If $S=\left\{m \in R: x^2-2(1+3 m) x+7(3+2 m)=0\right.$ has distinct roots $\}$, then the number of elements…
If $S=\left\{m \in R: x^2-2(1+3 m) x+7(3+2 m)=0\right.$ has distinct roots $\}$, then the number of elements in $S$ is
$2$
$3$
$4$
infinite
Solution
Given, equation $x^2-2(1+3 m) x+7(3+2 m)=0$
Here, $a>0$ and $D=b^2-4 a c>0$
Expression is always positive, if roots are distinct then $D>0$,
$[-2(1+3 m)]^2-4 \times 7(3+2 m)>0$
$\begin{aligned} & \Rightarrow 4\left[1+9 m^2+6 m\right]-84-56 m>0 \\ & \Rightarrow 36 m^2-32 m-80>0 \\ & \Rightarrow 9 m^2-8 m-20>0\end{aligned}$
$\begin{aligned} & \Rightarrow \quad(m-2)\left(m+\frac{10}{9}\right)>0 \\ & \Rightarrow \quad m \in\left(-\infty, \frac{-10}{9}\right) \cup[2, \infty)\end{aligned}$
$\therefore$ Integral value of $m$ are ...... $3 i, 1,2,3,4,5,6$. So, the number of element in $S$ is infinite.