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Putting the equation $x^2 - x \sin(x) - \cos (x)$ into Wolfram Alpha, I am surprised that it has a nice parabolic shape. Also, it has two complex roots.

Question

Is it possible to tell, in a simple way, that it has no real roots?

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    But it *does* have two real roots! And obviously so, since the expression is $-1$ when $x=0$, while it is approximately $x^2$ when $|x|$ is large.2012-09-17
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    It may be nice, but it's not as parabolic as a parabola :)2012-09-17
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    @rschwieb OP could have used the word "paraboloid", if mathematicians of the past hadn't already squandered that word on something else.2012-09-17
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    This is not the first time I have seen W|$\alpha$ produce a completely insane result.2012-09-17
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    @MJD *grin* at your first post and for the second post: What insane result? I put that into WA and it clearly showed two roots. I think the poster must have mistyped something.2012-09-17
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    @rschwieb: from the posted link: "Result: (no real solutions)"2012-09-17
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    @BenMillwood Whoa I totally overlooked the link. That *is* nuts! [This is what I saw.](http://www.wolframalpha.com/input/?i=x%5E2-x*sin%28x%29-cos%28x%29) Note the real roots are listed.2012-09-17
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    Oh wow I totally got trolled by Wolfram Alpha! Thanks guys, I should have checked. -___-2012-09-17

3 Answers 3

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$f(x) = x^2-x\sin x-\cos x$ is even, $f(0) < 0$ and $\lim_{x \to \infty} f(x) = \infty$, hence the equation has at least two real roots. Also, $f'(x) = x(2-\cos x)$ satisfies $f'(x) \geq x \geq 0$ for $x \geq 0$, hence these are the only real roots.

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$f(0)=-1$ and $f(2)=4-2\sin(2)-\cos(2)\geq 4-2-1=1 \,.$

Then, by IVT it has a root between $0$ and $2$. Similarly, it has a second root between $-2$ and $0$.

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    Doh...I should have known to do this. :S Well, thanks!2012-09-17
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I think you are confused by treating this as a standard quadratic which can only have two real roots or a pair of conjugate complex roots. But this is not. As others commented it is more appropriate to use calculus to detect the distribution of roots on the real line. In the complex case your equation become $$z^{2}-z\frac{e^{iz}-e^{-iz}}{2}-\frac{e^{iz}+e^{-iz}}{2}$$ this equation is trancendatal and probably do not have easy solutions.

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    I see. Thanks for the comment! +1 :)2012-09-17