Question

Find the square roots of the following perfect squares using the long division method:

a)$357096609$

b)$159.7696$

Answer 1

Hint: To find the square root of the long division method first separate the digits by commas from right to left once in two digits. Now we have to multiply a number by itself such that the product is less than 10. After that we have to bring the next two numbers and quotient to be multiplied by twice the number. We have to continue the process till the remainder gets zero.


Complete step by step solution
Given:

The numbers are $357096609$,$159.7696$.

(a)

We have to find the square root of $357096609$ using the long division method.


We will separate the digits by taking commas from right to left once in two digits.
$3,57,09,66,09$

Now we have to multiply a number by itself such that the product is less than $10$.

$1\mathop{\left){\vphantom{1\begin{array}{l}
03,57,09,66,09\\
01\\
 - - - - - - - - - - \\
02
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
03,57,09,66,09\\
01\\
 - - - - - - - - - - \\
02
\end{array}}}}
\limits^{\displaystyle \,\,\, 1}$

Now, we need to bring $57$down and double the dividend.

$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {18}}\\
28\left){\vphantom{1\begin{array}{l}
0257\\
224
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
0257\\
224
\end{array}}}
\end{array}$

Here, $8$ is common in both that means the multiplier should be the same.

$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {188}}\\
28\left){\vphantom{1\begin{array}{l}
0257\\
224
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
0257\\
224
\end{array}}}\\
368\left){\vphantom{1\begin{array}{l}
33\\
3309\\
2944
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
33\\
3309\\
2944
\end{array}}}
\end{array}$

Now double of $8$is $16$ so on adding $368$ and $16$we get $376$ so the dividend will be .

We know that the remainder for $3309$ and $2944$ is $365$ and we need to take $2$digits down, that is,
$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {1889}}\\
28\left){\vphantom{1\begin{array}{l}
0257\\
224
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
0257\\
224
\end{array}}}\\
368\left){\vphantom{1\begin{array}{l}
33\\
3309\\
2944
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
33\\
3309\\
2944
\end{array}}}\\
3769\left){\vphantom{1\begin{array}{l}
365\\
36566\\
33921
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
365\\
36566\\
33921
\end{array}}}
\end{array}$

Now, we know the double of $9$ is $18$ so now on adding $3769$ and $18$ we get $3778$so the dividend will be .

We know that the remainder for $36566$ and $33921$ is $2645$ and we need to take $2$digits down, that is,

$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
03,57,09,66,09\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {18897}}\\
28\left){\vphantom{1\begin{array}{l}
0257\\
224
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
0257\\
224
\end{array}}}\\
368\left){\vphantom{1\begin{array}{l}
33\\
3309\\
2944
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
33\\
3309\\
2944
\end{array}}}\\
3769\left){\vphantom{1\begin{array}{l}
365\\
36566\\
33921
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
365\\
36566\\
33921
\end{array}}}\\
37787\left){\vphantom{1\begin{array}{l}
2645\\
264509\\
264509\\
 - - - - - - \\
0
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
2645\\
264509\\
264509\\
 - - - - - - \\
0
\end{array}}}
\end{array}$

Hence, the square root of $\sqrt {357096609} = 18897$.

(b)

Now, let us find the square root for 159.7696.

We will separate the digits by taking commas from right to left once in two digits.
$01,59.76,96$

Now, we have to multiply a number by itself such that the product is less than $10$.
$1\mathop{\left){\vphantom{1\begin{array}{l}
01,59.7696\\
01\\
 - - - - - - - - - - \\
0
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
01,59.7696\\
01\\
 - - - - - - - - - - \\
0
\end{array}}}}
\limits^{\displaystyle \,\,\, 1}$

Now, we have to bring $59$down and double the dividend.
$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
01,59.76,96\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
01,59.76,96\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {12}}\\
22\left){\vphantom{1\begin{array}{l}
059\\
44
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
059\\
44
\end{array}}}
\end{array}$

We know that $2$ is common in both that means the multiplier should be the same.
$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
01,59.76,96\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
01,59.76,96\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {12.6}}\\
22\left){\vphantom{1\begin{array}{l}
059\\
44
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
059\\
44
\end{array}}}\\
246\left){\vphantom{1\begin{array}{l}
15\\
1576\\
1476
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
15\\
1576\\
1476
\end{array}}}
\end{array}$

Now the double of $6$ is $12$ so now on adding $246$and $12$ we get $252$so the dividend will be .

We know that the remainder for $1576$ and $1476$ is $100$and we need to take $2$ digits down we get,
$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
01,59.76,96\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
01,59.76,96\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {12.64}}\\
22\left){\vphantom{1\begin{array}{l}
059\\
44
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
059\\
44
\end{array}}}\\
246\left){\vphantom{1\begin{array}{l}
15\\
1576\\
1476
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
15\\
1576\\
1476
\end{array}}}\\
2524\left){\vphantom{1\begin{array}{l}
100\\
10096\\
10096
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
100\\
10096\\
10096
\end{array}}}
\end{array}$

Now, we know the double of $4$ is $8$. So, on adding $2524$ and we get $2528$ so the dividend will be .

We know that the remainder for $10096$and $10096$is $0$ and we need to take $2$digits down we get,
$\begin{array}{l}
1\mathop{\left){\vphantom{1\begin{array}{l}
01,59.76,96\\
01
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
01,59.76,96\\
01
\end{array}}}}
\limits^{\displaystyle \,\,\, {12.6400237}}\\
22\left){\vphantom{1\begin{array}{l}
059\\
44
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
059\\
44
\end{array}}}\\
246\left){\vphantom{1\begin{array}{l}
15\\
1576\\
1476
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
15\\
1576\\
1476
\end{array}}}\\
2524\left){\vphantom{1\begin{array}{l}
100\\
10096\\
10096
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
100\\
10096\\
10096
\end{array}}}\\
2524\left){\vphantom{1\begin{array}{l}
060\\
50.564
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
060\\
50.564
\end{array}}}\\
25.282\left){\vphantom{1\begin{array}{l}
9.436\\
943.6\\
935.3616169000001
\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}
9.436\\
943.6\\
935.3616169000001
\end{array}}}
\end{array}$

Hence, the square root of $\sqrt {159.7696} = 12.6400237$.


Note: Always be careful with long division because the last term of each step should be multiplied with the same number and placed in the quotient place. Also the dividend’s last digit must be doubled and take the sum of dividend and substitute in the next step.