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svetlana [45]
3 years ago
14

Solve for y x=y^2+4y

Mathematics
1 answer:
Lorico [155]3 years ago
4 0
x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\
1.\ \Delta0\\
\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\
y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\
y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\

-----------------------------------------------------

x=y^2+4y\\ y^2+4y-x=0\\
y^2+4y+4-4-x=0\\
(y+2)^2=x+4\\
y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\
y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\

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