The square root of 68 is $2\sqrt{17} \approx 8.246$. It is an irrational number, so its decimal never terminates or repeats.
Quick Answer:
Result: $\sqrt{68} = 2\sqrt{17}$
Notation: simplified radical $2\sqrt{17}$; decimal $\approx 8.246$
Method shown: prime factorisation (for the radical) and long division (for the decimal)
Approximate value: $8.246$ (to 3 decimal places)
Exact form: $2\sqrt{17}$ (not a whole number — 68 is not a perfect square)
Quick Reference Table
Expression | Simplified form | Decimal (3 dp) |
|---|---|---|
$\sqrt{64}$ | $8$ | $8.000$ |
$\sqrt{65}$ | $\sqrt{65}$ | $8.062$ |
$\sqrt{68}$ | $2\sqrt{17}$ | $8.246$ |
$\sqrt{72}$ | $6\sqrt{2}$ | $8.485$ |
$\sqrt{17}$ | $\sqrt{17}$ | $4.123$ |
$68^2$ | — | $4624$ |
Where the Square Root of 68 Appears
$\sqrt{68}$ is the length of the diagonal of a rectangle with sides 2 and 8, since $\sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68}$. It also turns up as the distance between the points $(0,0)$ and $(2,8)$ on a coordinate plane, because the distance formula reduces to the same square root.
What a Square Root Is
The square root of a number is the value that, multiplied by itself, gives that number. Because $8^2 = 64$ and $9^2 = 81$, the square root of 68 sits between 8 and 9. Since 68 is not a perfect square, its root is irrational — a full explanation lives on the square root page.
How to Compute the Square Root of 68
Method 1: Prime factorisation (for the simplified radical)
Break 68 into primes.
$68 = 2 \times 34$
$= 2 \times 2 \times 17$
$= 2^2 \times 17$
Take the pair of 2s outside the radical as a single 2:
$\sqrt{68} = \sqrt{2^2 \times 17} = 2\sqrt{17}$
Final answer: $\sqrt{68} = 2\sqrt{17}$.
Method 2: Estimation (for a quick decimal)
Find the two perfect squares 68 sits between.
$8^2 = 64$ and $9^2 = 81$, so $\sqrt{68}$ is between 8 and 9.
Since 68 is much closer to 64 than to 81, the value is a little above 8.2.
Estimate: about $8.25$, which matches the true value $8.246$.
Method 3: Long division (for the decimal to 3 dp)
Pair the digits and take the largest square $\le 68$, which is $64 = 8^2$; the first digit is 8, remainder 4.
Bring down $00$: divisor becomes 16_, and $162 \times 2 = 324 \le 400$, so the next digit is 2; remainder 76.
Continue: the next digit is 4, then 6.
Final answer: $\sqrt{68} \approx 8.246$.
Common Mistakes With Square Root of 68
Mistake 1: Simplifying to the wrong factor
Where it slips in: splitting 68 as $4 \times 17$ but forgetting that only $\sqrt{4}$ comes out. Don't do this: write $\sqrt{68} = 4\sqrt{17}$. The correct way: $\sqrt{4} = 2$, so $\sqrt{68} = 2\sqrt{17}$, not $4\sqrt{17}$.
Mistake 2: Calling 2√17 the decimal answer
Where it slips in: stopping at the radical when a decimal is asked for. Don't do this: report $\sqrt{68} = 2 \times 17 = 34$. The correct way: $2\sqrt{17} = 2 \times 4.123 = 8.246$. The 17 stays under the root; you multiply by its root, not by 17.
Mistake 3: Treating √68 as rational
Where it slips in: rounding early and calling 8.25 exact. Don't do this: write $\sqrt{68} = 8.25$ as if it terminates. The correct way: 68 is not a perfect square, so $\sqrt{68}$ is irrational; $8.246$ is a rounded approximation, and $2\sqrt{17}$ is the exact form.
Where to Go From Here
Try simplifying $\sqrt{72}$ and $\sqrt{65}$ using the same prime-factorisation step, then check your decimals against the table above. To build square-root skills with a teacher, explore Bhanzu's algebra tutor or math classes online.
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