$$(x + 1)^{1/2} =\frac{1}{(x − 2)^2}$$
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?
Best Answer
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 $\;(-1,0)\;,\;\;(0,2)\;,\;\;(2,3)\;$