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Solving by completing the square
X square+4x-4=0


Sagot :

[tex]x^2+4x-4=0 \\ x^2+4x+4-4-4=0 \\ (x+2)^2-8=0 \\ (x+2)^2=8 \\ x+2=\pm \sqrt{8} \\ x+2=\pm \sqrt{4 \times 2} \\ x+2=\pm2\sqrt{2} \\ x+2=-2\sqrt{2} \ \lor \ x+2=2\sqrt{2} \\ x=-2\sqrt{2}-2 \ \lor \ x=2\sqrt{2}-2 \\ \boxed{x=-2\sqrt{2}-2 \hbox{ or } x=2\sqrt{2}-2}[/tex]
[tex]x^2+4x-4=0 \\ x^2+4x+4-8=0\\ (x+2)^2=8\\ x+2=-\sqrt8 \vee x+2=\sqrt8\\ x=-2-2\sqrt2 \vee x=-2+2\sqrt2[/tex]
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