Question

exercițiile 2 și 3 va rog frumos.

Answer 1

$\displaystyle \mathtt{2.~~~~x^2+3x+1=0;~x_1^2+x_2^2=?}\\ \\ \mathtt{a=1,~b=3,~c=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathbf{S=x_1+x_2=- \frac{b}{a} }}\\ \\ \mathtt{S=- \frac{3}{1}=-3;~P= \frac{1}{1} =1~~~~~~~~~~~~~~~~~~~~~~~~~~\mathbf{P=x_1 \cdot x_2= \frac{c}{a} }}\\ \\ \mathtt{x_1^2+x_2^2=(-3)^2-2 \cdot 1=9-2=7~~~~~~~~~~~~~~\mathbf{x_1^2+x_2^2=S^2-2P} }\\ \\ \mathtt{x_1^2+x_2^2=7}$

$\displaystyle \mathtt{3.~~~~C_n^2-21=0,~n \in \mathbb{N},~n \geq 2}\\ \\ \mathtt{\mathbf{C_n^k= \frac{n!}{(n-k)! \cdot k!} }\Rightarrow C_n^2= \frac{n!}{(n-2)!\cdot 2!} }\\ \\ \mathtt{ \frac{n!}{(n-2)!\cdot 2!}-21=0 }\\ \\ \mathtt{\frac{(n-2)! \cdot (n-1) \cdot n}{(n-2)! \cdot 1 \cdot 2} -21=0}\\ \\ \mathtt{ \frac{(n-1) \cdot n}{2}-21=0 }\\ \\ \mathtt{ \frac{n^2-n}{2}-21=0 }\\ \\ \mathtt{n^2-n-42=0}\\ \\ \mathtt{a=1,~b=-1,~c=-42}\\ \\ \mathtt{\mathbf{\Delta=b^2-4ac}=(-1)^2-4 \cdot 1 \cdot (-42)=1+168=169\ \textgreater \ 0}\\ \\ \mathtt{\mathbf{x_1= \frac{-b- \sqrt{\Delta} }{2a} }= \frac{-(-1)- \sqrt{169} }{2 \cdot1}= \frac{1-13}{2}= \frac{-12}{2}=-6 \not \in \mathbb{N}}\\ \\ \mathtt{\mathbf{x_2= \frac{-b+ \sqrt{\Delta} }{2a} }= \frac{-(-1)+13}{2 \cdot 1} = \frac{1+13}{2}= \frac{14}{2}=7 \in \mathbb{N} }\\ \\ \mathtt{S=\{7\}}$