estimate the square root of 47

One way might be as follows:

The nearest square to 47 is 49; and 47 = 49 - 2 or 47 = 49*(1-2/49)

Now take the square root of this:

[49*(1-2/49)]^1/2 = 49^1/2*(1-2/49)^1/2 = 7*(1-2/49)^1/2.

Perform a binomial expansion of the term in brackets:  (1-2/49)^1/2 = 1 - (1/2)*2/49 + higher order terms.

The higher order terms get smaller and smaller, so drop them to get (1-2/49)^1/2 ≈ 1 - 1/49 = 48/49

Then 47^1/2 ≈ 7*48/49 

$${\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{48}}}{{\mathtt{49}}}} = {\frac{{\mathtt{48}}}{{\mathtt{7}}}} = {\mathtt{6.857\: \!142\: \!857\: \!142\: \!857\: \!1}}$$

or 47^1/2 ≈ 6.86

Compare this with:

$${\sqrt{{\mathtt{47}}}} = {\mathtt{6.855\: \!654\: \!600\: \!401\: \!044\: \!1}}$$

The approximate value above agrees to 2 decimal places.

Alan can it be more simple!even im stuck in this question badly!

Rosala....think of two perfect squares......one that is less than 47 and one that's greater than 47. The one that is less than 47 is 36, and the one that isgreater than 47 is 49. So, notice that √36 = 6 and √49 = 7. So, it seems as though the √47must lie between 6 and 7.....!!!

And,as Alan has shown, this is indeed true...!!!

Does that help????

I guess it's the binomial expansion that is giving you problems rosala.

Another way of approaching this is to use the following, iterative formula for square roots (I'll just present it without deriving it):

If we have x² = N then

x_n+1 = (1/2)(x_n + N/x_n)

where n is iteration number.

So, if N = 47 and we start with, say, x₀ = 6

then x₁ = (1/2)(6 +47/6) = 6.917  which is a first approximation

       x₂ = (1/2)(6.917 + 47/6.917) = 6.856 which is a better approximation.

If we start with x₀ = 7

then x₁ = (1/2)(7 + 47/7) = 6.857 which is a better first approximation.

Of course, this depends on your belief in the iterative formula x_n+1 = (1/2)(x_n + N/x_n).  I have just stated it rather than deriving it because the derivation involves calculus, which you probably haven't done yet (actually, there might be a non-calculus derivation, but I can't think of it off-hand!).

I don't know if this helps any! 

Also, the level of precision of the result is not specified in the question, so it could be that just 7 is an acceptable approximation!