Question

Find the value of \[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }}\] up to three decimal places. (Take \[\sqrt 2 = 1.414\] and \[\sqrt 5 = 2.236\]).

Answer 1

Hint:In this problem we have to find the value for \[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }}\] up to three decimal places. Here, the given is to put the values of \[\sqrt 2 = 1.414\] and \[\sqrt 5 = 2.236\], we will get the value of the given problem. We are going to solve this problem by taking common and term out of brackets and substitute the given values.

Complete step-by-step answer:
It is given that; \[\sqrt 2 = 1.414\]and \[\sqrt 5 = 2.236\]
We have to find the value of \[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }}\].
Now consider the given term, \[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }}\]
Taking common of the similar term we get,
\[\dfrac{{\sqrt 5 (\sqrt 2 - 1)}}{{2\sqrt 2 }}\]
Substitute, the values of \[\sqrt 2 = 1.414\] and \[\sqrt 5 = 2.236\]
\[\dfrac{{2.236(1.414 - 1)}}{{2 \times 1.414}}\]
Simplifying we get,
\[0.327335\]
We have to find the value up to three decimal places.
To find the value up to three decimal places, we have to take the value up to four decimal places.
Now, we have \[3\] at the fourth decimal place. Since, \[3\]is less than \[5,\] we will not add anything to the previous digit.
Hence, the value of \[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }}\] is \[0.327\].

Note:The problem can be solved by another way.
Let us find the value of \[\sqrt {10} \].
\[\sqrt {10} = \sqrt 5 \times \sqrt 2 \]
Substitute the values of \[\sqrt 2 = 1.414\]and \[\sqrt 5 = 2.236\], we get
\[ = 1.414 \times 2.236 = 3.161704\]
Now, we will put this the in the given problem we get,
\[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }} = \dfrac{{3.161704 - 2.236}}{{2 \times 1.414}} = 0.3273352\]
Taking up to three decimal places we get, the value of \[\dfrac{{\sqrt {10} - \sqrt 5 }}{{2\sqrt 2 }}\] is \[0.327\].We need to find the approximation up to \[n,\] we will take the digit of \[n + 1{\text{ }}th\] digit. If the digit at \[n + 1{\text{ }}th\] place is greater than equals to \[5\], we will add \[1\] to the digit at \[n\,th\] place. If the digit at \[n + 1{\text{ }}th\] place is less than \[5\], we will not add anything to the digit at \[n{\text{ }}th\] place.
Another approach question can be solved by rationalising i.e multiplying both numerator and denominator $\sqrt2$ and simplifying it we get the same answer.