What Is a Square Root?
A square root of a number is a value that multiplies by itself to give that number. Squaring and taking the square root are inverse operations, so $(\sqrt{243})^2 = 243$.
To write a root in simplest radical form, you remove the largest perfect-square factor. The full method sits in the guide to simplifying radical expressions, and the pattern of exact and inexact roots is set out in squares and square roots.
Where Does the Square Root of 243 Appear?
$\sqrt{243}$ appears as a length whenever a squared distance equals 243, for example the diagonal of a rectangle whose side-squares total 243, or an answer in a Pythagorean setup that lands on 243. It also shows up when simplifying powers of 3, since $243 = 3^5$, so its root sits neatly in the $\sqrt{3}$ family and keeps its exact form $9\sqrt{3}$ through a calculation instead of the rounded $\approx 15.588$.
Quick Reference Table
The table shows $\sqrt{243}$ inside the $\sqrt{3}$ family, where each radicand is a perfect square times 3. Only 225 and 256 give whole-number roots.
Number | Square Root | Type |
|---|---|---|
$\sqrt{225}$ | $15$ | Rational (perfect square) |
$\sqrt{243}$ | $9\sqrt{3} \approx 15.588$ | Irrational |
$\sqrt{256}$ | $16$ | Rational (perfect square) |
$\sqrt{12}$ | $2\sqrt{3} \approx 3.464$ | Irrational ($\sqrt{3}$ family) |
$\sqrt{27}$ | $3\sqrt{3} \approx 5.196$ | Irrational ($\sqrt{3}$ family) |
$\sqrt{48}$ | $4\sqrt{3} \approx 6.928$ | Irrational ($\sqrt{3}$ family) |
$\sqrt{3}$ | $\approx 1.732$ | Irrational |
Is the Square Root of 243 Rational or Irrational?
$\sqrt{243}$ is irrational. It simplifies to $9\sqrt{3}$, and since √3 is irrational, any whole-number multiple of it is irrational too.
The reason is the leftover factor. $243 = 3^5$, so four of the five 3s pair off into the perfect square $3^4 = 81$, but one 3 stays under the radical with no pair. The same argument used to prove that root 3 is irrational then shows $9\sqrt{3}$ has a non-terminating, non-repeating decimal.
How Do You Find the Square Root of 243?
Method 1: Prime factorization and simplification
Break 243 into prime factors.
$243 = 3^5$
Separate the perfect-square part from the leftover.
$\sqrt{243} = \sqrt{3^4} \times \sqrt{3}$
$\sqrt{3^4} = 3^2 = 9$.
$\sqrt{243} = 9\sqrt{3}$.
Final answer: $\sqrt{243} = 9\sqrt{3}$
Method 2: Largest perfect-square factor
Find the largest perfect square that divides 243.
$243 = 81 \times 3$, and 81 is a perfect square.
$\sqrt{243} = \sqrt{81} \times \sqrt{3}$
$\sqrt{81} = 9$, so $\sqrt{243} = 9\sqrt{3}$.
Final answer: $\sqrt{243} = 9\sqrt{3}$
Method 3: Decimal estimation
Estimate $\sqrt{3}$ and scale it.
$\sqrt{3} \approx 1.7321$, and $15^2 = 225$ confirms $\sqrt{243}$ is above 15.
$9 \times 1.7321 = 15.588$.
So $\sqrt{243} \approx 15.588$.
Final answer: $\sqrt{243} \approx 15.588$
Examples of the Square Root of 243
Example 1: The Mistake Worth Making Once
A student pulls out 9, writes $\sqrt{243} = 3\sqrt{27}$, and stops.
The radical is not fully simplified, because 27 still holds a perfect-square factor.
$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$, so $3\sqrt{27} = 3 \times 3\sqrt{3} = 9\sqrt{3}$.
Always remove the largest perfect square, or keep going until none remains.
Example 2: Simplifying by Prime Factorization
Simplify $\sqrt{243}$ from scratch.
$243 = 3^5$.
$\sqrt{243} = \sqrt{3^4} \times \sqrt{3} = 9\sqrt{3}$.
The simplest radical form is $9\sqrt{3}$.
Example 3: Checking the Decimal Value
Find $\sqrt{243}$ as a decimal.
$\sqrt{243} = 9\sqrt{3}$ and $\sqrt{3} \approx 1.7321$.
$9 \times 1.7321 = 15.588$.
So $\sqrt{243} \approx 15.588$.
Example 4: Verifying the Simplified Form
Confirm that $9\sqrt{3}$ squares back to 243.
$(9\sqrt{3})^2 = 9^2 \times (\sqrt{3})^2$.
$= 81 \times 3$.
$= 243$, which checks out.
Example 5: Solving x² = 243
Solve $x^2 = 243$ for $x$.
Take the square root of both sides.
$x = \pm\sqrt{243} = \pm 9\sqrt{3}$.
$x \approx 15.588$ or $x \approx -15.588$.
Common Mistakes
Mistake 1: Stopping Before the Radical Is Fully Simplified
Where it slips in: A student removes a small square factor and leaves another one inside.
Don't do this: Write $\sqrt{243} = 3\sqrt{27}$ and treat it as the final answer.
The correct way: Keep factoring until no perfect square remains inside: $\sqrt{243} = 9\sqrt{3}$.
Mistake 2: Merging the Coefficient Into the Radical
Where it slips in: After reaching $9\sqrt{3}$, a student folds the 9 back under the radical.
Don't do this: Write $9\sqrt{3} = \sqrt{27}$ by combining 9 and 3 incorrectly.
The correct way: The 9 is a coefficient outside the radical; $9\sqrt{3} \approx 15.588$, while $\sqrt{27} \approx 5.196$.
Mistake 3: Confusing √243 With 243²
Where it slips in: Reading quickly, a student squares 243 instead of rooting it.
Don't do this: Answer 59049 for $\sqrt{243}$; that is $243^2$, the opposite operation.
The correct way: The square root asks what number times itself gives 243, which is $9\sqrt{3} \approx 15.588$.
Conclusion
The square root of 243 is $9\sqrt{3}$, approximately 15.588, and is irrational.
$243 = 3^5$, so the perfect-square factor 81 pulls out as 9 and leaves $\sqrt{3}$.
It sits between 15 and 16, closer to 16, because 243 is near the perfect square 256.
Prime factorization and largest-square-factor methods both give $9\sqrt{3}$.
Keep the exact form $9\sqrt{3}$ through a calculation and convert to $\approx 15.588$ only at the end.
To master radical simplification with a teacher, explore Bhanzu's algebra tutor sessions or browse math classes online. Want a live Bhanzu trainer to walk through more radical problems? Book a free demo class.
For a formal treatment of the square-root operation, see the Wolfram MathWorld entry on square roots.
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