Question

Find the square root of 3 and also correct it to four decimal places.

Answer 1

Hint: In this question, we are given the number and we have to find its square root. Therefore, we can solve this question by the long division method to find the square root. Also, as we have to find it correct to four decimal places, we should find it up to 5 decimal places and then round it to four decimal places.

Complete step-by-step solution -
As we have to find the square root correct up to 4 decimal places, we have to write 3 as 3.0000000000 i.e. 10 zeros after the decimal place, as we take pairs of digits while calculating the square root.
In the long division method of finding the square root, we have to first find the square of a number which is closest to 3. We know that ${{1}^{2}}=1<3$ and ${{2}^{2}}=4>3$, therefore we should divide 3 by 1 first to obtain
$1\overset{1}{\overline{\left){\begin{align}
  & 3.0000000000 \\
 & -1 \\
 & 2.00 \\
\end{align}}\right.}}$
Thus, the digit before the decimal point in $\sqrt{3}$ should be 1……………………….(1.1)
We get the remainder as 2 and have taken down the next two digits after the decimal place. Now, we should double the divisor in the first step that is take the divisor as 2 and try to find out a number of the form 2x such that $2x\times x$ is lower than but closest to 200.
As $27\times 7=189$ and $28\times 8=224>200$, we should divide the remainder in the first step i.e. 200 by 27 to get
$27\overset{7}{\overline{\left){\begin{align}
  & \text{ }200 \\
 & -189 \\
 & \text{ 11} \\
\end{align}}\right.}}$
Thus, the digit at the first place after the decimal point in $\sqrt{3}$ should be 7……………………….(1.2)
Similarly, now the tens place in the new divisor should be of the form $20+2\times 7=34$ i.e. the new divisor should be of the form 34x such that $34x\times x$ is lower than but closest to 1100. Thus, we get
$343\overset{3}{\overline{\left){\begin{align}
  & \text{ 1100} \\
 & -1029 \\
 & \text{ 71} \\
\end{align}}\right.}}$
Thus, the digit at the second place after the decimal point in $\sqrt{3}$ should be 3……………………….(1.3)
Similarly, now the place in the new divisor before the units place should be of the form $340+2\times 3=346$ i.e. the new divisor should be of the form 346x such that $346x\times x$ is lower than but closest to 7100. Thus, we obtain
$3462\overset{2}{\overline{\left){\begin{align}
  & \text{ 7100} \\
 & -6924 \\
 & \text{ 176} \\
\end{align}}\right.}}$
Thus, the digit at the third place after the decimal point in $\sqrt{3}$ should be 2……………………….(1.4)
Now, the place in the new divisor before the units place should be of the form $3462+2\times 2=3466$ i.e. the new divisor should be of the form 3466x such that $3466x\times x$ is lower than but closest to 17600. However, as 34660 is greater than 17600, the only possible value of such x would be 0. Thus, we obtain
$34660\overset{0}{\overline{\left){\begin{align}
  & \text{ 17600} \\
 & \text{ }-0 \\
 & \text{ 17600} \\
\end{align}}\right.}}$
Thus, the digit at the fourth place after the decimal point in $\sqrt{3}$ should be 0……………………….(1.5)
Now, the place in the new divisor before the units place should be of the form $34660+2\times 0=34660$ i.e. the new divisor should be of the form 34660x such that $34660x\times x$ is lower than but closest to 1760000. Thus, x should be equal to 5. Then, we obtain
\[346605\overset{5}{\overline{\left){\begin{align}
  & \text{ 1760000} \\
 & -1733025 \\
 & \text{ 26975} \\
\end{align}}\right.}}\]
Thus, the digit at the fifth place after the decimal point in $\sqrt{3}$ should be 5……………………….(1.6)
Thus, from equations (1.1), (1.2), (1.3), (1.4), (1.5) and (1.6), we obtain
$\sqrt{3}\approx 1.73205$
Now, we know that if we want to round it to four decimal places, the last digit should be rounded off, also we know that if the last digit is equal to 5, the digit before it is increased by one if it is odd else is left unchanged. In this case, the digit before 5 is 0 which is even, therefore it should be left unchanged. Therefore, up to four decimal places,
$\sqrt{3}\approx 1.7320$
Which is the required answer.

Note: In this question, even though we are asked to find $\sqrt{3}$ correct up to 4 decimal places, we need to find out the value of the fifth decimal place and then round it off, otherwise if the fifth decimal place is greater than 5, the digit obtained in the fourth decimal place should be changed and thus only finding the value up to four decimal places will give us an incorrect answer.