The coefficients $\mathrm{a}, \mathrm{b}, \mathrm{c}$ in the quadratic equation $\mathrm{a} x^2+\mathrm{bx}+\mathrm{c}=0$ are from the set $\{1,2,3,4,5,6\}$. If the probability of this equation having one real root bigger than the other is p, then 216p equals :

<p>Equation is $a x^2+b x+c=0$</p> <p>$\mathrm{D}>0$ [for roots to be real & distinct]</p> <p>$\Rightarrow b^2-4 a c>0$</p> <p>For $b<2$ no value of $a$ & $c$ are possible</p> <p>$$\begin{aligned} & \text { For } b=3 \Rightarrow a c<\frac{9}{4} \\ & (a, c) \in\{(1,1),(1,2),(2,1)\} \Rightarrow 3 \text { cases } \end{aligned}$$</p> <p>For $b=4 \Rightarrow a c<4$</p> <p>$(a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3)\} \Rightarrow 5 \text { cases }$</p> <p>For $b=5 \Rightarrow a c<\frac{25}{4}$</p> <p>$$\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1)\}=14 \text { cases } \end{aligned}$$</p> <p>For $b=6 \Rightarrow a c<9$</p> <p>$$\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1),(2,4),(4,2)\}=16 \text { cases } \end{aligned}$$</p> <p>Total cases $=3+5+14+16=38$ cases</p> <p>$\Rightarrow$ Probability, $p=\frac{38}{216}$</p> <p>$\Rightarrow 216 p=38$</p>