$\sqrt{50} = 5\sqrt{2} \approx 7.0710678$ — irrational, non-terminating, non-repeating. The simplified radical form $5\sqrt{2}$ is what most textbook problems expect; the decimal $7.0711$ is what a calculator returns.
Result: $\sqrt{50} = 5\sqrt{2} \approx 7.0711$
Notation: Simplified radical form $5\sqrt{2}$; decimal approximation $7.0711$ (4 d.p.).
Method shown: Prime factorization (Quick), repeated subtraction (Standard, with a Wrong-Path-First detour), long division (Stretch).
Approximate value: $7.07106781187$ (11 d.p.).
Exact form: $5\sqrt{2}$ — cannot be reduced further, since $2$ is prime.
Quick Reference Table — Square Roots Near 50
$n$ | $\sqrt{n}$ (exact form) | $\sqrt{n}$ (4 d.p.) | Nearest perfect square |
|---|---|---|---|
$45$ | $3\sqrt{5}$ | $6.7082$ | $49$ |
$48$ | $4\sqrt{3}$ | $6.9282$ | $49$ |
$49$ | $7$ | $7.0000$ | $49$ (itself) |
$50$ | $\boldsymbol{5\sqrt{2}}$ | $\boldsymbol{7.0711}$ | $49$ |
$52$ | $2\sqrt{13}$ | $7.2111$ | $49$ |
$54$ | $3\sqrt{6}$ | $7.3485$ | $49$ |
$56$ | $2\sqrt{14}$ | $7.4833$ | $49$ |
$60$ | $2\sqrt{15}$ | $7.7460$ | $64$ |
$64$ | $8$ | $8.0000$ | $64$ (itself) |
$\sqrt{50}$ sits between $\sqrt{49} = 7$ and $\sqrt{64} = 8$ — closer to $7$ because $50$ is one away from $49$, and $14$ away from $64$.
Where √50 shows up
$\sqrt{50}$ is the diagonal of a $5 \times 5$ square — by Pythagoras, $\sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$. The same value appears as the hypotenuse of any right triangle with two legs of length $5$, and as the distance between the points $(0, 0)$ and $(5, 5)$ on a coordinate plane. In physics, $\sqrt{50}$ shows up in the magnitude of a vector with equal $x$ and $y$ components of $5$ units.
What "square root of 50" means
The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{50}$, the positive $x$ with $x^2 = 50$. Because $7^2 = 49$ and $8^2 = 64$, $\sqrt{50}$ lives between $7$ and $8$.
The simplified-radical form $5\sqrt{2}$ extracts the part of $50$ that is a perfect square (the $25$) and leaves the rest under the radical. The decimal form $7.0711$ comes from numerically approximating the irrational tail.
Is √50 rational or irrational?
$\sqrt{50}$ is irrational. Reason: a number is a perfect square if and only if every prime in its factorisation appears to an even power. $50 = 2^{1} \cdot 5^{2}$ — the $5$ pairs up neatly into $5^2$, but the $2$ stands alone with exponent $1$. That lone $2$ is the reason $\sqrt{50}$ doesn't collapse to an integer.
The decimal $7.0710678118\ldots$ neither terminates nor repeats. Like every irrational, it can be approximated but never written out exactly in decimal form.
How to find √50 — three methods
Method 1 — Prime factorization (Quick)
Break $50$ into prime factors, then pair them up.
$$50 = 2 \cdot 5 \cdot 5 = 2 \cdot 5^{2}$$
Pull the pair of $5$s out of the radical:
$$\sqrt{50} = \sqrt{2 \cdot 5^{2}} = 5\sqrt{2}$$
Final answer: $\sqrt{50} = 5\sqrt{2} \approx 7.0711$.
This is the method most useful when the question asks for simplified radical form. It takes one line once you see the pair of $5$s.
Method 2 — Repeated subtraction (Standard, Wrong-Path-First)
A common student instinct is to try subtracting odd numbers from $50$ — the trick that works for perfect squares. Let us walk it through and let it break.
Start at $50$. Subtract consecutive odd numbers: $50 - 1 = 49$, $49 - 3 = 46$, $46 - 5 = 41$, $41 - 7 = 34$, $34 - 9 = 25$, $25 - 11 = 14$, $14 - 13 = 1$. After seven subtractions, the remainder is $1$ — not $0$.
