Suppose $X \sim \mathcal{U}(-1, 3)$. What is le probability that the equation $y^2 + 4Xy + X = 0$ possesses real roots?
Suppose $a=1$, $b=4X$ and $c=X$, then $b^2-4ac = (4X)^2 - 4X = 4X(4X-1) > 0$. Hence, $X>0$ or $4X-1>0$.
We have $P(X>0 \cup X>\frac{1}{4}) = (1-P(X \le 0)) + (1-P(X\le \frac{1}{4})) = 1.4375 > 1$. It is clear this answer is wrong.
In the answer sheet, it says the answer is $P(X \ge \frac{1}{4}) + P(X \le 0) = \frac{15}{16}$.
Where am I wrong