Question

exercitiul 3 , va rog urgent . dau coroana

Answer 1

$\displaystyle \mathtt{2- \sqrt{2-x}=x}\\ \\ \mathtt{- \sqrt{2-x} =x-2}\\ \\ \mathtt{\left(- \sqrt{2-x}\right)^2=(x-2)^2 }\\ \\ \mathtt{2-x=x^2-4x+4} \\ \\ \mathtt{-x^2+4x-x-4+2=0}\\ \\ \mathtt{-x^2+3x-2=0}\\ \\ \mathtt{x^2-3x+2=0}\\ \\ \mathtt{a=1,~b=-3,~c=2}$

$\displaystyle \mathtt{x_1= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-3)- \sqrt{1} }{2 \cdot 1}= \frac{3-1}{2}= \frac{2}{2}=1 }\\ \\ \mathtt{ x_2= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-3)+ \sqrt{1} }{2 \cdot 1}= \frac{3+1}{2}= \frac{4}{2}=2 } \\ \\ \mathtt{S=\{1;2\}}$