Square Root of Perfect Squares by Long Division

This lesson shows how to find the square root of a perfect square using the long division method, working through several three and four digit examples by hand.

Squares to memorise first

Before starting, it helps to know the squares of the numbers from one to ten by heart:

$1^2=1,\quad 2^2=4,\quad 3^2=9,\quad 4^2=16,\quad 5^2=25$

$6^2=36,\quad 7^2=49,\quad 8^2=64,\quad 9^2=81,\quad 10^2=100$

These let you pick the nearest perfect square at each step and guess the digits of the answer.

Example: $\sqrt{361}$

Group the digits in pairs from the right. The number $361$ splits as $3 \mid 61$, so the left group is the single digit $3$.

First digit. $3$ is not a perfect square. The largest perfect square not exceeding $3$ is $1$, and $\sqrt{1}=1$. Write $1$ as the first digit. Then $1\times 1=1$, and $3-1=2$. Bring down the next group to get $261$.

Second digit. Double the quotient so far: $2\times 1 = 2$. We need a digit $d$ so that $\overline{2d}\times d$ ends in $1$ and is at most $261$. Since the units digit is $1$, try $d=9$ (because $9\times 9 = 81$ ends in $1$). Then $29\times 9 = 261$, which matches exactly, leaving remainder $0$.

$\sqrt{361}=19$

Check: $19^2 = 361$.

Example: $\sqrt{1024}$

Group as $10 \mid 24$.

First digit. $10$ is not a perfect square. The largest perfect square not exceeding $10$ is $9$, and $\sqrt 9 = 3$. Write $3$. Then $3\times 3 = 9$ and $10-9=1$. Bring down $24$ to get $124$.

Second digit. Double the quotient: $2\times 3 = 6$. The last digit is $4$, so the next digit could be $2$ or $8$ (since $2^2=4$ and $8^2=64$). Try $2$: $62\times 2 = 124$, which matches, remainder $0$.

$\sqrt{1024}=32$

Check: $32^2 = 1024$.

Example: $\sqrt{3364}$

Group as $33 \mid 64$.

First digit. $33$ is not a perfect square. The largest perfect square not exceeding $33$ is $25$, and $\sqrt{25}=5$. Write $5$. Then $5\times 5 = 25$ and $33-25=8$. Bring down $64$ to get $864$.

Second digit. Double the quotient: $2\times 5 = 10$. The last digit is $4$, so try $2$ or $8$. Testing $2$ gives $102\times 2 = 204$, too small. Testing $8$ gives $108\times 8 = 864$, which matches, remainder $0$.

$\sqrt{3364}=58$

Check: $58^2 = 3364$.

Example: $\sqrt{8649}$

Group as $86 \mid 49$.

First digit. $86$ is not a perfect square. The largest perfect square not exceeding $86$ is $81$, and $\sqrt{81}=9$. Write $9$. Then $9\times 9 = 81$ and $86-81=5$. Bring down $49$ to get $549$.

Second digit. Double the quotient: $2\times 9 = 18$. The last digit is $9$, so try $3$ or $7$ (since $3^2=9$ and $7^2=49$). Testing $3$ gives $183\times 3 = 549$, which matches, remainder $0$.

$\sqrt{8649}=93$

Check: $93^2 = 8649$.

Example: $\sqrt{5776}$

Group as $57 \mid 76$.

First digit. $57$ is not a perfect square. The largest perfect square not exceeding $57$ is $49$, and $\sqrt{49}=7$. Write $7$. Then $7\times 7 = 49$ and $57-49=8$. Bring down $76$ to get $876$.

Second digit. Double the quotient: $2\times 7 = 14$. The last digit is $6$, so try $4$ or $6$ (since $4^2=16$ and $6^2=36$). Testing $4$ gives $144\times 4 = 576$, too small. Testing $6$ gives $146\times 6 = 876$, which matches, remainder $0$.

$\sqrt{5776}=76$

Check: $76^2 = 5776$.

Key takeaways

• Split the number into pairs of digits starting from the right; the leftmost group may have a single digit.
• For the first digit, take the largest whole number whose square does not exceed the leftmost group.
• For each later digit, double the quotient so far, then find the digit $d$ for which the new divisor times $d$ matches the brought-down number, using the units digit as a clue.