The Answer At A Glance
Quick Answer:
Result: $\sqrt{8} = 2\sqrt{2} \approx 2.8284271$
Notation: Exact (simplest radical) form $2\sqrt{2}$; decimal approximation $2.8284$.
Method shown: Prime factorization to simplify, then long division to confirm the decimal.
Approximate value: $2.8284$ (4 d.p.)
Exact form: $2\sqrt{2}$ — $8 = 2^3$, so one pair of $2$s comes out of the root.
Quick Reference Table — Square Roots From 1 to 20
$n$ | $\sqrt{n}$ (exact) | $\sqrt{n}$ (4 d.p.) |
|---|---|---|
$1$ | $1$ | $1.0000$ |
$2$ | $\sqrt{2}$ | $1.4142$ |
$3$ | $\sqrt{3}$ | $1.7321$ |
$4$ | $2$ | $2.0000$ |
$5$ | $\sqrt{5}$ | $2.2361$ |
$6$ | $\sqrt{6}$ | $2.4495$ |
$7$ | $\sqrt{7}$ | $2.6458$ |
$8$ | $\boldsymbol{2\sqrt{2}}$ | $\boldsymbol{2.8284}$ |
$9$ | $3$ | $3.0000$ |
$10$ | $\sqrt{10}$ | $3.1623$ |
$12$ | $2\sqrt{3}$ | $3.4641$ |
$16$ | $4$ | $4.0000$ |
$18$ | $3\sqrt{2}$ | $4.2426$ |
$20$ | $2\sqrt{5}$ | $4.4721$ |
$\sqrt{8}$ sits between $\sqrt{4} = 2$ and $\sqrt{9} = 3$ — much closer to $3$, since $8$ is nearly $9$.
Where √8 Appears
$\sqrt{8}$, written $2\sqrt{2}$, is the diagonal of a square whose side length is $2$ — Pythagoras gives $\sqrt{2^2 + 2^2} = \sqrt{8}$. It is exactly twice the diagonal of a unit square ($\sqrt{2}$), which is why $2\sqrt{2}$ turns up whenever a $45^\circ$ direction is scaled. The same value appears as the distance between the points $(0,0)$ and $(2,2)$ on a coordinate grid.
What "square root of 8" Means
The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{8}$, it is the positive $x$ with $x^2 = 8$.
Because $2^2 = 4$ and $3^2 = 9$, the answer lands between $2$ and $3$ — and $(2\sqrt{2})^2 = 4 \cdot 2 = 8$ confirms the simplified form is exact.
Is The Square Root of 8 Rational or Irrational?
$\sqrt{8}$ is irrational. Its prime factorisation is $8 = 2^3$ — the prime $2$ appears three times, an odd power, so $8$ is not a perfect square.
Simplifying to $2\sqrt{2}$ does not change that: $\sqrt{2}$ is itself irrational, and an integer times an irrational number stays irrational. The decimal $2.8284271\ldots$ never terminates and never repeats.
How To Find √8 — Two Methods
Method 1 — Prime factorization (the simplification)
Break $8$ into primes: $8 = 2 \times 2 \times 2 = 2^2 \times 2$.
A pair of equal factors under a root comes out as a single factor: $\sqrt{2^2 \times 2} = 2\sqrt{2}$.
Then $2\sqrt{2} = 2 \times 1.41421 = 2.82842$.
Final answer: $\sqrt{8} = 2\sqrt{2} \approx 2.8284$.
Method 2 — Long division (digit by digit)
Write $8$ as $8.000000$ and pair the digits after the decimal point.
Step 1. The largest integer whose square is at most $8$ is $2$ ($2^2 = 4$). Subtract: $8 - 4 = 4$. Bring down $00$ to get $400$.
Step 2. Double the quotient $2$ to get $4$. Find $d$ with $(40 + d)\cdot d \leq 400$. Here $d = 8$ gives $48 \cdot 8 = 384$. Subtract: $400 - 384 = 16$. Bring down $00$ to get $1600$.
