Square root and Approximations

Square root and Approximations

To know about approximating square roots, first we should learn perfect squares because they are interrelated. Here is a table of perfect squares and square roots.

$\dpi{120} 1^{2}=1$	$\dpi{120} \sqrt{1} =1$
$\dpi{120} 2^{2}=4$	$\dpi{120} \sqrt{4} =2$
$\dpi{120} 3^{2}=9$	$\dpi{120} \sqrt{9} =3$
$\dpi{120} 4^{2}=16$	$\dpi{120} \sqrt{16} =4$
$\dpi{120} 5^{2}= 25$	$\dpi{120} \sqrt{25} =5$
$\dpi{120} 6^{2}= 36$	$\dpi{120} \sqrt{36} =6$
$\dpi{120} 7^{2}= 49$	$\dpi{120} \sqrt{49} =7$
$\dpi{120} 8^{2}= 64$	$\dpi{120} \sqrt{64} =8$
$\dpi{120} 9^{2}= 81$	$\dpi{120} \sqrt{81} =9$
$\dpi{120} 10^{2}=100$	$\dpi{120} \sqrt{100} =10$
$\dpi{120} 11^{2}= 121$	$\dpi{120} \sqrt{121}= 11$
$\dpi{120} 12^{2}=144$	$\dpi{120} \sqrt{144}=12$
$\dpi{120} 13^{2}=169$	$\dpi{120} \sqrt{169} = 13$
$\dpi{120} 14^{2}=196$	$\dpi{120} \sqrt{196}= 14$
$\dpi{120} 15^{2}= 225$	$\dpi{120} \sqrt{225}=15$
$\dpi{120} 16^{2}=256$	$\dpi{120} \sqrt{256}=16$
$\dpi{120} 17^{2}=289$	$\dpi{120} \sqrt{289} = 17$
$\dpi{120} 18^{2}=324$	$\dpi{120} \sqrt{324} = 18$
$\dpi{120} 19^{2}= 361$	$\dpi{120} \sqrt{361} = 19$
$\dpi{120} 20^{2}=400$	$\dpi{120} \sqrt{400} = 20$

This way you can learn the square roots upto 50. Lets discuss the process of approximating square roots now.

There are many ways to approximate the square roots. The one which we are going to discuss, is the easiest and quicker one. Here are the steps:

1)First find the nearest perfect squares, in which the sqrt lie between.

2) Lets assume the sqrt number to be approximated as S, smaller perfect square as N and larger perfect square as N+1 such that

$\dpi{120} N^{2}< S < (N+1)^{2}$

Then the approximation A is given by the following formula.

$\dpi{120} \mathbf{A=N+\left ( \frac{S-N^{2}}{2N+1} \right )}$

Lets work on some examples to understand the process

Example1. Approximate √34 to one decimal place.

Solution: √34 lie between √25 and √36 which are nearest perfect squares so,

$\dpi{120} \sqrt{25}< \sqrt{34}< \sqrt{36}$

$\dpi{120} 5^{2}< 34 < 6^{2}$

Here S= 34 , N=5 , N+1=6

Using formula, $\dpi{120} A=N+\left ( \frac{S-N^{2}}{2N+1} \right )$

$\dpi{120} A=5+\left ( \frac{34-5^{2}}{2*5+1} \right ) = 5 +\frac{9}{11} = 5.8$

Example2. Approximate √70 to one decimal place.

Solution: √70 lie between √64 and √81 which are the nearest perfect squares so,

$\dpi{120} \sqrt{64}< \sqrt{70}< \sqrt{81}$

$\dpi{120} 8^{2}< 70 < 9^{2}$

Here S= 70 , N=8 , N+1=9

Using formula, $\dpi{120} A=N+\left ( \frac{S-N^{2}}{2N+1} \right )$

$\dpi{120} A=8+\left ( \frac{70-8^{2}}{2*8+1} \right ) = 8 +\frac{6}{17} = 8.4$

The method discussed above gives you sqrt approximation upto one decimal place.

Second method: If you are on time constraint For SAT, ACT or any other test and you are given choices then we can fairly guess the right answer by finding the nearest perfect squares and distance of sqrt from the nearest perfect square.

For example we need to approximate √70 which lie between perfect squares 64 and 81.

$\dpi{120} \sqrt{64}< \sqrt{70}< \sqrt{81}$

$\dpi{120} 8< \sqrt{70}< 9$

Distance between 70 and its nearest perfect square 64 is 6 which is less then halfway (72.5) that means √70 would be approximated to a value less than 8.5 but closer to 8.5. Based on this observation we can guess the right answer out of the following options.

a) 7.8

b) 8.6

c) 8.4

d) 8.9

We are pretty sure about the right answer to be 8.4 because Only this option is less than 8.5 and closer to it.

To approximate √74 we choose 8.6 because 74 is more than halfway 72.5 and closer to it.

To approximate √ 79 we choose 8.9 because 79 is more than halfway 72.5 and more closer to 81 so it would be approximated closer to 9.

So when we are on a time constraint and have multiple choices , we can use above estimation process.

First method gives you the exact approximation upto one decimal place , second method also gives a pretty good estimation within seconds.

You can always check your results by using calculators.

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