We need to solve that following equation for $x$ , where , $0<x<1$ ,
$(1-x)^{20}+20x(1-x)^{19}=0.025$ ,
The solution says , we need to use the "iterative methods" to find the value of this $x$ ,
Now my question is , since , $0<x<1$ , can't we just write $(1-x)^{20} \approx (1-20x)$ ??
On applying this to the main equation :
$(1-19x)(1+19x)=0.025$ , => $1-19x^{2}=0.025$ , which gives $x\approx 0.051$
Isn't this the correct approach ? As the given solution is $x=0.249$ ..
Since you are supposed to use an iterative method, let me suppose that you heard about Newton method : starting from guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, let us consider $$f(x)=(1-x)^{20}+20x(1-x)^{19}-0.025$$ After simplification, the derivative is $$f'(x)=-380 (1-x)^{18} x$$ which immediately shows a problem if we start at $x_0=0$.
So, we need another starting point. Even if, as already said in answers and comments, what you used is not very good, at least it gives a number. So, let us try using $x_0=0.05$. Newton method will generate the following estimates $$x_1=0.144187$$ $$x_2=0.195074$$ $$x_3=0.229416$$ $$x_4=0.245540$$ $$x_5=0.248634$$ $$x_6=0.248733$$ which is the solution for six significant figures.
Edit
Let me point out that, in almost the same spirit as in your attempt, using the binomial expansion and stopping at the first terms, you would have found $$f(x)=\frac{39}{40}-190 x^2+\cdots$$ from which a "solution" $$x_0=\frac{1}{20}\sqrt{\frac{39}{19}}\approx 0.072$$ which would have been a little better as a starting point for the iterative process.
You don't have a correct expansion. If you expand your expression properly you get $-361x^2$. But even this is not the correct second order approximation
If you factor as $(1-x)^{19}(1+19x)$ and regard $x$ as small, you get the binomial expansion
$$\left(1-19x+\frac {19\cdot 18}2 x^2-\frac {19\cdot 18 \cdot 17}6 x^3+\dots\right)\left (1+19x\right)$$
And the term in $x^2$ is $(19\cdot9-19^2)x^2=-190x^2$ taking both contributions. This doesn't help the accuracy - others have commented on why this is unlikely with $x$ relatively large - you need more terms of the expansion and a way of extimating the error (crudely look at the size of the first term you've dropped - certainly if that is large, the approximation is likely to be poor).
No you cannot do this. The approximation you mention is only true for very small $x$ (compared to $1$), so $x <<1$. If we take $x$ to be close to one, it is easy to see that $(1-x)^{20}$ is small and positive, while $1-20 x$ is negative and close to $-19$.
$(1-x)^{20} \approx (1-20x)$ is only true for very small values of $x$, not for values on $(0,1)$. For example:
$$(1-0.5)^{20} = 9.5\cdot 10^{-7}$$
and
$$1-20\cdot 0.5 = -9$$
so the numbers are nowhere near close.
