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I have tried to figure this one out but I just can't seem to get it. I have the answer but I don't understand the method/thinking behind this. Please explain how to:

Find the number of solutions of the equation: $$2\sqrt{3} \arctan\sqrt x =\sqrt{(1 + x)}$$

I'm pretty sure the method doesn't matter but the given solution is here: pdf solution

If someone could explain I would be vey greatfull.

I'm not yet allowed to add images but this is the link generated by THIS SITE

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    its \sqrt{x} in between $ $ – 2017-02-02
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    I'm sorry, when you wrote $\arctan(x)^{1/2}$, did you mean $\sqrt{\arctan(x)}$ or $\arctan(\sqrt x)$? I might've made the wrong call when editing. – 2017-02-02
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    its 2sqrt3 arctan sqrtx – 2017-02-02
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    @Arthur I think you changed the equation when you edited it. I believe Rose is trying to solve $2\sqrt{3}\arctan\sqrt{x}=\sqrt{1+x}$. – 2017-02-02
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    Do you know Calculus? The solution provided in the pdf file uses Calculus... – 2017-02-02
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    @TimThayer I realized that. It's just that when it says `arctan(x)^(1/2)` instead of, say, `arctan(x^(1/2))`, it's difficult to tell. – 2017-02-02
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    I'm still learning calculus but I know it a bit. (This is my university question) The problem is I have basic high school math so this is a big jump for me – 2017-02-02
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    Does the solution contain a closed form or just the numerical value ? – 2017-02-02
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    Just the number of solutions, so in this case it's 2 – 2017-02-02
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    The link to the PDF is somewhat unfriendly: normally, to provide a PDF file, one just has the URL of the file itself rather than a "?download=" query. Trying to view this on my phone, I just see a blank screen. Usually we prefer that the essential parts of any external resource are reproduced in the text of the question; but if that's too much, is there a better link to the file? – 2017-02-02
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    Hmm https://www.mini.pw.edu.pl/~figurny/www/?Dydaktyka:MEIL_Calculus_1 its the "problem #2 solution" at the bottom of the page – 2017-02-02
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    I'm not going to read through an external pdf. Put the relevant parts in the post as text. – 2017-02-02
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    I really don't know what is so scary about a pdf but since I reaaalllly need help I took a screenshot but I'm new to the site so still not allowed to include a photo but there is a link to it. NOT AN EXTERNAL PDF. – 2017-02-02
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    There's nothing scary about a PDF, the problem is with the way the website delivers it. Thousands of web sites deliver PDFs in the most convenient way, but whoever set up _that_ site decided to make other people's lives more difficult than they needed to be. The screenshot works much better. – 2017-02-02
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    Okay, now can anyone explain how to get the solution? Pretty please – 2017-02-02

1 Answers 1

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There are several steps.

First, we notice that $2\sqrt{3} \arctan\sqrt x =\sqrt{1 + x}$ is exactly equivalent to the statement $$(2\sqrt{3} \arctan\sqrt x) - \sqrt{1 + x} = 0.$$ The left side of that equation is a function of $x.$ So if we define $$f(x) = (2\sqrt{3} \arctan\sqrt x) - \sqrt{1 + x},$$ then the solutions of the original equation are just the values of $x$ that satisfy $f(x) = 0.$ That lets us start applying techniques that are generally useful for counting the solutions to equations of the form $f(x) = 0.$

What the answer sheet shown in the question concludes is that $f(0) < 0,$ $f(3) > 0,$ and $f(x) < 0$ for large values of $x$; moreover, $f$ is continuous, so somewhere between $x=0$ and $x=3$ there is at least one solution (the graph of $f(x)$ crosses the $x$-axis), and there is at least one more solution for some $x$ greater than $3.$

In order to show that there is only one solution in each of those regions (so only two solutions altogether), and also to decide that we should look at $f(x)$ when $x=3,$ the answer sheet computes the derivative of $f,$ $$ f'(x) = \frac{\sqrt3}{(1+x)\sqrt x} - \frac{1}{2\sqrt{1+x}}. $$ It then does some algebra to figure out where $f'(x)$ is positive and where it is negative. It determines that $f'(x)$ is positive whenever $(x+4)(x-3) < 0.$ In general, that inequality is equivalent to $-4 < x < 3,$ but since $f$ and $f'$ are not defined for $x < 0,$ we only need to worry about what happens for $x \geq 0$; so the facts we need to know are that $f'(x)>0$ when $0\leq x<3,$ $f'(3) = 0,$ and $f'(x) < 0$ when $x > 3.$ The answer sheet states these facts graphically by presenting a rough sketch of the graph of $f'(x).$ These facts also imply that $f$ will have a maximum value at $x=3,$ so $f(3)$ is a good place to look in order to find a positive value of $f(x).$ And indeed $f(3)> 0.$

So we know that $f(x)=0$ at least once between $x=0$ and $x=3,$ and because $f(x)$ is increasing over that entire interval, it can only have one zero in that interval. That is, we have exactly one solution (no more, no less) for some $x$ in the interval $0 < x < 3.$ We also know that $f(x)=0$ at least once for $x>3,$ and because $f(x)$ is decreasing over that entire interval, it can only have one zero in that interval. So we have exactly one solution for some $x$ such that $x > 3.$ Therefore there are exactly two solutions.

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    Thank you! This was helpful. – 2017-02-04