There are two primary methods to approach this problem:
The prime factorization method involves breaking down a number into its prime factors. For 256, we repeatedly divide by the smallest prime number, which is 2:
$256 \div 2 = 128$
$128 \div 2 = 64$
$64 \div 2 = 32$
$32 \div 2 = 16$
$16 \div 2 = 8$
$8 \div 2 = 4$
$4 \div 2 = 2$
$2 \div 2 = 1$
Now, list all the factors and group them in pairs:
$\sqrt{256} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2)}$
Take one number from each pair out of the square root sign:
$\sqrt{256} = 2 \times 2 \times 2 \times 2 = 16$
So, using prime factorization, we find that the square root of 256 is 16.
The long division method is another systematic way to find a square root, especially useful for larger numbers. Here’s how it works for 256:
Group the digits: Starting from the right, group the digits in pairs. For 256, we have ‘2’ and ’56’.
Find the largest square less than or equal to the first group: The first group is ‘2’. The largest square less than or equal to 2 is $1^2 = 1$. Place ‘1’ in the quotient and subtract it from 2.
Bring down the next pair: Bring down ’56’, making the new number 156.
Double the quotient and find the next digit: Double the current quotient (1 becomes 2) and find a digit ‘x’ such that $2x \times x$ is less than or equal to 156. In this case, $26 \times 6 = 156$.
Calculate the remainder: Subtract 156 from 156, which gives a remainder of 0.
The quotient at the top is your answer: 16.
Therefore, $\sqrt{256} = 16$.
For a three-digit number like 289, we group it as 2 and 89.
Step 1: The largest square less than 2 is $1^2 = 1$. Subtract 1 from 2, leaving 1.
Step 2: Bring down the ’89’ to make the number 189.
Step 3: Double the quotient (1 becomes 2) and find the digit ‘7’, as $27 \times 7 = 189$.
Result: $\sqrt{289} = 17$.
For a four-digit number, we group it into two pairs: 10 and 24.
Step 1: The largest square less than 10 is $3^2 = 9$. Subtracting 9 from 10 leaves 1.
Step 2: Bring down ’24’ to get 124.
Step 3: Double the quotient (3 becomes 6). We find that $62 \times 2 = 124$.
Result: $\sqrt{1024} = 32$.
Even for very large six-digit numbers, the method remains the same. Group the digits as 24, 30, and 49.
First digit: The largest square less than 24 is $4^2 = 16$. $24 – 16 = 8$.
Second digit: Bring down ’30’ to get 830. Double 4 to get 8. $89 \times 9 = 801$. $830 – 801 = 29$.
Third digit: Bring down ’49’ to get 2949. Double the current quotient 49 to get 98. $983 \times 3 = 2949$.
Result: $\sqrt{243049} = 493$.
| Number | Pairs | Final Square Root |
| 289 | 2, 89 | 17 |
| 1024 | 10, 24 | 32 |
| 243049 | 24, 30, 49 | 493 |
Try to find the square root for the following four numbers:
$\sqrt{97344}$
$\sqrt{286225}$
$\sqrt{120409}$
$\sqrt{291600}$
If you got stuck, don’t worry! Here are the step-by-step breakdowns for each problem.
The Logic: Since the number ends in 4, the square root must end in either 2 or 8. Since $300^2 = 90,000$ and $320^2 = 102,400$, we know the answer is between 300 and 320.
The Result: 312
The Logic: Any perfect square ending in 25 must have a square root ending in 5. This makes the calculation much faster! We look for a number where $x^2 \approx 2862$. Since $500^2 = 250,000$, we start there.
The Result: 535
The Logic: This number ends in 9, so the root ends in 3 or 7. It sits just above $300^2$ ($90,000$) and below $400^2$ ($160,000$).
The Result: 347
The Logic: This one is the most satisfying to solve. You can simplify this by looking at $\sqrt{2916} \times \sqrt{100}$. Since $\sqrt{100} = 10$, you just need to find the square root of 2916 and add a zero to the end.
The Result: 540