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This problem

$\sqrt{1-x^2} + \sqrt{3+x^2} = 2$

has the solution $x = 1$ and $x = -1$.

However, I always get stuck like this:

  1. $1-x^2 + 3+x^2 = 4$
  2. $4 = 4$

How do I isolate that darn unknown?

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I believe it's the right time to search for your disappeared $2\sqrt{1-x^2}\sqrt{3+x^2}$ term. – Sangchul Lee May 25 '12 at 22:23
up vote 4 down vote accepted

$(a+b)^2\neq a^2+b^2$ in general. In your case $\sqrt{1-x^2}+\sqrt{3+x^2}=2$ leads to $1-x^2+2\sqrt{1-x^2}\sqrt{3+x^2}+3+x^2=4$, which gives you $2\sqrt{1-x^2}\sqrt{3+x^2}=0$. Since $\sqrt{3+x^2}\neq0$ for all real $x$, so $\sqrt{1-x^2}=0$. Hence $x=\pm1$.

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Hadn't realized that particular rule about squaring. Thanks. – Miroslav Cetojevic May 26 '12 at 18:05

Square both sides of original equation: $$1-x^2+3+x^2+2\sqrt{(1-x^2)(3+x^2)}=4\Longrightarrow\sqrt{(1-x^2)(3+x^2)}=0\Longrightarrow x =\pm1$$ since $\,3+x^2=0\,$has no real solutions

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