$\sqrt{20} = 2\sqrt{5} \approx 4.4721$ — irrational, non-terminating, non-repeating. The simplified radical form $2\sqrt{5}$ is the answer most algebra textbooks expect; the decimal $4.4721$ is what a calculator returns.
Result: $\sqrt{20} = 2\sqrt{5} \approx 4.4721$
Notation: Simplified radical form $2\sqrt{5}$; decimal approximation $4.4721$ (4 d.p.).
Method shown: Prime factorization (Quick), repeated subtraction with a Wrong-Path-First detour (Standard), long division (Stretch).
Approximate value: $4.47213595499958$ (14 d.p.).
Exact form: $2\sqrt{5}$ — cannot be simplified further, since $5$ is prime.
Quick Reference Table — Square Roots Near 20
$n$ | $\sqrt{n}$ (exact form) | $\sqrt{n}$ (4 d.p.) | Nearest perfect square |
|---|---|---|---|
$16$ | $4$ | $4.0000$ | $16$ (itself) |
$18$ | $3\sqrt{2}$ | $4.2426$ | $16$ |
$19$ | $\sqrt{19}$ | $4.3589$ | $16$ |
$20$ | $\boldsymbol{2\sqrt{5}}$ | $\boldsymbol{4.4721}$ | $16$ |
$21$ | $\sqrt{21}$ | $4.5826$ | $25$ |
$24$ | $2\sqrt{6}$ | $4.8990$ | $25$ |
$25$ | $5$ | $5.0000$ | $25$ (itself) |
$27$ | $3\sqrt{3}$ | $5.1962$ | $25$ |
$28$ | $2\sqrt{7}$ | $5.2915$ | $25$ |
$\sqrt{20}$ sits between $\sqrt{16} = 4$ and $\sqrt{25} = 5$ — closer to $4$ because $20$ is $4$ away from $16$ and $5$ away from $25$.
Where √20 shows up
$\sqrt{20}$ appears as the hypotenuse of a right triangle with legs $2$ and $4$ — by Pythagoras, $\sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$. The same value is the distance from $(0, 0)$ to $(2, 4)$ on a coordinate plane. The golden ratio's defining equation, $\phi = \tfrac{1 + \sqrt{5}}{2}$, hides $\sqrt{5}$ inside it — and $\sqrt{20} = 2\sqrt{5}$ is just twice that constant.
What "square root of 20" means
The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{20}$, the positive $x$ with $x^2 = 20$. Because $4^2 = 16$ and $5^2 = 25$, $\sqrt{20}$ lives between $4$ and $5$.
The simplified-radical form $2\sqrt{5}$ extracts the perfect-square factor (the $4$) from inside $20$ and leaves the rest under the radical. The decimal $4.4721$ comes from numerically approximating the irrational $\sqrt{5}$ that remains.
Is √20 rational or irrational?
$\sqrt{20}$ is irrational. A number is a perfect square if and only if every prime in its factorisation appears to an even power. $20 = 2^{2} \cdot 5^{1}$ — the $2$ pairs into $2^2$, but the $5$ stands alone with exponent $1$. That lone $5$ is the reason $\sqrt{20}$ doesn't collapse to an integer.
The decimal $4.4721359549\ldots$ neither terminates nor repeats. Even the simplified form $2\sqrt{5}$ carries that irrationality — $\sqrt{5}$ itself is irrational, and a rational ($2$) times an irrational stays irrational.
How to find √20 — three methods
Method 1 — Prime factorization (Quick)
Factor $20$ into primes, then pull out the pair.
$$20 = 2 \cdot 2 \cdot 5 = 2^{2} \cdot 5$$
The pair of $2$s leaves the radical as a single $2$:
$$\sqrt{20} = \sqrt{2^{2} \cdot 5} = 2\sqrt{5}$$
Final answer: $\sqrt{20} = 2\sqrt{5} \approx 4.4721$.
This is the method every algebra problem expects when it says "express in simplified radical form."
Method 2 — Repeated subtraction (Standard, with a Wrong-Path-First detour)
A common student instinct: try subtracting consecutive odd numbers from $20$. That trick works on perfect squares — let us run it and watch what happens.
Start at $20$. Subtract: $20 - 1 = 19$, $19 - 3 = 16$, $16 - 5 = 11$, $11 - 7 = 4$, $4 - 9 = -5$. The fourth subtraction left $4$, and the fifth overshoots into the negatives. The remainder never lands on zero.
