Question

Determinati x apartine R astfel incat / x 1 1 /
1 x 1
1 1 2 = 12

Answer 1

$\displaystyle \mathtt{\left|\begin{array}{ccc}\mathtt x&\mathtt1&\mathtt1\\\mathtt1&\mathtt x&\mathtt1\\\mathtt1&\mathtt1&\mathtt2\end{array}\right|=12~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \in \mathbb{R}}$

$\displaystyle\mathtt{ \left|\begin{array}{ccc}\mathtt x&\mathtt1&\mathtt1\\\mathtt1&\mathtt x&\mathtt1\\\mathtt1&\mathtt1&\mathtt2\end{array}\right|=x \cdot x \cdot 2+1 \cdot 1 \cdot 1+1 \cdot 1 \cdot 1-1 \cdot x \cdot 1-1 \cdot 1 \cdot 2-x \cdot 1 \cdot 1=}\\ \\ \mathtt{=2x^2+1+1-x-2-x=2x^2-2x}$

$\displaystyle \mathtt{2x^2-2x=12}\\ \\ \mathtt{2x^2-2x-12=0|:2}\\ \\ \mathtt{x^2-x-6=0}\\ \\ \mathtt{a=1,~b=-1,~c=-6}\\ \\ \mathtt{\Delta=b^2-4ac=(-1)^2-4 \cdot 1 \cdot (-6)=1+24=25\ \textgreater \ 0}\\ \\ \mathtt{x_1= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-1)- \sqrt{25} }{2 \cdot 1}= \frac{1-5}{2}= \frac{-4}{2}=-2 \in \mathbb{R}}\\ \\ \mathtt{x_2= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-1)+ \sqrt{25} }{2 \cdot 1}= \frac{1+5}{2}= \frac{6}{2}=3 \in \mathbb{R} }\\ \\ \mathtt{x_1=-2;~x_2=3}$