Question

Resolva a equação Cx,2=15

Answer 1

Cx,2 = 15

$\frac{x!}{2!.(x-2)!} = \frac{ x.(x-1).(x-2)!}{2!.(x-2)!} = \frac{ x^{2} - x}{2}$

$\frac{ x^{2} - x}{2}=15 =\ \textgreater \ x^{2} -x=30 =\ \textgreater \ x^{2} -x-30$

Δ = b2 - 4.a.c 
Δ = -12 - 4 . 1 . -30 
Δ = 1 - 4. 1 . -30 
Δ = 121

Há 2 raízes reais.

x = (-b +- √Δ)/2a

x' = (--1 + √121)/2.1   
x'' = (--1 - √121)/2.1

x' = 12 / 2   
x'' = -10 / 2

x' = 6   
x'' = -5

Números negativos não convêm, portanto x = 6

Answer 2

$C_{(x,2)}=15\\\\C_{(x,2)}= \dfrac{x!}{2!\cdot(x-2)!} \\\\\\15=\dfrac{x!}{2\cdot(x-2)!}\\\\\\30= \dfrac{x\cdot(x-1)\cdot{(x-2)!}}{(x-2)!}\\\\\\30=x^2-x\\\\x^2-x-30=0\\\\a=1\,\,\,\,\,\,\,\,\,\,b=-1\,\,\,\,\,\,\,\,\,c=-30\\\\\\\triangle=b^2-4ac\\\triangle=(-1)^2-4\cdot(1)\cdot(-30)\\\triangle=1+120\\\triangle=121\\\\\\x=\dfrac{-b \frac{+}{-} \sqrt{\triangle}}{2a}\\\\\\x=\dfrac{-(-1)\frac{+}{-}\sqrt{121}}{2}\\\\\\x= \dfrac{1 \frac{+}{-}11 }{2}\\\\\\x'= \dfrac{1+11}{2}\\\\\\x'=6$

$x"= \dfrac{1-11}{2}\\\\\\x"=-5$

x = -5 não serve como solução para esse caso.

$Logo\\\\\boxed{x=6}$