Square Root of 20 — Simplify, Methods, Value

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Square Root of 20 — Simplify, Methods, Value

Algebra

TL;DR

The square root of 20 is ( 2\sqrt{5} ) in simplified radical form and approximately 4.4721 as a decimal — irrational, because the prime 5 inside the radical has no partner. This article walks through what ( 20\sqrt{20} ) equals, three ways to compute it (prime factorization, repeated subtraction, long division), where it shows up in geometry, and the slips that cost the most marks.

Result: ( 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.

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 ( n\sqrt{n} ) (exact form) ( n\sqrt{n} ) (4 d.p.) Nearest perfect square
16 4 4.0000 16
18 3( \sqrt{2} ) 4.2426 16
19 19( \sqrt{19} ) 4.3589 16
20 2\sqrt{5} 4.4721 16
21 21( \sqrt{21} ) 4.5826 25
24 26( \sqrt{6} ) 4.8990 25
25 5 5.0000 25
27 33( \sqrt{3} ) 5.1962 25
28 27( \sqrt{7} ) 5.2915 25

( 20\sqrt{20} ) sits between 16 and 25 — closer to 16 because 20 is 4 away from 16 and 5 away from 25.

Where √20 shows up

( 20\sqrt{20} ) appears as the hypotenuse of a right triangle with legs 2 and 4 — by Pythagoras, ( 2^2 + 4^2 = 4 + 16 = 20 = \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 = \frac{1 + \sqrt{5}}{2} ), hides ( \sqrt{5} ) inside it — and ( 20 = 2\sqrt{20} ) 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 ( 20\sqrt{20} ), the positive ( x ) with ( x^2=20 ). Because ( 4^2 = 16 ) and ( 5^2 = 25 ), ( 20 ) lives between 16 and 25.

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?

( 20\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\cdot5 ) — the 2 pairs into 2, but the 5 stands alone with exponent 1. That lone 5 is the reason ( 20\sqrt{20} ) doesn't collapse to an integer.

The decimal 4.4721359549… 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\cdot2\cdot5=2^2\cdot5 )

The pair of 2s leaves the radical as a single 2:

( 20=2^{2}\cdot5=2\sqrt{5} )

Final answer: ( 20=2\sqrt{5} \approx 4.4721 ).

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.

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 ( 20\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): ( 20=2\sqrt{5} ).

Method 3 — Long division (Stretch)

For a decimal approximation, run the long-division algorithm on 20.000000.

Step 1. Largest integer with square ≤20 is 4 (( 4^2 = 16 )). Subtract: ( 20−16=4 ). Bring down 000000: 400.

Step 2. Double 4: 8. Find ( d ) with ( (80+d)\cdot d≤400 ). ( d=4 ) gives ( 84 \cdot 4 = 336 ). Subtract: ( 400−336=64 ). Bring down 000000: 6400.

Step 3. Double 4.4: 8.8. Find ( d ) with ( (880+d)\cdot d≤6400 ). ( d=7 ) gives ( 887\cdot 7 = 6209 ). Subtract: ( 6400−6209=191 ). Bring down 000000: 19,100.

Step 4. Double 4.47: 8.94. Find ( d ) with ( (8940+d)\cdot d≤19,100 ). ( d=2 ) gives ( 8942⋅2=17,884 ). Subtract: ( 19,100−17,884=1216 ). Bring down 000000: 121,600.

Continuing produces ( 20≈4.4721 ).

Final answer: ( 20≈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\cdot2\cdot5 ) and pulls every factor out: ( 20=2\ imes2\ imes5=20 ).

Don't do this: ( 20=2\ imes2\ imes5 ).

The correct way: Only the paired factors leave the radical, and they leave as a single number — not as both members of the pair. The pair of 2s becomes a single 2 outside; the unpaired 5 stays under. So ( 20=2\sqrt{5} ).

2. Reading "simplify" as "evaluate"

Where it slips in: A test question says "express ( 20 ) in simplest radical form." The student writes 4.47.

Don't do this: ( 20≈4.47 ).

The correct way: ( 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 ( 20=5\sqrt{20} ) and checks by multiplying ( 5\cdot5=25 ) — which is not 20 — but doesn't recognize 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 = 20 ) — the check works.

Conclusion

A practical next step

  1. Simplify ( 45 ) using prime factorization. (Answer should be ( 3\sqrt{5} ).
  2. Compute ( 20 ) to two decimal places by long division without copying the worked example.
  3. Find the distance from (0,0) to (2,4) on a coordinate plane in exact and decimal form.