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How do I solve this equation?
(-4x+12)^1/2 = x

Sagot :

Konrad509idn

[tex](-4x+12)^{1/2}=x\\\sqrt{-4x+12}=x\\\\D:-4x+12\geq0\wedge x\geq0\\D:4x\leq12\wedge x\geq0\\D:x\leq3\wedge x\geq0\\D:x\in\langle0,3\rangle\\\\\sqrt{-4x+12}=x\\-4x+12=x^2\\x^2+4x-12=0\\x^2-2x+6x-12=0\\x(x-2)+6(x-2)=0\\(x+6)(x-2)=0\\x=-6 \vee x=2\\\\-6\not \in D\\\\\boxed{x=2}[/tex]

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