Question

How do you find a one-decimal place approximation for $\sqrt {37} $?

Answer 1

Hint: Here we will try to simplify the given square root and then find the approximate value of the root by using the Babylonian method. After doing some simplification, round-off the number to 1 decimal. The result we get will be the required answer.

Complete step-by-step solution:
For the given question, we can write it as,
$ \Rightarrow \sqrt {37} $
Now we will find the rational approximations for $\sqrt {37} $ using the Babylonian method as follows:
We know that given a rational approximation in the form $\dfrac{p}{q}$ to $\sqrt n $ , we can find a better-improved approximation by calculating the value of $\dfrac{{{p^2} + n{q^2}}}{{2pq}}$.
Now, in our given problem the number 37 is close to the number 36 which has a square root value of 6 therefore for the first approximation we will use $\dfrac{p}{q}$ as $\dfrac{6}{1}$.
Here $n = 37$.
Now the next approximation can be calculated as:
$ \Rightarrow \dfrac{{{6^2} + 37 \times {1^2}}}{{2 \times 6 \times 1}}$
On simplifying we get:
$ \Rightarrow \dfrac{{36 + 37}}{{12}}$
This can be written as:
$ \Rightarrow \dfrac{{73}}{{12}}$
Now to get a more accurate answer, we will use $\dfrac{p}{q}$ as $\dfrac{{73}}{{12}}$ in the next approximation.
Now the next approximation can be calculated as:
$ \Rightarrow \dfrac{{{{73}^2} + 37 \times {{12}^2}}}{{2 \times 73 \times 12}}$
On simplifying we get:
$ \Rightarrow \dfrac{{5329 + 5328}}{{1752}}$
This can be written as:
$ \Rightarrow \dfrac{{10657}}{{1752}}$
Divide numerator by denominator,
$ \Rightarrow 6.083$
Now round-off the number to 1 decimal,
$\therefore 6.1$

Hence, the one-decimal place approximation for $\sqrt {37} $ is 6.1.

Note: It is to be remembered that the solution from the Babylonian method is an approximate value of the square root and not the actual value.
The actual answer might vary from the approximate value when calculated using a calculator.
The square root of a number can also be represented as a power in the form $\sqrt x $.