Square Root of 25 — Value, Method, and Examples

The Square Root of 25 Is 5

The square root of 25 is 5. It is a perfect square: $5 \times 5 = 25$, so $\sqrt{25} = 5$ with no decimal tail and no rounding.

This is one of the first square roots worth knowing by heart — it anchors the $3$–$4$–$5$ triangle, the most-used right triangle in school geometry.

Quick Answer

Quick Answer:

Result: $\sqrt{25} = 5$

Notation: radical form $\sqrt{25}$; exponent form $25^{1/2}$.

Method shown: prime factorisation, with cross-checks by repeated subtraction and long division.

Rational or irrational: rational — $5$ can be written as $\tfrac{5}{1}$.

Exact form: $5$ (an integer; no radical remains).

Quick Reference Table — Square Roots of Nearby Perfect Squares

Every value in this table is a whole number — these ten are the perfect squares from $1$ to $100$, and knowing them on sight makes most square-root work fast.

Where the square root of 25 appears

The $\sqrt{25} = 5$ result sits at the centre of the $3$–$4$–$5$ right triangle: legs of $3$ and $4$ give a hypotenuse of $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. That single triple shows up in the Pythagorean theorem, in carpentry for squaring a corner, and in the distance between the points $(0,0)$ and $(3,4)$. It also appears whenever a quadratic has a discriminant of $25$, since $\sqrt{25} = 5$ keeps the roots rational.

What "square root of 25" Means

A square root of a number $n$ is a value $x$ for which $x^2 = n$. For $25$, the value is $5$, since $5^2 = 25$.

The radical symbol $\sqrt{;}$ asks for the principal (non-negative) root, so $\sqrt{25} = 5$. The equation $x^2 = 25$ has two solutions, $x = 5$ and $x = -5$, but the symbol $\sqrt{25}$ on its own names only the positive one.

How To Find The Square Root of 25

Method 1 — Prime factorisation

Break $25$ into prime factors, then pair them.

$$25 = 5 \times 5$$

One factor leaves each pair, so $\sqrt{25} = 5$. This is the cleanest method for any perfect square.

Final answer: $\sqrt{25} = 5$.

Method 2 — Repeated subtraction of odd numbers

Subtract successive odd numbers from $25$ and count the steps until you reach $0$.

$25 - 1 = 24$, $;24 - 3 = 21$, $;21 - 5 = 16$, $;16 - 7 = 9$, $;9 - 9 = 0$.

Five subtractions reached zero, so $\sqrt{25} = 5$. (This works because $n^2$ is the sum of the first $n$ odd numbers.)

Method 3 — Long division

Pair the digits: $\overline{25}$. The largest integer whose square is at most $25$ is $5$, and $5^2 = 25$ exactly. Subtracting leaves a remainder of $0$, so the root terminates.

Final answer: $\sqrt{25} = 5$.

Examples of Square Root of 25

Example 1

Evaluate $\sqrt{25}$ directly.

$5^2 = 25$, so $\sqrt{25} = 5$. A clean integer — no approximation needed.

Example 2

A student is asked: does $x^2 = 25$ mean $x = \sqrt{25}$?

Wrong attempt. Writing $x = \sqrt{25} = 5$ as the only answer drops a solution. Check: $(-5)^2 = 25$ as well, so $-5$ also satisfies the equation.

Correct. The equation $x^2 = 25$ has two roots, $x = \pm 5$. The symbol $\sqrt{25}$ names only the principal root, $5$. The two ideas are different — solving an equation is not the same as evaluating a radical.

Example 3

Simplify $3\sqrt{25}$.

$\sqrt{25} = 5$, so $3\sqrt{25} = 3 \times 5 = 15$.

Example 4

Find the hypotenuse of a right triangle with legs $3$ and $4$.

$c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. The $3$–$4$–$5$ triangle.

Example 5

Solve $\sqrt{25x^2}$ for $x > 0$.

$\sqrt{25x^2} = \sqrt{25}\cdot\sqrt{x^2} = 5x$. The square root distributes over a product (unlike over a sum).

Where Students Trip Up on the Square Root of 25

Mistake 1: Writing √25 as ±5

Where it slips in: A student confuses solving $x^2 = 25$ with evaluating the radical $\sqrt{25}$.

Don't do this: $\sqrt{25} = \pm 5$.

The correct way: $\sqrt{25} = 5$. The radical returns only the principal (positive) root; the $\pm$ belongs to equation-solving, not to the symbol itself.

Mistake 2: Confusing the square root with halving

Where it slips in: Treating $\sqrt{25}$ as $25 \div 2$.

Don't do this: $\sqrt{25} = 12.5$.

The correct way: $\sqrt{25}$ asks "what times itself gives $25$?" — the answer is $5$, not half of $25$. A quick check, $5 \times 5 = 25$, catches the error.

Mistake 3: Calling √25 irrational

Where it slips in: Lumping every square root in with $\sqrt{2}$ or $\sqrt{3}$.

Don't do this: Assuming $\sqrt{25}$ never terminates.

The correct way: $25$ is a perfect square, so $\sqrt{25} = 5$ is a whole number — clearly rational, since $5 = \tfrac{5}{1}$.

Conclusion

• The square root of 25 is $5$, an exact whole number, because $25$ is a perfect square ($5 \times 5 = 25$).

• $\sqrt{25}$ names only the principal root, $5$; the equation $x^2 = 25$ is what gives $\pm 5$.

• Prime factorisation is the fastest route: $25 = 5 \times 5$, so $\sqrt{25} = 5$.

• Because $5$ is an integer, $\sqrt{25}$ is rational, not irrational.

• $\sqrt{25} = 5$ is the heart of the $3$–$4$–$5$ right triangle.

A Practical Next Step

1. Evaluate $\sqrt{16}$ and $\sqrt{36}$ from memory, then check by squaring.

2. Solve $x^2 = 25$ and list both solutions — then write $\sqrt{25}$ and explain why it gives only one.

3. Use $\sqrt{25} = 5$ to find the hypotenuse of a $3$–$4$ legged triangle.

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