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:
- $1-x^2 + 3+x^2 = 4$
- $4 = 4$
How do I isolate that darn unknown?
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:
- $1-x^2 + 3+x^2 = 4$
- $4 = 4$
How do I isolate that darn unknown?
$(a+b)^2\neq a^2+b^2$ 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 $x=\pm1$.
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