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$$(x + 1)^{1/2} =\frac{1}{(x − 2)^2}$$

Ok, I know how to show how an equation has at least $1$ real roots or exactly $1$ real roots, but for this equation, I know there is indeed at least $34 real roots. But I don't know, how to show it. This are my solutions so far:

Let $f(x)$ be the above equations.

$$f(0) = 3 , \ f(2) = -1$$

Since $f(x)$ is continuous from $[0 , 2]$, by intermediate value theorem, $f(c) = 0$ for some $c\in (0,2)$.

$f(x)$ has at least $1$ real roots.

Suppose that $f$ has at least two real roots $c_1 , c_2 \in \Bbb{R}$ , where $c_1 < c_2 $, i.e . $f(c_1) = 0$ , $f(c_2)=0 \Rightarrow f(c_1) = f(c_2)$

Since $f(x)$ is continous on $[c_1,c_2]$ and differentiable on $(c_1,c_2)$, by Rolle's theorem, there exists $d \in (c_1,c_2)$ such that $f'(d) = 0$. I go on to show that there exists value of $x$ such that $f'(d) = 0$. Hence I conclude that there are at least $2$ real roots.

How do I go on from here and show that there are at least $3$ real roots?

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For $\;x\ge-1\;$:

$$\sqrt{x+1}=\frac1{(x-2)^2}\iff f(x):=(x-2)^4(x+1)-1=0$$

And

$$f(-1)<0)\;,\;\;f(0)>0\;,\;\;f(2)<0\;,\;f(3)>0$$

Thus you have zeros at least in $\;(-1,0)\;,\;\;(0,2)\;,\;\;(2,3)\;$

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    $\begingroup$ You are perfectly correct, indeed. The problem is that squaring introduces other roots; fortunately, they are complex. Cheers :-) $\endgroup$ – Claude Leibovici Sep 27 '14 at 15:35
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Hint

If the function is $$f(x)=\sqrt{x+1}-\frac{1}{(x-2)^2}$$ you noticed the discontinuity at $x=2$ and the fact that the function is only defined if $x \geq -1$. You can notice that $f(-1)=-\frac{1}{9} <0$, $f(0)=\frac{3}{4}>0$,$f(3)=1 >0$; moreover, $$f(2+\epsilon)=\sqrt{3+\epsilon }-\frac{1}{\epsilon ^2} <0 $$ $$f(2-\epsilon)=\sqrt{3-\epsilon }-\frac{1}{\epsilon ^2}<0$$ So, you have three roots (at least).

For sure, you could build a function after squaring the pieces; this leads, as Timbuc showed, to a pentic equation $$g(x)=x^5-7 x^4+16 x^3-8 x^2-16 x+15$$ which has the same roots plus two extra (which are complex). You must know that this can be a consequence of that.

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  • $\begingroup$ By squaring both side, you can have a new function that has the same real roots as the old function , where the new function is continuous at x = 2, but why is it the new function holds only for x more than or equals to 1? I understand that holds for the old function but not for the new function. $\endgroup$ – Joe Sep 27 '14 at 15:45
  • $\begingroup$ This is exactly the problem : squaring, removing the fractions, $\cdots$ can change the domain where the function is defined. This is why you must be very careful; the problem is not the same with $f(x)$ and $g(x)$; the only thing they have in common is that three of their roots are the same. $\endgroup$ – Claude Leibovici Sep 27 '14 at 15:48
  • $\begingroup$ Hi claude! Thanks for replying! I understand squaring the function will change the domain as well as introduce new roots, which in this case luckily it is complex roots. So meaning to say, if the old function is defined x more than or equals to -1, then the new function (squared) will also be defined from x more than or equals to -1? Why is that so? For the pentic equation, i have calculated that at x = -2, g(x) = -257. (Sorry edit : its x more than or equals to -1) $\endgroup$ – Joe Sep 27 '14 at 15:53
  • $\begingroup$ Whatever you do with $f(x)$ or any of its possible transform, you must remember that the domain starts at $x=-1$. Don't look behind ! Cheers :-) $\endgroup$ – Claude Leibovici Sep 27 '14 at 16:02

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