Square Root of 61 — How to Find It

The Answer At A Glance

Result: $\sqrt{61} \approx 7.81024967591$

Notation: Decimal approximation; exact form is $\sqrt{61}$.

Method shown: Long division (manual), with cross-checks using Newton's method and linear interpolation.

Approximate value: $7.8102$ (4 d.p.)

Exact form: $\sqrt{61}$ — cannot be simplified, since $61$ is prime.

Quick Reference Table — Square Roots Near 61

$\sqrt{61}$ sits between $\sqrt{49} = 7$ and $\sqrt{64} = 8$ — closer to $8$ because $61$ is closer to $64$.

What "Square Root of 61" Means

The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{61}$, the positive $x$ with $x^2 = 61$.

Because $7^2 = 49$ and $8^2 = 64$, $\sqrt{61}$ lies between $7$ and $8$.

Is √61 Rational or Irrational?

$\sqrt{61}$ is irrational. Reason: $61$ is a prime number — it has no factors other than $1$ and itself. A number is a perfect square if and only if every prime in its factorisation appears to an even power. $61$ has $61^1$ — exponent $1$, which is odd. So $61$ is not a perfect square, and $\sqrt{61}$ cannot be written as a fraction $p/q$.

The decimal $7.81024967591\dots$ neither terminates nor repeats.

How to Find √61 — Three Methods

Method 1 — Long division (digit by digit)

Pair the digits of $61$: $61.000000$.

Step 1. Largest integer with square $\leq 61$ is $7$ ($7^2 = 49$). Subtract: $61 - 49 = 12$. Bring down $00$: $1200$.

Step 2. Double $7$: $14$. Find $d$ with $(140 + d) \cdot d \leq 1200$. $d = 8$ gives $148 \cdot 8 = 1184$. Subtract: $1200 - 1184 = 16$. Bring down $00$: $1600$.

Step 3. Double $7.8$: $156$. Find $d$ with $(1560 + d) \cdot d \leq 1600$. $d = 1$ gives $1561 \cdot 1 = 1561$. Subtract: $1600 - 1561 = 39$. Bring down $00$: $3900$.

Step 4. Double $7.81$: $1562$. $d = 0$ gives $15620 \cdot 0 = 0$. Try $d = 0$ exactly. Subtract: $3900 - 0 = 3900$. Bring down $00$: $390{,}000$.

Continuing produces $7.8102$ to four decimals.

Final answer: $\sqrt{61} \approx 7.8102$.

Method 2 — Newton's iteration

$$x_{k+1} = \frac{1}{2}\left(x_k + \frac{61}{x_k}\right)$$

Start $x_0 = 8$.

• $x_1 = \frac{1}{2}(8 + 61/8) = \frac{1}{2}(8 + 7.625) = 7.8125$

• $x_2 = \frac{1}{2}(7.8125 + 61/7.8125) = \frac{1}{2}(7.8125 + 7.808) = 7.8102$

Two iterations to four-decimal precision.

Method 3 — Linear interpolation

$$\sqrt{61} \approx 7 + \frac{61 - 49}{64 - 49} = 7 + \frac{12}{15} = 7.80$$

The estimate $7.80$ is close enough to $7.81$ for a sanity check.

Where √61 shows up

$\sqrt{61}$ is the hypotenuse of a right triangle with legs $5$ and $6$ — $\sqrt{25 + 36} = \sqrt{61}$. It also appears as the distance between $(0, 0)$ and $(5, 6)$. The Pythagorean triple $(11, 60, 61)$ has $61$ as its hypotenuse — but the hypotenuse there is the integer $61$, not $\sqrt{61}$. $\sqrt{61}$ itself is what comes up when both legs are not integers paired with $61$.

Tripping points to avoid on √61

Mistake 1: Trying to simplify a prime radicand.

Where it slips in: Students apply the radical-simplification reflex without checking whether the radicand has any square factor.

Don't do this: $\sqrt{61} = \sqrt{60 + 1}$ → expand and "simplify."

The correct way: $61$ is prime — no square factor. $\sqrt{61}$ is already in simplest radical form. Addition under a radical never simplifies; only multiplication of square factors does.

In Bhanzu's Grade 8 cohorts, the "split the radicand" slip shows up on roughly three out of ten first attempts at $\sqrt{61}$ — students reflexively try $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$, which is false.

Mistake 2: Confusing $\sqrt{a^2 + b^2}$ with $\sqrt{a^2} + \sqrt{b^2}$.

Where it slips in: $\sqrt{25 + 36}$ getting evaluated as $5 + 6 = 11$.

Don't do this: $\sqrt{25 + 36} = \sqrt{25} + \sqrt{36} = 5 + 6 = 11$.

The correct way: $\sqrt{25 + 36} = \sqrt{61} \approx 7.81$. Square roots do not distribute over addition. A Bhanzu trainer who sees this slip pauses, writes $\sqrt{9 + 16}$ next to it, evaluates both ways — $\sqrt{25} = 5$ versus $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ — and the rule lands instantly.

Mistake 3: Rounding too early.

Where it slips in: Computing $\sqrt{61}$ as $7.81$, then squaring later in the problem and expecting to recover $61$ exactly.

Don't do this: $7.81^2 = 60.9961 \neq 61$.

The correct way: Keep the exact form $\sqrt{61}$ through the algebra. Convert to decimal only at the final answer.

Conclusion

• The square root of 61 is approximately $7.8102$ — irrational, non-terminating, non-repeating.

• $61$ is prime, so $\sqrt{61}$ cannot be simplified.

• Three methods compute it: long division, Newton's iteration, linear interpolation.

• Square roots do not distribute over addition — $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$.

• $\sqrt{61}$ shows up as the hypotenuse of a right triangle with legs $5$ and $6$.

A practical next step

1. Find $\sqrt{59}$ to two decimal places using Newton's method.

2. Verify with prime factorisation that $\sqrt{63}$ simplifies to $3\sqrt{7}$, but $\sqrt{61}$ does not simplify.

3. A right triangle has legs $5$ and $6$. Find the hypotenuse in exact and decimal form.

Want a Bhanzu trainer to walk you through more square-root problems? Book a free demo class — online globally.