Step 3. Double $2.8$ to get $56$. Find $d$ with $(560 + d)\cdot d \leq 1600$. Here $d = 2$ gives $562 \cdot 2 = 1124$. Subtract: $1600 - 1124 = 476$.
Continuing produces $2.8284\ldots$, matching $2\sqrt{2}$.
Final answer: $\sqrt{8} \approx 2.8284$.
What are the most common mistakes with √8?
Mistake 1: Leaving √8 unsimplified
Where it slips in: A student computes the decimal but stops before simplifying the radical, so an exam answer reads "$\sqrt{8}$" where "$2\sqrt{2}$" was wanted.
Don't do this: Treating $\sqrt{8}$ as fully simplified just because it's a single root symbol.
The correct way: Check for a square factor first — $8 = 4 \times 2$, so $\sqrt{8} = 2\sqrt{2}$.
Mistake 2: Pulling out the wrong factor
Where it slips in: Rushing the prime factorisation and taking out the whole $4$ instead of the square root of $4$.
Don't do this: $\sqrt{8} = \sqrt{4 \times 2} = 4\sqrt{2}$.
The correct way: $\sqrt{4 \times 2} = \sqrt{4},\sqrt{2} = 2\sqrt{2}$. Only the root of the square factor leaves the radical.
Mistake 3: Splitting the root over addition
Where it slips in: When $\sqrt{8}$ appears as $\sqrt{4 + 4}$ inside the diagonal calculation.
Don't do this: $\sqrt{4 + 4} = \sqrt{4} + \sqrt{4} = 2 + 2 = 4$.
The correct way: $\sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.828$. Square roots distribute over multiplication, never over addition.
Examples of Square Root of 8
Example 1
Simplify $\sqrt{8}$ to radical form.
$\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \approx 2.828$. The square factor $4$ leaves the root as $2$.
Example 2 (Wrong path first)
Find the diagonal of a square with side $2$.
Wrong attempt. A student writes the diagonal as $\sqrt{4 + 4} = \sqrt{4} + \sqrt{4} = 4$.
Why it breaks. A diagonal of $4$ would be longer than two full sides ($2 + 2 = 4$) laid end to end — but a straight diagonal is always shorter than that path.
Correct. $\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \approx 2.828$.
Example 3
Add $\sqrt{8} + \sqrt{2}$.
$\sqrt{8} = 2\sqrt{2}$, so $\sqrt{8} + \sqrt{2} = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2} \approx 4.243$. Once both are in $\sqrt{2}$ form, they add like $2x + x = 3x$.
Example 4
Rationalise $\dfrac{4}{\sqrt{8}}$.
$\dfrac{4}{\sqrt{8}} = \dfrac{4}{2\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2} \approx 1.414$.
Example 5
A square garden has area $8$ square metres. Find its side length.
Side $= \sqrt{8} = 2\sqrt{2} \approx 2.83$ m. Even an "exact" area can give an irrational side — perfectly normal for non-perfect-square areas.
Conclusion
The square root of 8 is $2\sqrt{2}$, approximately $2.828$ — irrational but simplifiable.
$8 = 2^3 = 4 \times 2$, so the square factor $4$ leaves the root as $2$.
Prime factorization simplifies $\sqrt{8}$; long division confirms the decimal.
Only the root of a square factor comes out — $\sqrt{4 \times 2} = 2\sqrt{2}$, not $4\sqrt{2}$.
$2\sqrt{2}$ is the diagonal of a square with side $2$.
A practical next step
Simplify $\sqrt{18}$ and $\sqrt{32}$ into the form $a\sqrt{2}$, then check each by squaring.
Show by prime factorisation that $\sqrt{8} = 2\sqrt{2}$ but $\sqrt{6}$ does not simplify.
A square has area $50$ m². Find its side length in exact and decimal form.
Want a live Bhanzu trainer to walk through more square-root problems? Book a free demo class — online globally.
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