How many real solutions of this equation $$\sqrt{x}+\sqrt{1-x^2}=\sqrt{2-3 x-4 x^2}?$$ I posted my question at here https://mathematica.stackexchange.com/questions/51316/how-can-i-get-the-exact-real-solution-of-this-equation In my opinion, there is one solution is $\frac{1}{9}\left(\sqrt{34}-5\right).$
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$\begingroup$ you are right. since $0 \le x\le 1$. there is only one positive solution. the other solution is negative. $\endgroup$– mikeJun 23, 2014 at 7:23
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1$\begingroup$ @mike Real solutions doesn't necessarily mean that the range of the square root is real. You could have a real solution that gives two equal complex values. oops sorry my bad eyes, I read "possible" where you said "positive" $\endgroup$– DanielVJun 23, 2014 at 7:25
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1$\begingroup$ If you allow only the real square root function, valid for nonnegative arguments, there is only one solution. But if you allow the square roots of negative numbers, there are more. I think the former is more reasonable, but the crux of the matter is that the problem as stated is ambiguous. $\endgroup$– Harald Hanche-OlsenJun 23, 2014 at 7:27
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1$\begingroup$ It depends on what you mean by $\sqrt\,$ of a negative number. The complex square root is ambiguous because it has branches: You can define $\sqrt{-1}$ either as $i$ or as $-i$ and both are valid extensions of the real square root to negatives. One could also define a weird branch in which $\sqrt{-1}=i$, $\sqrt{-4}=-2i$ and $\sqrt{-3}$ is undefined (the branch cut goes through there). For this reason, usually "the" square root function is defined only on the positives, where it is unambiguous, and extensions are denoted by different notation. $\endgroup$– Yoni RozensheinJun 23, 2014 at 7:33
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$\begingroup$ The nice thing is that this particular problem doesn't depend on the particular branch of the square root chosen. You could define $\sqrt{-1} = -i$ and since the square roots are being added, it won't change the solution set. It's oddly unambiguous despite the ambiguity. $\endgroup$– DanielVJun 23, 2014 at 7:45
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2 Answers
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$$\sqrt{1 - x^2} + \sqrt{x} = \sqrt{-4x^2 - 3x + 2}$$ $$2\sqrt{x}\sqrt{1 - x^2} - x^2 + x + 1 = -4x^2 - 3x + 2$$ $$2\sqrt{x}\sqrt{1 - x^2} = -3x^2 - 4x + 1$$ $$4x - 4x^3 = 9x^4 + 24x^3 + 10x^2 - 8x + 1$$
Has solutions: $$x \in \left\{-\frac{\sqrt{34} + 5}{9}, \frac{\sqrt{34} - 5}{9}, -\sqrt{2} - 1, \sqrt{2} - 1\right\}$$
Checking the four values, 2 are valid:
$$x \in \left\{\frac{\sqrt{34} - 5}{9}, -\sqrt{2} - 1\right\}$$
Note that the second value extends the range of the square root into complex numbers.
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If you expand and eliminate the square roots in the obvious way, you get a quartic which factorises as $$(9x^2+10x-1)(x^2+2x-1)\ .$$ This has four real roots, but they may not all work in your original equation - you will need to check.
