The Square Root of 65 is About 8.0623
The square root of 65 is approximately 8.0623, and it is irrational — the decimal never terminates and never repeats. The exact form is simply $\sqrt{65}$, which cannot be simplified because $65 = 5 \times 13$ has no repeated prime factor.
Because $8^2 = 64$ and $9^2 = 81$, $\sqrt{65}$ sits just past $8$ — barely, since $65$ is only one more than the perfect square $64$.
Quick Answer
Result: $\sqrt{65} \approx 8.06225774829$
Notation: decimal approximation; exact form is $\sqrt{65}$.
Method shown: long division (manual), cross-checked by estimation and Newton's iteration.
Approximate value: $8.0623$ (4 d.p.)
Exact form: $\sqrt{65}$ — cannot be simplified, since $65 = 5 \times 13$ has no square factor.
Quick Reference Table — Square Roots Near 65
$n$ | $\sqrt{n}$ (exact) | $\sqrt{n}$ (4 d.p.) |
|---|---|---|
$60$ | $2\sqrt{15}$ | $7.7460$ |
$61$ | $\sqrt{61}$ | $7.8102$ |
$62$ | $\sqrt{62}$ | $7.8740$ |
$63$ | $3\sqrt{7}$ | $7.9373$ |
$64$ | $8$ | $8.0000$ |
$65$ | $\boldsymbol{\sqrt{65}}$ | $\boldsymbol{8.0623}$ |
$66$ | $\sqrt{66}$ | $8.1240$ |
$67$ | $\sqrt{67}$ | $8.1854$ |
$68$ | $2\sqrt{17}$ | $8.2462$ |
$81$ | $9$ | $9.0000$ |
$\sqrt{65}$ falls between $\sqrt{64} = 8$ and $\sqrt{81} = 9$, very close to $8$ because $65$ is just above the perfect square $64$.
Where The Square Root of 65 Appears
$\sqrt{65}$ is the hypotenuse of a right triangle with legs $4$ and $7$ — $\sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$ — and also of one with legs $1$ and $8$, since $\sqrt{1 + 64} = \sqrt{65}$. By the Pythagorean theorem, it is the distance between the points $(0, 0)$ and $(4, 7)$. The integer $65$ is itself the hypotenuse of two different Pythagorean triples — $(16, 63, 65)$ and $(33, 56, 65)$ — but in those the hypotenuse is the whole number $65$, not $\sqrt{65}$; the radical is what appears when the legs do not pair to a perfect square.
What "square root of 65" Means
A square root of a non-negative number $n$ is the value $x$ for which $x^2 = n$. For $\sqrt{65}$, it is the positive $x$ with $x^2 = 65$.
Since $8^2 = 64$ and $9^2 = 81$, $\sqrt{65}$ lies between $8$ and $9$, and the radical symbol returns the principal (positive) value.
Is √65 Rational or Irrational?
$\sqrt{65}$ is irrational. Here is the reason: $65 = 5 \times 13$, two distinct primes, each appearing to the first power. A number is a perfect square only when every prime in its factorisation appears to an even power. With $5^1 \times 13^1$, both exponents are odd, so $65$ is not a perfect square and $\sqrt{65}$ cannot be written as a fraction $p/q$.
The decimal $8.06225774829\dots$ neither terminates nor repeats — the signature of an irrational number.
How to Find √65 — Three Methods
Method 1 — Long division (digit by digit)
Set up $65.000000$ in digit-pairs.
Step 1. Largest integer with square $\le 65$ is $8$ ($8^2 = 64$). Subtract: $65 - 64 = 1$. Bring down $00$: dividend $100$.
Step 2. Double the quotient $8$: $16$. Find $d$ with $(160 + d)\cdot d \le 100$. Here $d = 0$ ($1600 \times 0 = 0$), so the next digit is $0$. Bring down $00$: dividend $10000$.
Step 3. Quotient so far $8.0$; double to $160$. Find $d$ with $(1600 + d)\cdot d \le 10000$. $d = 6$ gives $1606 \times 6 = 9636$. Subtract: $10000 - 9636 = 364$. Bring down $00$: dividend $36400$.
Step 4. Quotient $8.06$; double to $1612$. Find $d$ with $(16120 + d)\cdot d \le 36400$. $d = 2$ gives $16122 \times 2 = 32244$. Subtract: $36400 - 32244 = 4156$.
Continuing produces $8.0623$ to four decimals.
Final answer: $\sqrt{65} \approx 8.0623$.
