kim4norm1 kim4norm1
  • 24-08-2014
  • Mathematics
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x=square root of 880-8x

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konrad509
konrad509 konrad509
  • 24-08-2014
[tex]x=\sqrt{880-8x}\\ D:880-8x\geq0 \wedge x\geq0\\ D:8x\leq880 \wedge x\geq0\\ D:x\leq 110 \wedge x\geq0\\ D:x\in\langle0,110\rangle\\ x^2=880-8x\\ x^2+8x-880=0\\ x^2+8x+16-896=0\\ (x+4)^2=896\\ x+4=8\sqrt{14} \vee x+4=-8\sqrt{14}\\ x=-4+8\sqrt{14} \vee x=-4-8\sqrt{14}\\ \\ x=-4-8\sqrt{14}\not \in D \Rightarrow \boxed{x=-4+8\sqrt{14}}[/tex]
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