$\sqrt{4} = 2$ — exact, rational, integer. Unlike $\sqrt{50}$ or $\sqrt{20}$, $4$ is a perfect square, so the answer doesn't need a radical, a decimal, or an approximation symbol.
Result: $\sqrt{4} = 2$ (principal square root); both $2$ and $-2$ satisfy $x^2 = 4$.
Notation: Integer; radical form $\sqrt{4}$ simplifies to $2$.
Method shown: Prime factorization (Quick), repeated subtraction (Standard, with a Wrong-Path-First detour on the $\pm$ question), long division (Stretch).
Approximate value: No approximation needed — the answer is exact.
Exact form: $2$.
Quick Reference Table — Small Perfect Squares
$n$ | $\sqrt{n}$ (exact) | Perfect square? | Rational / irrational |
|---|---|---|---|
$1$ | $1$ | Yes | Rational |
$2$ | $\sqrt{2} \approx 1.4142$ | No | Irrational |
$3$ | $\sqrt{3} \approx 1.7321$ | No | Irrational |
$4$ | $\boldsymbol{2}$ | Yes | Rational |
$5$ | $\sqrt{5} \approx 2.2361$ | No | Irrational |
$9$ | $3$ | Yes | Rational |
$16$ | $4$ | Yes | Rational |
$25$ | $5$ | Yes | Rational |
$36$ | $6$ | Yes | Rational |
$49$ | $7$ | Yes | Rational |
$\sqrt{4}$ is the second-smallest non-trivial perfect-square root, sitting between $\sqrt{1} = 1$ and $\sqrt{9} = 3$.
Where √4 lives
$\sqrt{4} = 2$ shows up wherever a square has area $4$ — its side is $2$. The Pythagorean triple $(3, 4, 5)$ uses $\sqrt{4^2} = 4$ as one leg; the equation $x^2 = 4$ has solutions $x = \pm 2$, which is what algebra students meet the first time they solve a quadratic. In physics, a $2$-second free fall covers $\tfrac{1}{2} g t^2$ — and on Earth, $t = \sqrt{4} = 2$ seconds is the time to fall about $20$ metres.
What "square root of 4" means
The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{4}$, the positive $x$ with $x^2 = 4$ — which is $2$, because $2 \cdot 2 = 4$.
The $\sqrt{\phantom{x}}$ symbol by convention denotes the principal (non-negative) square root. The equation $x^2 = 4$ has two solutions, $x = 2$ and $x = -2$, but the expression $\sqrt{4}$ refers only to $2$. Students who blur this distinction lose marks on quadratic-equation problems — covered in the mistakes section.
Is √4 rational or irrational?
$\sqrt{4} = 2$ is rational. A number is a perfect square if and only if every prime in its factorisation appears to an even power. $4 = 2^{2}$ — the $2$ has exponent $2$, which is even, so $4$ is a perfect square and its square root is the integer $2$.
Every integer is rational ($2 = 2/1$), so $\sqrt{4}$ is rational. This is the cleanest case — no decimal tail, no radical, no approximation. The decimal expansion of $\sqrt{4}$ is just $2.000\ldots$.
How to find √4 — three methods
Method 1 — Prime factorization (Quick)
Factor $4$ into primes, then pair them up.
$$4 = 2 \cdot 2 = 2^{2}$$
The pair of $2$s leaves the radical as a single $2$:
$$\sqrt{4} = \sqrt{2^{2}} = 2$$
Final answer: $\sqrt{4} = 2$.
For a perfect square, prime factorization always produces an integer — every prime appears in a complete pair, and every pair leaves the radical as one copy.
Method 2 — Repeated subtraction (Standard, with a Wrong-Path-First on $\pm$)
A student new to square roots might write $\sqrt{4} = \pm 2$, copying the rule they learned for solving $x^2 = 4$. Let us see why that is wrong for the square root expression, even though it is right for the equation.
Run repeated subtraction on $4$. Subtract consecutive odd numbers: $4 - 1 = 3$, $3 - 3 = 0$. After two subtractions, the remainder is zero — so $\sqrt{4} = 2$. The method gives one number, not two.
