The Answer At A Glance
Result: $\sqrt{14} \approx 3.7416574$
Notation: Decimal approximation; exact form is $\sqrt{14}$.
Method shown: Long division by hand, cross-checked with estimation between perfect squares.
Approximate value: $3.7417$ (4 d.p.)
Exact form: $\sqrt{14}$ — cannot be simplified, since $14 = 2 \times 7$ has no square factor.
Quick Reference Table — Square Roots From 5 to 24
$n$ | $\sqrt{n}$ (exact) | $\sqrt{n}$ (4 d.p.) |
|---|---|---|
$5$ | $\sqrt{5}$ | $2.2361$ |
$9$ | $3$ | $3.0000$ |
$10$ | $\sqrt{10}$ | $3.1623$ |
$11$ | $\sqrt{11}$ | $3.3166$ |
$12$ | $2\sqrt{3}$ | $3.4641$ |
$13$ | $\sqrt{13}$ | $3.6056$ |
$14$ | $\boldsymbol{\sqrt{14}}$ | $\boldsymbol{3.7417}$ |
$15$ | $\sqrt{15}$ | $3.8730$ |
$16$ | $4$ | $4.0000$ |
$18$ | $3\sqrt{2}$ | $4.2426$ |
$20$ | $2\sqrt{5}$ | $4.4721$ |
$24$ | $2\sqrt{6}$ | $4.8990$ |
$\sqrt{14}$ sits between $\sqrt{9} = 3$ and $\sqrt{16} = 4$, a little past the midpoint because $14$ is closer to $16$ than to $9$.
Where √14 Appears
$\sqrt{14}$ is the hypotenuse of a right triangle with legs $\sqrt{5}$ and $3$, since $\sqrt{(\sqrt{5})^2 + 3^2} = \sqrt{5 + 9} = \sqrt{14}$. It is also the space diagonal of a box measuring $1 \times 2 \times 3$ — Pythagoras in three dimensions gives $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$. That same sum, $1 + 4 + 9 = 14$, is the sum of the first three square numbers, which is why $\sqrt{14}$ quietly appears in vector-length problems on a $(1, 2, 3)$ direction.
What "square root of 14" Means
The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{14}$, it is the positive $x$ with $x^2 = 14$.
Because $3^2 = 9$ and $4^2 = 16$, the answer must land between $3$ and $4$ — and $3.742^2 = 14.0026\ldots$, which confirms the value.
Is The Square Root of 14 Rational or Irrational?
$\sqrt{14}$ is irrational. Its prime factorisation is $14 = 2 \times 7$ — two distinct primes, each appearing once, so no prime sits at an even power and $14$ is not a perfect square.
Because there is no repeated factor to pull out, $\sqrt{14}$ also cannot be simplified into a smaller radical. The decimal $3.7416574\ldots$ never terminates and never repeats, which is the signature of an irrational number.
How To Find √14 — Two Methods
Method 1 — Long division (digit by digit)
Write $14$ as $14.000000$ and pair the digits.
Step 1. The largest integer whose square is at most $14$ is $3$ ($3^2 = 9$). Subtract: $14 - 9 = 5$. Bring down $00$ to get $500$.
Step 2. Double the quotient $3$ to get $6$. Find $d$ with $(60 + d)\cdot d \leq 500$. Here $d = 7$ gives $67 \cdot 7 = 469$. Subtract: $500 - 469 = 31$. Bring down $00$ to get $3100$.
Step 3. Double $3.7$ to get $74$. Find $d$ with $(740 + d)\cdot d \leq 3100$. Here $d = 4$ gives $744 \cdot 4 = 2976$. Subtract: $3100 - 2976 = 124$.
Continuing produces $3.7416\ldots$
Final answer: $\sqrt{14} \approx 3.7417$.
Method 2 — Estimation between perfect squares
Since $3^2 = 9$ and $4^2 = 16$, start at $3.7$: $3.7^2 = 13.69$, a little low. Try $3.74$: $3.74^2 = 13.9876$. Try $3.742$: $3.742^2 = 14.002564$. Each refinement closes in on $14$, settling at $3.742$ for everyday use — the average (Babylonian) method by inspection.
What Are The most Common Mistakes With √14?
Mistake 1: Trying to simplify a non-square radicand
Where it slips in: A student factorises $14 = 2 \times 7$ and then tries to take one factor out of the root.
Don't do this: $\sqrt{14} = \sqrt{2},\sqrt{7} = 2\sqrt{7}$ or $7\sqrt{2}$.
The correct way: A factor only leaves the radical when it appears as a pair. Here $2$ and $7$ each appear once, so nothing comes out — $\sqrt{14}$ is already simplest.
Mistake 2: Splitting the root over addition
Where it slips in: When $\sqrt{14}$ appears as $\sqrt{5 + 9}$ inside a Pythagoras calculation.
Don't do this: $\sqrt{5 + 9} = \sqrt{5} + \sqrt{9} = 2.236 + 3 = 5.236$.
The correct way: $\sqrt{5 + 9} = \sqrt{14} \approx 3.742$. Square roots distribute over multiplication, never over addition.
Mistake 3: Confusing −√14 with √−14
Where it slips in: Reading a minus sign as if it lives inside or outside the radical interchangeably.
Don't do this: Treating $-\sqrt{14}$ and $\sqrt{-14}$ as the same thing.
The correct way: $-\sqrt{14} \approx -3.742$ is a real number; $\sqrt{-14}$ is not real (it is the imaginary number $i\sqrt{14}$). The sign's position changes everything.
Examples of Square Root of 14
Example 1
Confirm that $\sqrt{14}$ does not simplify.
$14 = 2 \times 7$, two single primes, so $\sqrt{14}$ stays as is — about $3.742$.
Example 2 (Wrong path first)
Find the space diagonal of a $1 \times 2 \times 3$ box.
Wrong attempt. A student adds the edges: diagonal $= 1 + 2 + 3 = 6$.
Why it breaks. The straight diagonal through a box is always shorter than walking along three edges; a value of $6$ equals that full edge path, which a direct diagonal can never reach.
Correct. $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \approx 3.742$.
Example 3
Evaluate $\sqrt{14} \times \sqrt{14}$.
$\sqrt{14} \times \sqrt{14} = 14$. A square root undoes the square — the radical disappears exactly.
Example 4
Rationalise $\dfrac{7}{\sqrt{14}}$.
Multiply top and bottom by $\sqrt{14}$: $\dfrac{7}{\sqrt{14}} = \dfrac{7\sqrt{14}}{14} = \dfrac{\sqrt{14}}{2} \approx 1.871$.
Example 5
A square has area $14$ square centimetres. Find its side length.
Side $= \sqrt{14} \approx 3.74$ cm. The area is a whole number, yet the side is irrational — normal when the area isn't a perfect square.
Conclusion
The square root of 14 is approximately $3.742$ — irrational, non-terminating, non-repeating.
$14 = 2 \times 7$ has no repeated prime, so $\sqrt{14}$ cannot be simplified.
Long division and estimation both reach the value by hand.
A factor leaves a radical only when it appears as a pair — single primes stay inside.
$\sqrt{14}$ is the space diagonal of a $1 \times 2 \times 3$ box, from $1 + 4 + 9 = 14$.
A Practical Next Step
Find $\sqrt{13}$ to three decimal places by long division and check by squaring.
Show that $\sqrt{12}$ simplifies to $2\sqrt{3}$ while $\sqrt{14}$ does not simplify.
A box measures $2 \times 3 \times 4$. Find its space diagonal in exact and decimal form.
Want a live Bhanzu trainer to walk through more square-root problems? Book a free demo class — online globally.
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