2 votes 2 votes Consider the quadratic equation $x^2+\dfrac{x}{2}+c=0$, where $c$ is chosen uniformly randomly from the interval $[0,1]$. What is the probability that the given quadratic equation has a real solution? The solutions of $a x^2+b x+c=0$ are given by $x=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2a}$.$1 / 2$$1 / 4$$1 / 8$$1 / 16$ Probability goclasses2024-mockgate-14 probability uniform-distribution 2-marks + – GO Classes asked Feb 5 • edited Feb 7 by Sachin Mittal 1 GO Classes 410 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes $\frac{1}{16}$ since a real solution occurs precisely when $b^2-4 c=\frac{1}{4}-4 c \geq 0$, i.e., $0 \leq c \leq \frac{1}{16}$, which is $\frac{1}{16}^{\text {th }}$ fraction of the interval $[0,1]$ over which $c$ ranges. GO Classes answered Feb 5 • edited Feb 5 by Lakshman Bhaiya GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.