Finding the root $\sqrt3$ of $f(x)=(x^2-2)(x^2-3)(x^2-5)$ with halving technique Assume that the function $f(x)=(x^2-2)(x^2-3)(x^2-5)$ is given.
Question :  
Find an interval $I$ such that $\sqrt2,\sqrt3,\sqrt5 \in I$ and also, when you apply halving technique on this interval, it finds $\sqrt3$ as a root of $f$. 
What is halving technique?
It says choose an interval like $[0,3]$ ( for example )
$f(3) \gt 0$ and $f(0) \lt 0$ and $f(\frac{0+3}{2})=f(1.5) \gt 0$ ... So a root exists in the interval $[0,1.5]$  
I tried to find an interval which has all the roots but only finds root $\sqrt3$ when using halving technique.  But every interval i used , found other roots ( $\sqrt5$ and $\sqrt2$ ) . 
 A: You need an interval that isolates $\sqrt 3$.One such interval is $[1.5, 2]$.
If you insist on starting with a interval that contains all three positive roots, then it is very unlikely that bisection will find $\sqrt 3$, as this paper argues:

George Corliss, "Which root does the bisection algorithm find?", SIAM Review, 19 (2): 325–327 (1977), doi:10.1137/1019044

It says that there is zero probability of finding even-numbered roots.
The bracketing intervals for your $f$ are $[a,b]$ with $f(a)<0$ and $f(b)>0$ and so $b>\sqrt 5$ and $a \in (-\sqrt5, -\sqrt 3) \cup (-\sqrt 2, \sqrt 2)$. In these intervals, $\sqrt 3$ is the fourth or the second root.

The only way you'll find $\sqrt 3$ is if you land on it during bisection. This will happen in the first iteration if you start with $[-2,2\sqrt3+2]$, but that requires you to know $\sqrt 3$...
A: Unless you start with an interval such as $[0,2\sqrt3]$, for which the halving technique (eventually) lands exactly on $\sqrt3$, it cannot be done.  That is, if $f(x_n)\not=0$ for all $x_n$ in your sequence of halving-technique midpoints $x_1,x_2,x_3,\ldots$, where $x_1={a+b\over2}$ with $a\lt\sqrt2$, $b\gt\sqrt5$, and $f(a)f(b)\lt0$, then $x_n$ cannot converge to $\sqrt3$.
The reason why is that the halving technique preserves the positive/negative "orientation" of its initial interval.  That is, the sign of the function evaluated at the left hand endpoint never changes, and likewise for the right hand endpoint.  So if your initial interval has $b\gt\sqrt5$, for which $f(b)$ is positive, all your halving-technique intervals will have right hand endpoints with a positive function value.  But in order to converge on the root $\sqrt3$, that orientation would need to reverse, which is not possible.  In fact, if you start with $b\gt\sqrt5$, the halving technique can only converge to $\sqrt5$, $\sqrt2$, or $-\sqrt3$, which are the roots where the function changes from negative to positive.
