The set of all values of $k$ for which the inequality $x^2-(3 \mathrm{k}+1)$ $x+4 \mathrm{k}^2+3…
- $\left(-\frac{13}{7}, 1\right)$
- $\left(-1, \frac{13}{7}\right)$
- $\left(-\infty,-\frac{13}{7}\right) \cup(1, \infty)$
- $(-\infty,-1) \cup\left(\frac{13}{7}, \infty\right)$
Solution
So, $1\gt0$ and $b^2-4 a c \lt 0$ $\begin{aligned} & \Rightarrow\{-(3 k+1)\}^2-4 \times 1 \times\left(4 k^2+3 k-3\right) \lt 0 \\ & \Rightarrow(7 k+13)(k-1)\gt0 \end{aligned}$

So, $x \in\left(-\infty, \frac{-13}{7}\right) \cup(1, \infty)$.
Asked in: AP EAMCET 2024 (23 May Shift 1)