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Solve for y x=y^2+4y

Sagot :

konrad509
[tex] x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\ 1.\ \Delta<0 \Rightarrow y\in\emptyset\\\\ 2. \ \Delta=0\\ y=-\frac{4}{2\cdot1}=-2\\\\ 3.\ \Delta>0\\ \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}\\[/tex]

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[tex]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}\\[/tex]

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