What Is A Square Root?
The square root of a number $n$ is a value $r$ with $r^2 = n$, the number that multiplies by itself to give $n$. For a refresher on the basics, see what is a square root.
No whole number squares to 212, since $14^2 = 196$ and $15^2 = 225$. So $\sqrt{212}$ lies between 14 and 15, and its exact value is the radical $2\sqrt{53}$, a form produced by simplifying radical expressions.
Where Does √212 Appear?
$\sqrt{212}$ shows up as the diagonal of a $14 \times 4$ rectangle, because the Pythagorean theorem gives $\sqrt{14^2 + 4^2} = \sqrt{196 + 16} = \sqrt{212}$. So a right triangle with legs 14 and 4 has a hypotenuse of exactly $2\sqrt{53} \approx 14.56$ units. The simplified form $2\sqrt{53}$ is what you keep in geometry and algebra, because it holds the value exactly instead of rounding.
Quick Reference Table
Number $n$ | $\sqrt{n}$ (approx.) | Simplest Radical Form |
|---|---|---|
196 | 14 | 14 |
200 | 14.142 | $10\sqrt{2}$ |
208 | 14.422 | $4\sqrt{13}$ |
212 | 14.560 | $2\sqrt{53}$ |
216 | 14.697 | $6\sqrt{6}$ |
220 | 14.832 | $2\sqrt{55}$ |
225 | 15 | 15 |
242 | 15.556 | $11\sqrt{2}$ |
245 | 15.652 | $7\sqrt{5}$ |
256 | 16 | 16 |
Is The Square Root Of 212 Rational Or Irrational?
$\sqrt{212}$ is irrational. It cannot be written as a fraction of two integers, and its decimal never terminates or repeats, the defining trait of the irrational numbers.
The cause is that 212 is not a perfect square. It has a perfect-square factor of 4, but the leftover 53 is prime, so the value keeps an irrational $\sqrt{53}$ inside it. The contradiction proof that a prime root like this is irrational is worked out in prove that root 3 is irrational, and it carries over to $\sqrt{53}$ unchanged.
How Do You Simplify The Square Root Of 212?
Break 212 into its prime factors, then remove each matching pair:
$$212 = 2^2 \times 53$$
$$212 = (2 \times 2) \times 53 = 4 \times 53$$
$$\sqrt{212} = \sqrt{4} \times \sqrt{53} = 2\sqrt{53}$$
The pair of 2s leaves the radical as a factor of 2, and 53 stays inside because it is prime. So $2\sqrt{53}$ is the simplest radical form.
How Do You Find √212?
For the decimal value, long division works on any number that is not a perfect square.
Is 212 A Perfect Square?
No. Its last digit is 2, and no perfect square ends in 2, 3, 7, or 8. That single digit rules 212 out immediately, so you know to expect an irrational decimal.
Square Root Of 212 By Long Division
Step 1: The largest square not exceeding 2 is 1, so the first quotient digit is 1, and $2 - 1 = 1$.
Step 2: Bring down 12 to make 112, and double the quotient 1 to get 2.
Step 3: Find a digit $d$ with $(20 + d) \times d \le 112$; here $d = 4$ gives $24 \times 4 = 96$, leaving 16.
Step 4: Add a decimal point, bring down a pair of zeros to make 1600, and double 14 to get 28.
Step 5: Find $d$ with $(280 + d) \times d \le 1600$; here $d = 5$ gives $285 \times 5 = 1425$, leaving 175.
Step 6: Bring down another pair of zeros to make 17500, double 145 to get 290, and $d = 6$ gives $2906 \times 6 = 17436$.
$$\sqrt{212} \approx 14.56$$
Further steps refine the value to $14.560$. Since $2\sqrt{53} = 2 \times 7.280 = 14.560$, the two methods agree, and you can speed up estimates with the nearest-square method in square root tricks or review the routine in how to do long division.
Examples Of √212
Example 1
Write $\sqrt{212}$ in simplest radical form.
$$212 = 4 \times 53$$
$$\sqrt{212} = \sqrt{4} \times \sqrt{53} = 2\sqrt{53}$$
Final answer: $2\sqrt{53}$.
Example 2
A student writes $\sqrt{212} = \sqrt{196 + 16} = \sqrt{196} + \sqrt{16} = 14 + 4 = 18$. Is that correct?
Test the proposed answer by squaring it:
$$18^2 = 324$$
That is 324, not 212, so the step is wrong. The error is splitting the root across a sum, but $\sqrt{a + b}$ is not $\sqrt{a} + \sqrt{b}$.
Factor 212 into a product instead:
$$\sqrt{212} = \sqrt{4 \times 53} = 2\sqrt{53} \approx 14.560$$
That value sits correctly between 14 and 15. Final answer: $2\sqrt{53} \approx 14.560$.
Example 3
Find the decimal value of $\sqrt{212}$ from its simplest form.
$$\sqrt{53} \approx 7.280$$
$$2 \times 7.280 = 14.560$$
Final answer: about $14.560$.
Example 4
Between which two whole numbers does $\sqrt{212}$ lie?
$$14^2 = 196$$
$$15^2 = 225$$
Since $196 < 212 < 225$, the root sits between 14 and 15. Final answer: between 14 and 15.
Example 5
Simplify $\sqrt{212} + \sqrt{53}$.
$$\sqrt{212} = 2\sqrt{53}$$
$$2\sqrt{53} + \sqrt{53} = 3\sqrt{53}$$
Because both terms share the radical $\sqrt{53}$, they add like terms. Final answer: $3\sqrt{53}$.
Common Mistakes
Mistake 1: Splitting the root across addition
Where it slips in: Rewriting 212 as $196 + 16$ to lean on the nearby perfect square.
Don't do this: Writing $\sqrt{212} = \sqrt{196} + \sqrt{16} = 18$.
The correct way: Roots distribute over multiplication, not addition. Factor 212 into $4 \times 53$, which gives the exact $2\sqrt{53}$.
Mistake 2: Trying to reduce √53 further
Where it slips in: After correctly reaching $2\sqrt{53}$, hunting for more to pull out.
Don't do this: Claiming $\sqrt{53}$ simplifies or that it equals a neat decimal.
The correct way: The memorizer who expects every radical to keep shrinking gets stuck here. 53 is prime, so $\sqrt{53}$ has no square factor, and $2\sqrt{53}$ is already final.
Mistake 3: Leaving √212 unsimplified
Where it slips in: Reporting only a rounded decimal when the exact form is asked for.
Don't do this: Writing $\sqrt{212} = 14.56$ and stopping.
The correct way: Check for a perfect-square factor first. Since $212 = 4 \times 53$, the exact answer is $2\sqrt{53}$, and the decimal 14.560 is only an approximation of it.
Conclusion
The square root of 212 is $2\sqrt{53}$, about 14.560, found by pulling the perfect square 4 out of the radical or by long division. To build confidence with radicals alongside a teacher, explore Bhanzu's algebra tutor, get help with algebra, or join math classes online. Ready to practise live? Book a free demo class.
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