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$$\sqrt{x+3}+\sqrt{x-1} = -1;\tag{1}$$ $$ \mathbb{D}=[1,\infty)\tag{2}$$ $$x+3=(-1-\sqrt{x-1})^2\tag{3}$$ $$x+3=1+2\sqrt{x-1}+x-1\tag{4}$$ $$x_{1/2} = \pm\sqrt{\frac 3 2}+1\tag{5}$$ $$\mathbb{D} -> x=\sqrt{\frac 3 2}+1\tag{6} $$ According to the solution there shouldn't be any value for x.

Where is my mistake?

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2 Answers 2

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If you look eqution (1), you see that left hand of equation is always positive, thus there isn't no solution for this equation.

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When you square both sides, going from (1) to (3), you bring in extra solutions that aren't right for equation (1).

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