Square Root of 12 - Value, Simplification, Examples

The square root of 12 simplifies to $\boxed{2\sqrt{3}} \approx 3.464$. Unlike $\sqrt{64}$ (which is the integer 8), $\sqrt{12}$ is irrational — but its radical form simplifies because 12 has a perfect-square factor of 4.

Quick Answer:

• Result: $\sqrt{12} = 2\sqrt{3} \approx 3.464$

• Notation: $2\sqrt{3}$ (simplest radical form), or $3.464$ to three decimal places

• Method shown: Factoring out a perfect square

• Exact form: $2\sqrt{3}$ — irrational but simplifiable because $12 = 4 \times 3$

• Irrational? Yes — $\sqrt{3}$ is irrational, so $2\sqrt{3}$ is too

Quick Reference Table

Where $\sqrt{12}$ Appears

$\sqrt{12} = 2\sqrt{3}$ shows up as the height of an equilateral triangle with side length 4. For an equilateral triangle of side $s$, the height is $h = \frac{s\sqrt{3}}{2}$ — and with $s = 4$, $h = \frac{4\sqrt{3}}{2} = 2\sqrt{3} = \sqrt{12}$. It also appears as the diagonal of a $2 \times \sqrt{8}$ rectangle by the Pythagorean theorem, and in trigonometry as $4 \cos(30°) = 2\sqrt{3}$.

Why $\sqrt{12}$ Simplifies — and $\sqrt{10}$ Doesn't

A square root simplifies when the number under the radical has a perfect-square factor greater than 1. Compare:

• $12 = \mathbf{4} \times 3$ — has perfect-square factor 4, so $\sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$.

• $10 = 2 \times 5$ — no perfect-square factor greater than 1, so $\sqrt{10}$ doesn't simplify.

The general rule:

$$\sqrt{a \cdot b^2} = b\sqrt{a}$$

Pull every perfect-square factor out, leaving only the non-square part inside.

How to Simplify $\sqrt{12}$ — Step by Step

Step 1: Find the largest perfect-square factor of 12.

Factors of 12: $1, 2, 3, 4, 6, 12$. The perfect squares among these: $1, 4$. The largest perfect-square factor is $\mathbf{4}$.

Step 2: Rewrite 12 as $4 \times 3$.

$$\sqrt{12} = \sqrt{4 \times 3}$$

Step 3: Use the product property of square roots.

$$\sqrt{4 \times 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$$

Step 4: Check — can $\sqrt{3}$ be simplified further? Factors of 3: just $1$ and $3$. No perfect-square factor greater than 1. So $2\sqrt{3}$ is simplest form.

$$\boxed{\sqrt{12} = 2\sqrt{3}}$$

The Decimal Value — Long Division

If you need a decimal approximation: $\sqrt{3} \approx 1.7320508\ldots$, so $\sqrt{12} = 2\sqrt{3} \approx 3.4641016\ldots$

To compute $\sqrt{12}$ directly by long division:

Step 1: Greatest integer whose square is $\leq 12$: that's $3$ ($3^2 = 9$). First digit of quotient: $3$.

Step 2: $12 - 9 = 3$. Bring down two zeros: $300$. Double the quotient: $6$. Find $d$ such that $(60 + d)d \leq 300$. Try $d = 4$: $64 \times 4 = 256$. Try $d = 5$: $65 \times 5 = 325$ — too big. So $d = 4$. Quotient: $3.4$.

Step 3: $300 - 256 = 44$. Bring down zeros: $4400$. Doubled quotient $34 \times 2 = 68$. Find $d$ such that $(680 + d)d \leq 4400$. Try $d = 6$: $686 \times 6 = 4116$. Try $d = 7$: $687 \times 7 = 4809$ — too big. So $d = 6$. Quotient: $3.46$.

Step 4: Continue. Quotient grows to $3.464$.

$$\sqrt{12} \approx 3.464$$

Is the Square Root of 12 Rational or Irrational?

$\sqrt{12}$ is irrational. It cannot be written as a fraction $\frac{p}{q}$ of two integers, and its decimal expansion ($3.4641016\ldots$) neither terminates nor repeats.

Why. A whole number has a rational square root only if it's a perfect square. 12 sits between $3^2 = 9$ and $4^2 = 16$ — not a perfect square. So $\sqrt{12}$ is irrational.

The simplified form makes this visible. $\sqrt{12} = 2\sqrt{3}$. The $2$ is rational, but $\sqrt{3}$ is irrational (3 is not a perfect square). A rational number times an irrational number is always irrational (unless the rational is $0$). So $2\sqrt{3}$ — and therefore $\sqrt{12}$ — is irrational.

Proof sketch via $\sqrt{3}$. Suppose $\sqrt{3} = \frac{p}{q}$ in lowest terms. Then $3q^2 = p^2$, so $3$ divides $p^2$, which means $3$ divides $p$. Write $p = 3k$. Then $3q^2 = 9k^2$, so $q^2 = 3k^2$, which means $3$ divides $q^2$, so $3$ divides $q$. But $p$ and $q$ both divisible by 3 contradicts lowest terms. So $\sqrt{3}$ is irrational — and so is $2\sqrt{3} = \sqrt{12}$. $\blacksquare$

Practical takeaway. Any decimal value you write for $\sqrt{12}$ — $3.46$, $3.464$, $3.4641016$ — is an approximation. The exact value lives only in the radical form. Algebraically, leave it as $2\sqrt{3}$ unless a decimal is specifically asked for.

Common Mistakes With Square Root of 12

Mistake 1: Not simplifying $\sqrt{12}$ to $2\sqrt{3}$

Where it slips in: Final answers in geometry and algebra problems.

Don't do this: Leaving $\sqrt{12}$ as the final answer.

The correct way: Always check for perfect-square factors. $12 = 4 \times 3$, so $\sqrt{12} = 2\sqrt{3}$. In most textbooks and exams, unsimplified radicals lose marks even when the numerical value is correct.

Mistake 2: Pulling out a non-perfect-square factor

Where it slips in: Trying to simplify by factoring incorrectly.

Don't do this: $\sqrt{12} = \sqrt{6 \times 2} = \sqrt{6}\sqrt{2}$ — this doesn't simplify because neither 6 nor 2 is a perfect square.

The correct way: You must factor 12 with at least one perfect-square factor. $12 = 4 \times 3$ works because 4 is a perfect square. Stick with the perfect-square pairing.

Mistake 3: Writing $2\sqrt{3}$ as $\sqrt{6}$ or $\sqrt{2\cdot 3}$

Where it slips in: Confusing the multiplication outside the radical with multiplication inside.

Don't do this: $2\sqrt{3} = \sqrt{6}$. (Wrong.)

The correct way: $2\sqrt{3} = \sqrt{4} \cdot \sqrt{3} = \sqrt{4 \times 3} = \sqrt{12}$. The 2 outside the radical becomes 4 inside if you put it under the radical. Check by squaring: $(2\sqrt{3})^2 = 4 \cdot 3 = 12$, and $(\sqrt{6})^2 = 6$ — different numbers.