If $X$ denotes the set of real numbers $p$ for which the equation $\displaystyle x^{2}=p(x+p)$ has its roots greater than $p$, then $X$ is equal to
- $\displaystyle \left ( -2,-\frac{1}{2} \right )$
- $\displaystyle \left ( -\frac{1}{2},\frac{1}{4} \right )$
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Null set
- $(-\infty,0)$
x^2 - px - p^2 = 0. Roots = [p +/- sqrt(p^2 + 4p^2)] / 2 = p(1 +/- sqrt(5)) / 2. We need roots > p. p(1 + sqrt(5)) / 2 > p => (1 + sqrt(5))/2 > 1 (True for p > 0) or (1 + sqrt(5))/2 < 1 (False for p < 0). p(1 - sqrt(5)) / 2 > p => (1 - sqrt(5))/2 > 1 (False). Checking conditions shows no real p satisfies the condition.
Rewrite the given equation in the standard quadratic form x^2 - px - p^2 = 0. For the roots to be real and greater than p, the discriminant D = p^2 + 4p^2 must be non-negative, which simplifies to 5p^2 >= 0, a condition that is true for all real p. Let the roots be alpha and beta; since their product is -p^2, which is always less than or equal to zero, the roots must have opposite signs or one must be zero (if p=0). If p is positive, one root is positive and the other is negative, so the negative root cannot possibly be greater than the positive number p. If p is negative, the positive root must be greater than p, but the negative root will be less than p because the difference between the roots is sqrt(5)*|p|, which is greater than |p|. Since neither a positive nor a negative p allows both roots to simultaneously satisfy the condition of being greater than p, the set of valid real numbers is empty, making it a null set.