2 * (sum of product of roots taken two at a time). By Vieta's formulas, the sum of the roots is -p and the sum of the products of the roots taken two at a time is q, so the sum of the squares of the roots is p^2 - 2q. Since the square of any real number is non-negative, having all three real roots requires p^2 - 2q to be greater than or equal to zero. Therefore, if p^2 - 2q < 0, it is impossible for all three roots to be real, meaning the equation must have one real and two imaginary roots.