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Given equation: $$\sqrt{4 - x + 4\sqrt{-x}} = 4 - \sqrt{4 - x - 4\sqrt{-x}},$$ how can one prove, that this equation is true only for $x \in [-4, 0]$?

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3 Answers 3

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We can set $t:=\sqrt{-x}$ (so that $t\ge0$) and get

$$\sqrt{t^2+4t+4}=4-\sqrt{t^2-4t+4},$$

$$t+2=4-|t-2|,$$

$$t-2=-|t-2|,$$

$$t-2\le0.$$ Hence $$0\le t\le2,$$ $$-4\le x\le0.$$

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squaring the equation $$\sqrt{4-x+4\sqrt{-x}}+\sqrt{4-x-4\sqrt{-x}}=4$$ we obtain $$4-x+4\sqrt{-x}+4-x-4\sqrt{-x}+2\sqrt{(4-x)^2-16(-x)}=16$$ simplifying we get $$8-2x+2\sqrt{(x+4)^2}=16$$ can you proceed?

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  • $\begingroup$ It will simplify to the $x + 4$ = $|x + 4|$. For $x \geq -4$ we got terms on both sides canceled, so it means that equation is true $\forall x$ in this interval. However, $x$ has to be lesser than $0$, because so is required by the term $\sqrt{-x}$. Henceforth, solution is interval $[-4, 0]$. I am right? :-) $\endgroup$ Aug 20, 2016 at 19:41
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Put $y = -x \implies \sqrt{4+y+4\sqrt{y}} = 4 - \sqrt{4+y-4\sqrt{y}}\implies 2+\sqrt{y} = 4 - |2-\sqrt{y}|=2+\sqrt{y}$ , and this true for all $y \ge 0$. Thus is true for all $x \le 0$, and $-4 \le x$. The other case which leads to $2+\sqrt{y} = 4 - (\sqrt{y}-2)\implies y = 4\implies x = -4$, and this is part of the solutions covered in the first case.

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  • $\begingroup$ After the second $\;\implies\;$ , how did you get that equality? How did you square the right side? $\endgroup$
    – DonAntonio
    Aug 20, 2016 at 19:33
  • $\begingroup$ Still I can't see it...I understand $\;4+y\pm4\sqrt y=(2\pm \sqrt y)^2\;$ , but how do you dispose of the sum on the right hand? $\endgroup$
    – DonAntonio
    Aug 20, 2016 at 19:36
  • $\begingroup$ You repeated $2+\sqrt{y}$ on both sides. $2+\sqrt{y} = 4 - |2-\sqrt{y}|=2+\sqrt{y}$. $\endgroup$
    – user236182
    Aug 20, 2016 at 19:45

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