Question

Solve the following equation:
$12C_1^x+C_2^{x+4}=162,xϵN$

Answer 1

$12\frac{x!}{(x-1)!}+\frac{(x+4)!}{2!(x+4-2)!}=162$

$12\left(x\right)+\frac{(x+4)!}{2\times (x+2)!}=162$

$2\left(12x\right)+(x+3)(x+4)=162\times 2$

$24x+x^2+7x+12=324$

$x^2+31x-312=0$

$x=\frac{-31\pm \sqrt{961+1248}}{2}$

$=\frac{-31\pm \sqrt{2209}}{2}$

$=\frac{-31\pm 47}{2}$

$=\frac{-31+47}{2}=8$

$=\frac{-31-47}{2}=-39$

Since $x$ have only positive value as $C$ is not defined for negative value.

$x=8$