That non-zero remainder is the signal: $50$ is not a perfect square. The method only finds a clean integer root when subtraction ends exactly at zero (try it on $49$ — seven subtractions, remainder $0$, so $\sqrt{49} = 7$).
The rescue is prime factorization. $50 = 2 \cdot 25$, so $\sqrt{50} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}$. The repeated-subtraction step still earned its keep — it told us within seven moves that we'd need the radical form, not an integer.
Final answer: $\sqrt{50} = 5\sqrt{2}$.
Method 3 — Long division (Stretch)
For a decimal approximation to four places, run the long-division algorithm on $50.000000$.
Step 1. Largest integer with square $\leq 50$ is $7$ ($7^2 = 49$). Subtract: $50 - 49 = 1$. Bring down $00$: $100$.
Step 2. Double $7$: $14$. Find $d$ with $(140 + d) \cdot d \leq 100$. $d = 0$ gives $140 \cdot 0 = 0$. Subtract: $100 - 0 = 100$. Bring down $00$: $10{,}000$.
Step 3. Double $7.0$: $140$. Find $d$ with $(1400 + d) \cdot d \leq 10{,}000$. $d = 7$ gives $1407 \cdot 7 = 9849$. Subtract: $10{,}000 - 9849 = 151$. Bring down $00$: $15{,}100$.
Step 4. Double $7.07$: $1414$. Find $d$ with $(14{,}140 + d) \cdot d \leq 15{,}100$. $d = 1$ gives $14{,}141 \cdot 1 = 14{,}141$. Subtract: $15{,}100 - 14{,}141 = 959$. Bring down $00$: $95{,}900$.
Continuing produces $\sqrt{50} \approx 7.0710$ to four decimals, and $7.0711$ rounded.
Final answer: $\sqrt{50} \approx 7.0711$.
Tripping points to avoid on √50
1. Treating $\sqrt{50}$ as $25$ — confusing the operation with division by 2
Where it slips in: Students rushing through the problem read "square root" as "half" and answer $25$.
Don't do this: $\sqrt{50} = 50 \div 2 = 25$.
The correct way: Square root asks: what number multiplied by itself gives $50$? Not $25 \cdot 25 = 625$. The rusher archetype hits this slip on the first attempt; the fix is to check by squaring back — if $x^2 \neq 50$, $x$ is wrong.
2. Stopping at $\sqrt{50} \approx 7.07$ when the question asks for radical form
Where it slips in: A test question says "express in simplest radical form." The student returns $7.07$ or $7.0711$.
Don't do this: $\sqrt{50} = 7.07$ — written as the final answer when the question asked for an exact form.
The correct way: $\sqrt{50} = 5\sqrt{2}$. The decimal is an approximation; $5\sqrt{2}$ is exact. Read the question's verb — "evaluate" expects decimal, "simplify" expects the radical.
3. Pulling out the wrong factor under the radical
Where it slips in: Students see $50 = 2 \cdot 25$ but try to pull both out: $\sqrt{50} = 2 \cdot 5 = 10$.
Don't do this: $\sqrt{2 \cdot 25} = 2 \cdot 5 = 10$.
The correct way: Only the perfect-square factor leaves the radical, and it leaves as its square root — $\sqrt{25} = 5$, not $25$. The $2$ stays under, because $2$ has no integer square root. So $\sqrt{50} = 5 \cdot \sqrt{2} = 5\sqrt{2}$.
Conclusion
The square root of 50 simplifies to $5\sqrt{2}$ — exact form — and approximates to $7.0711$ as a decimal.
$\sqrt{50}$ is irrational; the prime factor $2$ inside the radical has no partner.
Three methods reach the answer: prime factorization (one line), repeated subtraction (signals the irrationality), long division (gives the decimal).
Only perfect-square factors leave the radical, and they leave as their square root — not as themselves.
$\sqrt{50}$ is the diagonal of any $5 \times 5$ square, which is why the Pythagorean theorem reproduces it constantly.
A practical next step
Show that $\sqrt{72}$ simplifies to $6\sqrt{2}$ using prime factorization.
Find $\sqrt{50}$ to two decimal places using long division (without copying the worked example).
A square has side length $5$ cm. Find its 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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