Square Root of 8 — Value, Simplified Form, and Examples

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

$\sqrt{8}$ 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

1. Simplify $\sqrt{18}$ and $\sqrt{32}$ into the form $a\sqrt{2}$, then check each by squaring.

2. Show by prime factorisation that $\sqrt{8} = 2\sqrt{2}$ but $\sqrt{6}$ does not simplify.

3. A square has area $50$ m². Find its side length in exact and decimal form.

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