Square Root of 269 - Value, Simplest Form, and Examples

What Is A Square Root?

The square root of a number $n$ is the value $r$ that satisfies $r^2 = n$. So the square root of 269 is the number that, multiplied by itself, returns 269.

No integer does this: $16^2 = 256$ is too small and $17^2 = 289$ is too big. That places $\sqrt{269}$ between 16 and 17, and because 269 is not a perfect square, the value is not a whole number.

Where Does √269 Appear?

$\sqrt{269}$ is the length of the diagonal of a rectangle whose sides are 10 and 13, because $10^2 + 13^2 = 100 + 169 = 269$. Anywhere a right triangle has legs 10 and 13, its hypotenuse measures exactly $\sqrt{269}$ units.

Between which two whole numbers does √269 lie? It sits between 16 and 17, since $16^2 = 256$ and $17^2 = 289$, and 269 is closer to 256, so the value lands just above 16.4. This kind of quick bracketing is the fastest sanity check before you commit to a full calculation, and it is the first move taught in square root tricks.

Quick Reference Table

Is The Square Root Of 269 Rational Or Irrational?

$\sqrt{269}$ is irrational, meaning it cannot be written as a fraction $\frac{p}{q}$ of two integers, and its decimal expansion neither ends nor repeats.

Quick reasoning. A whole number has a rational square root only when it is a perfect square. 269 is a prime number, so its only factors are ${1, 269}$, neither of which is a perfect square greater than 1. That guarantees an irrational number as the root.

Formal proof (by contradiction). Suppose $\sqrt{269}$ were rational. Then write it in lowest terms:

$$\sqrt{269} = \frac{p}{q}$$

$$269 = \frac{p^2}{q^2}$$

$$p^2 = 269,q^2$$

Since 269 is prime and divides $p^2$, it must divide $p$. Write $p = 269k$:

$$(269k)^2 = 269,q^2$$

$$269,k^2 = q^2$$

Now 269 divides $q^2$, so it divides $q$ as well. That makes 269 a common factor of both $p$ and $q$, which contradicts the lowest-terms assumption. So no such fraction exists, and $\sqrt{269}$ is irrational.

How Do You Find √269? (Long Division And Prime Factorization)

Prime factorization first. Break 269 into prime factors to see whether anything can leave the radical:

$$269 = 269 \text{ (prime)}$$

There are no repeated prime factors, so no pair can be pulled out. That is why $\sqrt{269}$ is already in its simplest radical form. The same reasoning behind simplifying radical expressions tells you a prime radicand can never be reduced.

Long division for the decimal. Group the digits and work two places at a time.

Step 1: The largest square not exceeding 2 is $1$ ($1^2 = 1$). First quotient digit is 1, remainder 1.

Step 2: Bring down 69 to get 169. Double the quotient to 2; find $d$ with $(20 + d)\times d \le 169$. Here $26 \times 6 = 156$, so $d = 6$. Quotient is 16, remainder 13.

Step 3: Bring down 00 to get 1300. Double 16 to 32; $324 \times 4 = 1296 \le 1300$, so $d = 4$. Quotient is 16.4, remainder 4.

Step 4: Continue in the same way to reach $16.401$.

$$\sqrt{269} \approx 16.401$$

Examples Of √269

Example 1

Estimate $\sqrt{269}$ to the nearest tenth without a calculator.

$16^2 = 256$ and $16.5^2 = 272.25$.

Since 269 is below 272.25 but above 256, the root is just under 16.5.

Testing $16.4^2 = 268.96$, which is very close to 269.

Final answer: $\sqrt{269} \approx 16.4$.

Example 2

A student simplifies $\sqrt{269}$ as $\sqrt{269} = \sqrt{9 \times 29.9}$. What went wrong?

The first instinct is to force a perfect square out of any radicand, so the student invents a factor 9.

But $9 \times 29.9 = 269.1$, not 269, and 29.9 is not even an integer, so the split is invalid.

The correct move is to factor 269 into integers only. Its factors are ${1, 269}$, with no perfect square among them.

Final answer: $\sqrt{269}$ cannot be simplified; it stays $\sqrt{269}$.

Example 3

Solve $x^2 = 269$.

$$x^2 = 269$$

$$x = \pm\sqrt{269}$$

$$x \approx \pm 16.401$$

Final answer: $x = \sqrt{269}$ or $x = -\sqrt{269}$.

Example 4

Find the diagonal of a rectangle with sides 10 and 13.

$$d = \sqrt{10^2 + 13^2}$$

$$d = \sqrt{100 + 169}$$

$$d = \sqrt{269} \approx 16.401$$

Final answer: the diagonal is $\sqrt{269} \approx 16.401$ units.

Example 5

Evaluate $3\sqrt{269}$ to two decimal places.

$$3 \times 16.40122 = 49.20366$$

Rounding gives $49.20$. Keeping $\sqrt{269}$ symbolic until this last step avoids the rounding drift that many learners introduce by converting to a decimal too early.

Final answer: $3\sqrt{269} \approx 49.20$.

Common Mistakes

Mistake 1: Treating 269 as if it had a perfect-square factor

Where it slips in: the moment a student sees a "simplify" instruction.

Don't do this: writing $\sqrt{269} = 2\sqrt{67.25}$ or any split with a non-integer inside.

The correct way: factor 269 into integers only. It is prime, so $\sqrt{269}$ is already simplest. In practice, the most common first-instinct error here is assuming every radical must reduce to something smaller.

Mistake 2: Reporting $\sqrt{269} = \pm 16.401$

Where it slips in: confusing the square-root symbol with the solutions of $x^2 = 269$.

Don't do this: attaching $\pm$ to the plain radical.

The correct way: $\sqrt{269}$ means the positive root only, about $16.401$. The two-sided answer $\pm\sqrt{269}$ belongs to the equation $x^2 = 269$, not to the symbol itself.

Mistake 3: Rounding before the final step

Where it slips in: multi-step problems where $\sqrt{269}$ appears in the middle.

Don't do this: replacing $\sqrt{269}$ with $16.4$ at the start and carrying that through every operation.

The correct way: keep $\sqrt{269}$ symbolic and substitute the decimal only at the end, so rounding error does not compound.

Conclusion

• The square root of 269 is about $16.401$, irrational, and sits between 16 and 17.

• Because 269 is prime, $\sqrt{269}$ is already in simplest radical form and cannot be reduced.

• Long division gives the decimal; prime factorization confirms there is nothing to pull out.

To build this skill with a teacher, explore Bhanzu's algebra tutor or a high school math tutor, and see the live sessions on math classes online. Ready to try one? Book a free demo class.

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