Have you ever tried to find the square root of a number like 4 or 9? It’s simple—you get 2 and 3. These are perfect squares. But what happens when you try to find the square root of 2, 7, or 10?
The world of mathematics gets a lot more interesting (and a bit more “infinite”) when we look at non-perfect square numbers.
A non-perfect square is any positive integer that cannot be created by multiplying an integer by itself. While $3 \times 3 = 9$ (perfect), there is no whole number that multiplies by itself to give you 8.
When you take the square root of a non-perfect square, the result is always an irrational number. But what does that actually mean in plain English?
Non-Terminating: The decimal goes on forever. It never hits a “dead end” or finishes at a specific digit.
Non-Recurring: Unlike fractions like $1/3$ (which is $0.333…$), the decimals in these square roots never settle into a repeating pattern. They are unpredictable and infinite.
Example: $\sqrt{2} \approx 1.41421356…$
Understanding that these roots are irrational is a foundational step in higher mathematics. It helps us categorize the number system and explains why we often use symbols (like $\sqrt{5}$) rather than decimals in geometry and engineering—it’s the only way to be 100% accurate!
| Number Type | Example | Square Root Result | Category |
| Perfect Square | 16 | 4 | Rational (Whole Number) |
| Non-Perfect Square | 17 | 4.123105… | Irrational (Non-terminating) |
To show that $\sqrt{2} \approx 1.414$, we follow the long division process:
Step 1: Start with the largest square less than 2, which is $1 \times 1 = 1$. This leaves a remainder of 1.
Step 2: Bring down a pair of zeros to make it 100. Double the quotient (1) to get 2, and find a digit $x$ such that $2x \times x$ is less than 100 ($24 \times 4 = 96$).
Step 3: Repeat the process. Bringing down more zeros eventually gives us the digits 4, 1, and 4.
Result: Therefore, $\sqrt{2} = 1.414$ (approx).
The process for $\sqrt{3} \approx 1.732$ follows the same logic:
Step 1: The largest square less than 3 is 1 ($1 \times 1 = 1$), leaving a remainder of 2.
Step 2: Bring down zeros to make 200. Using the divisor 2, we find $27 \times 7 = 189$, leaving 11.
Step 3: Continuing the division with further pairs of zeros results in the next digits: 3 and 2.
Result: Therefore, $\sqrt{3} = 1.732$ (approx).
Building on our look at irrational numbers, let’s dive into the Long Division Method. Since non-perfect squares result in decimals that never end, we use this specific process to find an accurate approximation.
Unlike perfect squares, where the answer is a clean whole number, finding the square root of a number like 2 or 3 requires a more detailed approach. Here is how you can calculate these values step-by-step.
To show that $\sqrt{2} \approx 1.414$, we use the following steps:
Initial Square: We start by finding the largest perfect square less than 2, which is $1 \times 1 = 1$.
First Remainder: Subtracting 1 from 2 leaves a remainder of 1.
Bringing Down Zeros: We add a decimal point and bring down a pair of zeros to make the remainder 100.
Developing the Quotient: By doubling our current quotient (1) and finding a digit $x$ such that $2x \times x \leq 100$, we find that $24 \times 4 = 96$.
Continuing the Chain: Subtracting 96 from 100 leaves 4; we bring down another pair of zeros and repeat the process to find the subsequent digits.
Final Approximation: This results in $\sqrt{2} = 1.414$ (approx).
The process for showing $\sqrt{3} \approx 1.732$ follows a similar pattern:
Starting Point: Again, the largest square less than 3 is 1 ($1 \times 1 = 1$), leaving a remainder of 2.
Expansion: We bring down two zeros to make 200.
Finding the Divisor: We double our quotient to get 2 and find that $27 \times 7 = 189$.
Next Steps: Subtracting 189 from 200 leaves 11; bringing down more zeros allows us to find the digits 3 and 2.
Final Approximation: This gives us $\sqrt{3} = 1.732$ (approx).
For quick reference in your math problems, keep these three common approximations in mind:
$\sqrt{2} \approx 1.414$
$\sqrt{3} \approx 1.732$
$\sqrt{5} \approx 2.236$
Pro Tip: Because these are non-terminating, non-recurring decimals, we always write “(approx)” after the value to show it is an estimate!
How to Simplify Square Roots
The key to simplifying a square root is to find the largest perfect square (like 4, 9, 16, 25, etc.) that is a factor of the number under the radical. Once you find it, you can take its square root and move it outside the radical sign.
The general formula used is:
$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$
Common Radical Simplifications to Remember
Here is a handy reference list of frequently used simplifications. Grouping them by their remaining surd factor makes them easier to learn:
Group 1: Roots involving $\sqrt{2}$
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$
$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$
$\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$
Group 2: Roots involving $\sqrt{3}$
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
Group 3: Roots involving $\sqrt{5}$
$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$
$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$
$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$
$\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}$
Practical Application: Finding Decimal Values
Once you have simplified a surd, finding its decimal value becomes much easier if you know the approximate values of $\sqrt{2}$, $\sqrt{3}$, or $\sqrt{5}$.
Example:
If $\sqrt{3} \approx 1.732$, find the value of $\sqrt{75}$ correct to two decimal places.
Solution:
- Simplify the surd: First, find the perfect square factor of 75.
$$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$$- Substitute the known value: Replace $\sqrt{3}$ with $1.732$.
$$5 \times (1.732)$$- Calculate:
$$= 8.660$$Final Answer: The value of $\sqrt{75}$ is approximately 8.66.