Scratch that — let us be careful. The method gives the positive root because the count of subtractions is a count, not a signed value. The expression $\sqrt{4}$ by definition refers to the positive (principal) root. The equation $x^2 = 4$ is different — that asks for every number whose square is $4$, and there are two: $+2$ and $-2$. Same arithmetic, different question.
Final answer: $\sqrt{4} = 2$ (the principal root). The equation $x^2 = 4$ has solutions $x = \pm 2$.
Method 3 — Long division (Stretch)
Long division on a perfect square terminates immediately. Run it on $4.00$.
Step 1. Largest integer with square $\leq 4$ is $2$ ($2^2 = 4$). Subtract: $4 - 4 = 0$. Bring down $00$: $0$.
Step 2. Double $2$: $4$. Find $d$ with $(40 + d) \cdot d \leq 0$. $d = 0$. Subtract: $0 - 0 = 0$.
Every further step gives $d = 0$, so $\sqrt{4} = 2.000\ldots = 2$.
Final answer: $\sqrt{4} = 2$.
Long division is overkill for $\sqrt{4}$ — but it confirms the answer and demonstrates how the algorithm behaves on a perfect square versus a non-square like $\sqrt{20}$.
Where students lose the mark on √4
1. Confusing the square root expression with the quadratic equation.
Where it slips in: A student sees $\sqrt{4}$ and writes $\pm 2$, transferring the rule from $x^2 = 4 \Rightarrow x = \pm 2$.
Don't do this: $\sqrt{4} = \pm 2$ — written as the final answer to a square-root expression.
The correct way: $\sqrt{4} = 2$ (the principal, non-negative root). The $\pm$ shows up only when you solve $x^2 = 4$ — taking the square root of both sides introduces $\pm$ explicitly: $x = \pm\sqrt{4} = \pm 2$. The expression alone is single-valued. The memorizer archetype trips here — they remember "always $\pm$" without distinguishing where it applies.
2. Treating $4$ as if it were not a perfect square.
Where it slips in: A student reflexively writes $\sqrt{4} \approx 2.000$ or computes $\sqrt{4}$ on a calculator and reports the decimal.
Don't do this: $\sqrt{4} \approx 2.000$ — used as the final answer.
The correct way: $\sqrt{4} = 2$, exactly. The approximation symbol $\approx$ is reserved for irrational results. Using it on a perfect square hints to the marker that the student does not know the answer is exact — and in proof-based problems, exact form matters.
3. Misreading $\sqrt{4}$ as $4^{1/4}$ or $4 \div 2$.
Where it slips in: A rushing student reads the radical symbol as "divide by 2" or, worse, confuses it with the fourth root $4^{1/4}$.
Don't do this: $\sqrt{4} = 4 \div 2 = 2$ (right answer, wrong method) — or $\sqrt{4} = 4^{1/4} = \sqrt{2} \approx 1.414$.
The correct way: $\sqrt{4}$ means $4^{1/2}$. The defining property is $\sqrt{4} \cdot \sqrt{4} = 4$, which forces $\sqrt{4} = 2$. The Bhanzu Grade 7 algebra session opens with the squaring-back habit: every square-root answer gets multiplied by itself to confirm. $\sqrt{4} = 2$ checks because $2 \cdot 2 = 4$; the fourth-root answer would fail this check.
Wrapping Up
The square root of 4 is $2$ — the principal, non-negative root.
$\sqrt{4}$ is rational because $4$ is a perfect square: $4 = 2^2$, every prime to an even power.
The expression $\sqrt{4}$ resolves to a single value ($2$); the equation $x^2 = 4$ has two solutions ($\pm 2$). Don't confuse the expression with the equation.
Three methods all confirm $\sqrt{4} = 2$ — prime factorization in one line, repeated subtraction in two steps, long division terminating immediately.
Square-root answers should always be squared back to verify — $2 \cdot 2 = 4$ ✓.
A Practical Next Step
Show that $\sqrt{9} = 3$ using prime factorization, then verify by squaring back.
Solve $x^2 = 4$ and explain why the answer has two values while $\sqrt{4}$ has one.
Find $\sqrt{16}$ using repeated subtraction; confirm with prime factorization.
Want a live Bhanzu trainer to walk through more perfect-square problems? Book a free demo class — online globally.
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