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Solve the equation $$\sqrt{x+1}+\sqrt{2-x}+2=x^2+2x$$

I tried with squaring both sides, but I got confused, any help will be thankful.

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  • $\begingroup$ Why does this have the abstract-algebra tag? $\endgroup$ – Samir Khan Aug 12 '15 at 0:56
  • $\begingroup$ @SamirKhan Sorry didn't see it clear $\endgroup$ – Rnas Aug 12 '15 at 0:59
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You'll want just the terms with square roots on one side before you square both sides: $$\sqrt{x+1}+\sqrt{2-x}=x^2+2x -2$$

Squaring, then simplifying so that again only the terms with square roots are on the left, $$ (x+1) + 2\sqrt{x+1}\sqrt{2-x} + (2-x) = x^4+4x^3-8x+4\\ 2\sqrt{x+1}\sqrt{2-x} = x^4+4x^3-8x+1\\ $$

Squaring again, $$ 4(x+1)(2-x) = x^8+8x^7+16x^6-16x^5-62x^4+8x^3+64x^2-16x+1\\ x^8+8x^7+16x^6-16x^5-62x^4+8x^3+68x^2-20x-7=0 $$

At this point I would throw my hands in the air and ask Wolfram Alpha, which tells me that there are four solutions; checking them in the original system, only one of them works: $x \approx 1.31295$.

What is the context of this problem? It would be extremely difficult to solve this by hand.

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