Hold on — that overshoot is the diagnosis, not a failure. The method only lands at zero when $n$ is a perfect square. $20$ isn't one. After step four, we know $\sqrt{20}$ is between $4$ and $5$, which already matches the table. The rescue is to switch to prime factorization for the simplified form, or long division for the decimal.
Final answer (radical): $\sqrt{20} = 2\sqrt{5}$.
The second-guesser archetype hits this method hardest — they get to step five, see the negative, and assume they made an arithmetic mistake. They didn't. They proved $20$ isn't a perfect square.
Method 3 — Long division (Stretch)
For a decimal approximation, run the long-division algorithm on $20.000000$.
Step 1. Largest integer with square $\leq 20$ is $4$ ($4^2 = 16$). Subtract: $20 - 16 = 4$. Bring down $00$: $400$.
Step 2. Double $4$: $8$. Find $d$ with $(80 + d) \cdot d \leq 400$. $d = 4$ gives $84 \cdot 4 = 336$. Subtract: $400 - 336 = 64$. Bring down $00$: $6400$.
Step 3. Double $4.4$: $88$. Find $d$ with $(880 + d) \cdot d \leq 6400$. $d = 7$ gives $887 \cdot 7 = 6209$. Subtract: $6400 - 6209 = 191$. Bring down $00$: $19{,}100$.
Step 4. Double $4.47$: $894$. Find $d$ with $(8940 + d) \cdot d \leq 19{,}100$. $d = 2$ gives $8942 \cdot 2 = 17{,}884$. Subtract: $19{,}100 - 17{,}884 = 1216$. Bring down $00$: $121{,}600$.
Continuing produces $\sqrt{20} \approx 4.4721$ to four decimals.
Final answer: $\sqrt{20} \approx 4.4721$.
Slip-ups that cost marks on √20
1. Pulling all the prime factors out, not just the pairs
Where it slips in: A student factors $20 = 2 \cdot 2 \cdot 5$ and pulls every factor out: $\sqrt{20} = 2 \cdot 2 \cdot 5 = 20$.
Don't do this: $\sqrt{20} = 2 \cdot 2 \cdot 5$.
The correct way: Only the paired factors leave the radical, and each pair leaves as a single number — not as both members of the pair. The pair of $2$s becomes a single $2$ outside; the unpaired $5$ stays under. So $\sqrt{20} = 2\sqrt{5}$.
2. Reading "simplify" as "evaluate"
Where it slips in: A test question says "express $\sqrt{20}$ in simplest radical form." The student writes $4.47$.
Don't do this: $\sqrt{20} \approx 4.47$ — submitted as the final answer when the question asked for radical form.
The correct way: $\sqrt{20} = 2\sqrt{5}$. The decimal is an approximation that loses information; $2\sqrt{5}$ is exact. Always read the question's verb — "simplify" wants the radical, "evaluate" or "approximate" wants the decimal.
3. Multiplying instead of squaring when checking.
Where it slips in: A student answers $\sqrt{20} = 5$ and checks by multiplying $5 \cdot 5 = 25$ — which is not $20$ — but doesn't recognise the mismatch as a sign their answer is wrong.
Don't do this: Skip the check, or run the check and shrug at the wrong number.
The correct way: After every square-root answer, square it back. If the result isn't the original number, the answer is wrong. $(2\sqrt{5})^2 = 4 \cdot 5 = 20$ — the check works. This is the habit a Bhanzu Grade 9 trainer drills in the first week of the radicals unit: every answer gets squared back before it leaves the page. About four out of every ten students fix their own error in the squaring-back step before the trainer ever sees it.
Conclusion
The square root of 20 is $2\sqrt{5}$ in simplified radical form and $\approx 4.4721$ as a decimal.
$\sqrt{20}$ is irrational; the unpaired prime $5$ blocks an integer answer.
Three methods reach $\sqrt{20}$: prime factorization (one line), repeated subtraction (proves irrationality), long division (gives the decimal).
Only paired prime factors leave a radical, and they leave as a single copy — not as both.
Every square-root answer should be squared back to confirm — $(2\sqrt{5})^2 = 20$.
A practical next step
Simplify $\sqrt{45}$ using prime factorization. (Answer should be $3\sqrt{5}$.)
Compute $\sqrt{20}$ to two decimal places by long division without copying the worked example.
Find the distance from $(0, 0)$ to $(2, 4)$ on a coordinate plane in exact and decimal form.
Want a live Bhanzu trainer to walk through more radical-simplification problems? Book a free demo class — online globally.
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