The coefficients a,b,c in the quadratic equation ax^2+bx+c=0 are from the set 1,2,3,4,5,6.

Question

The coefficients $a,b,c$ in the quadratic equation $a{x}^{2}+bx+c=0$ 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 :

Accepted Answer

Equation is $a{x}^{2}+bx+c=0$

$D>0$ [for roots to be real \& distinct]

$\Rightarrow {b}^{2}-4ac>0$

For $b$ $$\& b=3\Rightarrow acFor $b=4\Rightarrow ac$(a,c)\in {(1,1),(1,2),(2,1),(3,1),(1,3)}\Rightarrow 5$$$

For $b=5\Rightarrow ac$\begin{gathered} (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 \end{gathered}$

For $b=6\Rightarrow ac$$$\begin{gathered} (a,c)\in {(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \ (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \ (6,1),(2,4),(4,2)}=16 \end{gathered}$$$

Total cases $=3+5+14+16=38$ cases

$\Rightarrow$ Probability, $p=\frac{38}{216}$

$\Rightarrow 216p=38

Solutions · 1