[tex]\it \sqrt{x} +x=y
\\ \\
\sqrt{x} \Rightarrow x\geq0 \Rightarrow \sqrt{x} +x \geq 0 \Rightarrow y\geq0[/tex]
[tex]\it Fie\ \sqrt{x} =t,\ (\ t\geq0\ ) \Rightarrow egalitatea\ din\ problem\breve{a}\ devine:
\\ \\
t+t^2=y \Rightarrow t^2+t-y=0
\\ \\ \\
t_{1,2} =\dfrac{-1\pm\sqrt{1+4y}}{2} \Rightarrow \begin{cases}\it t_1=\dfrac{-1-\sqrt{1+4y}}{2} \ \textless \ 0\ (nu\ convine)\\ \\
\it t_2=\dfrac{-1+\sqrt{1+4y}}{2} \ \textgreater \ 0\end{cases}[/tex]
Revenim asupra notației și obținem:
[tex]\it \sqrt{x} = \dfrac{-1+\sqrt{1+4y}}{2} \Rightarrow \sqrt{x} = \dfrac{\sqrt{1+4y}-1}{2}[/tex]
[tex]\it \Rightarrow (\sqrt{x})^2 = \left(\dfrac{\sqrt{1+4y}-1}{2} \right)^2 \Rightarrow x = \left(\dfrac{\sqrt{1+4y}-1}{2} \right)^2 [/tex]
