Question

Trebuie sa sa rezolv ecuatia

Answer 1

$\displaystyle \mathtt{3 \cdot C_n^2-4 \cdot C_n^1=10}\\ \\\mathtt{\left\{ \begin{array}{ccc}\mathtt{n \in N}\\\mathtt{n \geq 1}\\\mathtt{n \geq 2}\end{array}\right \Rightarrow x \in \{2,~3,~4,~5,...\}=D} \\ \\ \mathtt{\mathbf{C_n^k= \frac{n!}{(n-k)! \cdot k!} }\Rightarrow C_n^2= \frac{n!}{(n-2)! \cdot 2!} }\\ \\ \mathtt{\mathbf{C_n^1=n}}$
$\displaystyle \mathtt{3 \cdot \frac{n!}{(n-2)! \cdot 2!}-4 n=10 }\\ \\ \mathtt{3 \cdot \frac{(n-2)! \cdot (n-1) \cdot n}{(n-2)! \cdot 2}-4 n=10 }\\ \\ \mathtt{3 \cdot \frac{(n-1) \cdot n}{2}-4 n=10 }\\ \\ \mathtt{3 \cdot \frac{n^2-n}{2} -4n=10}\\ \\ \mathtt{ \frac{3n^2-3n}{2}-4n=10 }$

$\displaystyle \mathtt{3n^2-3n-8n=20}\\ \\ \mathtt{3n^2-3n-8n-20=0}\\ \\ \mathtt{3n^2-11n-20=0}\\ \\ \mathtt{a=3,~b=-11,~c=-20}\\ \\ \mathtt{\Delta=b^2-4ac=(-11)^2-4 \cdot 3 \cdot (-20)=121+240=361\ \textgreater \ 0}\\ \\ \mathtt{x_1= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-11)- \sqrt{361} }{2 \cdot 3}= \frac{11-19}{6}=- \frac{8}{6} =- \frac{4}{3}\not \in D }\\ \\ \mathtt{x_2= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-11)+ \sqrt{361} }{2 \cdot 3}= \frac{11+19}{6}= \frac{30}{6}=5 \in D }\\ \\ \mathtt{S=\{5\}}$