Question

Determinati numerele reale x si y stiind ca :
0,16x^2+0,8x+y^2-2y+2=0
unde este. de exemplu 0,16x^2. asta este. 0,16x la puterea 2

Answer 1

    
[tex]\displaystyle 0,16x^2+0,8x+y^2-2y+2=0 \\ 0,16x^2+0,8x+y^2-2y+1+1=0 \\ (0,16x^2+0,8x+1)+(y^2-2y+1)=0 \\ ((0,4x)^2+2 \cdot 0,4x \cdot 1+1^2)+(y^2+2 \cdot y \cdot (-1)+(-1)^2)=0 \\ (0,4x+1)^2+(y -1)^2=0 \\ \\ 0,4x+1=0 ~~\Longrightarrow~~x = \frac{-1}{0,4} = - \frac{10}{4}= \boxed{- \frac{5}{2}}~~~(radacina ~~dubla ) \\ \\ y-1 = 0 ~~\Longrightarrow~~ y=\boxed{1}~~~(radacina ~~dubla )[/tex]