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I have some difficulties with this problem:

Solve the equation: $4 \sqrt{1-x} = x+6-3\sqrt{1-x^2}+5\sqrt{1+x}$

I tried to let's $\sqrt{1-x} = a$ and $\sqrt{1+x}=b$ then try to solve equations but it seems difficult.

Can anyone help me deal with this problem or recommend any idea? Thank you

Jnote
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  • Since repeated squaring and rearranging will probably lead to a high degree polynomial anyway , I think, it is best to just solve this numerically. – Peter Feb 13 '23 at 17:48
  • Please edit your question by typing in the attempts you make until the point you get stuck, otherwise your question will be closed. Also, any source for this problem, any previous problems you've seen before that look similar? What if I suggested a trigonometric substitution like $x = \sin 2 \theta$ or something, is that allowed in your syllabus etc.? – Sarvesh Ravichandran Iyer Feb 13 '23 at 17:49
  • @Jnote, the only real solution is $x=-\dfrac{\sqrt3}2$. – Angelo Feb 13 '23 at 18:10
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    For what it's worth, after reading dezdichado's answer (which I also upvoted), it is unclear to me whether his answer represents the problem composer's intent. My first try would have been $$\cos(2\theta) = x \implies \sqrt{1+x} = \sqrt{2} ~|\cos(\theta)|, ~\sqrt{1-x} = \sqrt{2} ~|\sin(\theta)|.$$ Note that the only real solution is $\cos(2\pi/3).$ However, I was stymied in the middle of this try, so the problem composer's intent seems unclear to me. – user2661923 Feb 13 '23 at 18:32
  • Correction to last comment. The only real solution is $\cos(5\pi/6).$ – user2661923 Feb 14 '23 at 15:59

2 Answers2

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Just so you know, your approach can be completed: $$4a + 3ab = b^2+5b+5\implies a^2(4+3b)^2 = (b^2+5b+5)^2$$ and $b^2+a^2 = 2$ and therefore: $$(2-b^2)(4+3b)^2 = (b^2+5b+5)^2.$$

If you use WA, then it factors as: $$(b+1)(5b+7)(2b^2+2b-1) = 0$$ and since $b = \sqrt{1+x}\geq 0,$ the only real solution you will get is: $$x = b^2-1 = \left(\dfrac{\sqrt{3}-1}{2}\right)^2-1 = -\dfrac{\sqrt{3}}{2}.$$

dezdichado
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  • "If you use WA" : what is WA? – user2661923 Feb 13 '23 at 18:14
  • @user2661923 wolfram alpha or any other engine. Also, it's just degree $4$ equation an avid student can just figure out if it factors or not by rational root theorem etc. – dezdichado Feb 13 '23 at 18:19
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    Personally, I consider wolfram alpha somewhat out of bounds for this sort of problem. Problem composers don't normally test their student's facility with a Math engine. However, I do agree that the 4th degree equation does have two rational roots, which implies that the problem can be reduced to attacking a quadratic, using a reasonable approach. +1 : Very nice answer + explanatory comment. – user2661923 Feb 13 '23 at 18:23
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    See also the comment that I left, following the posted question. – user2661923 Feb 13 '23 at 18:37
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I multiplied the values out by hand and got $100x^4 -96x^3 -75x^2 + 72x$ with two visually obvious linear factors namely $x$ and $25x-24$ leaving $4x^2-3$. Easy to check only one root works.

What is the origin of this question? Anything interesting?

Laska
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