Method 2 — Estimation between perfect squares
$\sqrt{65}$ lies between $\sqrt{64} = 8$ and $\sqrt{81} = 9$. Linear interpolation gives
$$\sqrt{65} \approx 8 + \frac{65 - 64}{81 - 64} = 8 + \frac{1}{17} \approx 8.0588.$$
Close to the true $8.0623$ — good enough for a sanity check.
Method 3 — Newton's iteration
$$x_{k+1} = \frac{1}{2}\left(x_k + \frac{65}{x_k}\right)$$
Start at $x_0 = 8$.
$x_1 = \tfrac{1}{2}(8 + 65/8) = \tfrac{1}{2}(8 + 8.125) = 8.0625$
$x_2 = \tfrac{1}{2}(8.0625 + 65/8.0625) = \tfrac{1}{2}(8.0625 + 8.06202) = 8.0623$
Two iterations reach four-decimal precision.
Examples of Square Root of 65
Example 1
Estimate $\sqrt{65}$ to the nearest whole number.
$8^2 = 64$ and $9^2 = 81$, and $65$ is much nearer $64$, so $\sqrt{65} \approx 8$.
Example 2
A student is asked to simplify $\sqrt{65}$.
Wrong attempt. Reaching for the radical-simplification reflex, the student writes $\sqrt{65} = \sqrt{64 + 1} = \sqrt{64} + \sqrt{1} = 8 + 1 = 9$.
Correct. Square roots do not distribute over addition: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. Check $9^2 = 81 \neq 65$. Since $65 = 5 \times 13$ has no square factor, $\sqrt{65}$ is already in simplest form and is approximately $8.0623$.
Example 3
Find the hypotenuse of a right triangle with legs $4$ and $7$.
$c = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65} \approx 8.06$. Leave it as $\sqrt{65}$ for exact work.
Example 4
Simplify $\sqrt{65} \times \sqrt{65}$.
$\sqrt{65} \times \sqrt{65} = (\sqrt{65})^2 = 65$. A root times itself recovers the radicand exactly.
Example 5
Simplify $\sqrt{260}$ using $\sqrt{65}$.
$260 = 4 \times 65$, so $\sqrt{260} = \sqrt{4}\cdot\sqrt{65} = 2\sqrt{65} \approx 16.12$. The square factor $4$ comes out; $65$ stays under the radical.
Where Students Trip Up on √65
Mistake 1: Splitting the radical over a sum
Where it slips in: A student rewrites $\sqrt{65}$ as $\sqrt{64 + 1}$ and tries to "simplify" the sum.
Don't do this: $\sqrt{64 + 1} = \sqrt{64} + \sqrt{1} = 8 + 1 = 9$.
The correct way: Addition under a radical never breaks apart — only square factors of a product come out. $\sqrt{65}$ stays as $\sqrt{65} \approx 8.0623$.
Mistake 2: Trying to simplify a square-free radicand
Where it slips in: Applying the "pull out a factor" reflex without checking for a square factor.
Don't do this: Forcing $\sqrt{65}$ into a $k\sqrt{m}$ form.
The correct way: $65 = 5 \times 13$ — two distinct primes, no repeated factor. There is nothing to pull out, so $\sqrt{65}$ is already simplest.
Mistake 3: Rounding too early
Where it slips in: Replacing $\sqrt{65}$ with $8.06$ partway through a problem, then squaring later.
Don't do this: $8.06^2 = 64.9636 \neq 65$.
The correct way: Carry the exact form $\sqrt{65}$ through the algebra; convert to a decimal only at the final answer.
Conclusion
The square root of 65 is approximately $8.0623$ — irrational, non-terminating, non-repeating.
$65 = 5 \times 13$ has no square factor, so $\sqrt{65}$ is already in simplest radical form.
Three methods compute it: long division, estimation between $8$ and $9$, and Newton's iteration.
Square roots do not distribute over addition — $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$.
$\sqrt{65}$ is the hypotenuse of a right triangle with legs $4$ and $7$.
A Practical Next Step
Find $\sqrt{66}$ to two decimal places using Newton's method from $x_0 = 8$.
Show that $\sqrt{63}$ simplifies to $3\sqrt{7}$ but $\sqrt{65}$ does not — explain the difference in their factorisations.
A right triangle has legs $4$ and $7$. Find the hypotenuse in exact and decimal form.
Want a Bhanzu trainer to walk through more square-root problems? Book a free demo class — online globally.
Was this article helpful?
Your feedback helps us write better content