Quick Reference Table
Number $n$ | $\sqrt{n}$ (approx.) | Simplest Radical | Rational or Irrational |
|---|---|---|---|
196 | 14 | 14 | Rational |
225 | 15 | 15 | Rational |
242 | 15.556 | $11\sqrt{2}$ | Irrational |
245 | 15.652 | $7\sqrt{5}$ | Irrational |
256 | 16 | 16 | Rational |
260 | 16.125 | $2\sqrt{65}$ | Irrational |
Where Does √242 Appear?
$\sqrt{242}$ is the exact length of the diagonal of a square whose side is 11, because a square of side $s$ has a diagonal of $s\sqrt{2}$, and $11\sqrt{2} = \sqrt{242}$. It also appears in any right-triangle distance where both legs equal 11, since $11^2 + 11^2 = 242$.
What Is A Square Root?
A square root of a number $n$ is a value $r$ with $r^2 = n$. The square root of 242 is the number that, squared, returns 242.
No whole number works: $15^2 = 225$ and $16^2 = 256$. So $\sqrt{242}$ sits between 15 and 16.
Is The Square Root Of 242 Rational Or Irrational?
$\sqrt{242}$ is irrational. Even after simplifying to $11\sqrt{2}$, the leftover $\sqrt{2}$ is a classic irrational number, so the whole product is irrational.
The reason is in the factorisation. $$242 = 2 \times 11^2$$
The prime 2 appears an odd number of times, so 242 is not a perfect square, and its root is an irrational number. Every non-perfect-square in the square root 1 to 30 list behaves the same way.
How Do You Find √242? (Prime Factorisation And Long Division)
Prime factorisation is the fastest route to the exact form. Break 242 into primes, then pair them.
Step 1: Factor 242. $$242 = 2 \times 121$$
Step 2: Factor 121. $$121 = 11 \times 11$$
Step 3: Pull the pair of 11s out of the radical. $$\sqrt{242} = \sqrt{11^2 \times 2} = 11\sqrt{2}$$
This same simplifying radical expressions move works for any root hiding a perfect-square factor.
Long division gives the decimal. It runs like ordinary long division.
Step 1: Pair the digits. $$\overline{2}\ \overline{42}.\ \overline{00}$$
Step 2: The largest square at most 2 is 1. $$1^2 = 1$$
Step 3: Subtract and bring down 42. $$2 - 1 = 1 \rightarrow 142$$
Step 4: Double the quotient 1 to 2, then find $d$ with $(20 + d)\times d \le 142$. $$25 \times 5 = 125 \le 142$$
Step 5: The quotient is 15, remainder 17. $$242 - 225 = 17$$
Step 6: Continue with pairs of zeros. $$15.5,\ 15.55,\ 15.556$$
So $\sqrt{242} \approx 15.556$, matching $11\sqrt{2}$.
Examples Of √242
Example 1
Write $\sqrt{242}$ in simplest radical form.
Factor out the perfect square. $$242 = 121 \times 2$$ $$\sqrt{242} = \sqrt{121}\times\sqrt{2} = 11\sqrt{2}$$
The simplest radical form is $11\sqrt{2}$.
Example 2
A student simplifies $\sqrt{242}$ as $\sqrt{121} \times \sqrt{2} = 11 \times 2 = 22$. What went wrong?
Follow the wrong path. The error is treating $\sqrt{2}$ as if it equals 2.
Check the size. $$22^2 = 484$$
But 484 is nowhere near 242, so 22 cannot be $\sqrt{242}$.
The fix: only the perfect square 121 leaves the radical as the whole number 11. The 2 stays under the root. $$\sqrt{242} = 11\sqrt{2} \approx 15.556$$
Example 3
Estimate $\sqrt{242}$ to the nearest tenth.
Bracket it with perfect squares. $$15^2 = 225$$ $$16^2 = 256$$
242 is closer to 225, so the root is just past the halfway mark. $$\sqrt{242} \approx 15.6$$
Example 4
Find $(\sqrt{242})^2$ and then $(11\sqrt{2})^2$ to confirm they match.
Squaring the radical returns the number. $$(\sqrt{242})^2 = 242$$
Squaring the simplified form: $$(11\sqrt{2})^2 = 11^2 \times (\sqrt{2})^2 = 121 \times 2 = 242$$
Both give 242, confirming $\sqrt{242} = 11\sqrt{2}$.
Example 5
A square has an area of 242 square centimetres. Give its exact side length.
The side is the square root of the area. $$s = \sqrt{242} = 11\sqrt{2}$$ $$s \approx 15.56 \text{ cm}$$
The exact side is $11\sqrt{2}$ centimetres.
Common Mistakes
Mistake 1: Turning √2 into 2
Where it slips in: Right after pulling 121 out of the radical.
Don't do this: Writing $11\sqrt{2} = 11 \times 2 = 22$.
The correct way: $\sqrt{2} \approx 1.414$, so $11\sqrt{2} \approx 15.556$, and the learners who rush the last step land on 22 and never check it against $15^2$ and $16^2$.
Mistake 2: Leaving √242 unsimplified
Where it slips in: Stopping at $\sqrt{242}$ when an exam wants simplest radical form.
Don't do this: Reporting $\sqrt{242}$ when a perfect-square factor is available.
The correct way: Factor first, so $242 = 11^2 \times 2$, then write $11\sqrt{2}$.
Mistake 3: Mismatching the decimal and radical
Where it slips in: Rounding $11\sqrt{2}$ too soon in a longer calculation.
Don't do this: Replacing $11\sqrt{2}$ with 15.56 at the start and carrying the error forward.
The correct way: Keep $11\sqrt{2}$ exact until the final line, then round once.
Conclusion
The square root of 242 is $11\sqrt{2}$, about 15.556, and irrational.
The perfect square 121 factors out of 242, leaving $\sqrt{2}$ inside.
Prime factorisation gives the exact form; long division gives the decimal.
Never turn $\sqrt{2}$ into 2, the mistake that produces the wrong answer 22.
To build this skill with a teacher, try Bhanzu's algebra tutor, a high school math tutor, or structured algebra classes. Want a guided walkthrough? Book a free demo class.
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