Question

Solve for y x=y^2+4y

Answer 1

`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=-(4)/(2\cdot1)=-2\\\\ 3.\ \Delta>0\\ √(\Delta)=√(16+4x)=√(4(4+x))=2√(x+4)\\ y_1=(-4-2√(x+4))/(2\cdot1)=-2-√(x+4)\\ y_2=(-4+2√(x+4))/(2\cdot1)=-2+√(x+4)\\`

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`x=y^2+4y\\ y^2+4y-x=0\\ y^2+4y+4-4-x=0\\ (y+2)^2=x+4\\ y+2=√(x+4) \vee y+2=-√(x+4)\\ y=-2+√(x+4) \vee y=-2-√(x+4)\\`