Question

Va rooog pt 10 puncte
$\sqrt{x} + x = y$
scoateti pe x in functie de y

Answer 1

Fie rad x = t ⇒ x=t^2
t^2 +  t  -y = 0
t1,2=(-1 +/- rad(1+4y) /2
x1,2= (1 +/- 2rad(1+4y) + 1+4y) / 2 = 1+2y +/- rad(1+4y)

Answer 2

[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:


$\it \sqrt{x} = \dfrac{-1+\sqrt{1+4y}}{2} \Rightarrow \sqrt{x} = \dfrac{\sqrt{1+4y}-1}{2}$



$\it \Rightarrow (\sqrt{x})^2 = \left(\dfrac{\sqrt{1+4y}-1}{2} \right)^2 \Rightarrow x = \left(\dfrac{\sqrt{1+4y}-1}{2} \right)^2$