Square Root of 5 — Value, How to Find It, and Examples

The Answer At A Glance

Quick Answer:

Result: $\sqrt{5} \approx 2.2360680$

Notation: Decimal approximation; exact form is $\sqrt{5}$.

Method shown: Long division by hand, cross-checked with estimation between perfect squares.

Approximate value: $2.2361$ (4 d.p.)

Exact form: $\sqrt{5}$ — cannot be simplified, since $5$ is prime.

Quick Reference Table — Square Roots From 1 to 20

$\sqrt{5}$ sits between $\sqrt{4} = 2$ and $\sqrt{9} = 3$, close to $2.24$ because $5$ is only just past $4$.

Where √5 Appears

$\sqrt{5}$ is the diagonal of a $1 \times 2$ rectangle — Pythagoras gives $\sqrt{1^2 + 2^2} = \sqrt{5}$. Its most famous role is inside the golden ratio: $\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.618$, the number that the ratios of consecutive Fibonacci numbers ($1, 1, 2, 3, 5, 8, \ldots$) converge toward. So $\sqrt{5}$ quietly sits behind golden rectangles, pentagons, and the spiral patterns the golden ratio describes.

What "square root of 5" Means

The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{5}$, it is the positive $x$ with $x^2 = 5$.

Because $2^2 = 4$ and $3^2 = 9$, the answer must land between $2$ and $3$ — and $2.236^2 = 4.9997\ldots$, which confirms it.

Is The Square Root of 5 Rational or Irrational?

$\sqrt{5}$ is irrational. A whole number is a perfect square only when every prime in its factorisation appears to an even power; $5$ is prime, so it carries the single factor $5^1$ — an odd exponent — and is not a perfect square.

That means $\sqrt{5}$ cannot be written as a fraction $p/q$, and its decimal $2.2360680\ldots$ never terminates and never repeats. As of recent computations the digits have been calculated to trillions of places, with no pattern emerging — exactly what irrationality predicts.

How To Find √5 — Two Methods

Method 1 — Long division (digit by digit)

Write $5$ as $5.000000$ and pair the digits after the decimal point.

Step 1. The largest integer whose square is at most $5$ is $2$ ($2^2 = 4$). Subtract: $5 - 4 = 1$. Bring down $00$ to get $100$.

Step 2. Double the quotient $2$ to get $4$. Find $d$ with $(40 + d)\cdot d \leq 100$. Here $d = 2$ gives $42 \cdot 2 = 84$. Subtract: $100 - 84 = 16$. Bring down $00$ to get $1600$.

Step 3. Double $2.2$ to get $44$. Find $d$ with $(440 + d)\cdot d \leq 1600$. Here $d = 3$ gives $443 \cdot 3 = 1329$. Subtract: $1600 - 1329 = 271$.

Continuing produces $2.2360\ldots$

Final answer: $\sqrt{5} \approx 2.2361$.

Method 2 — Estimation between perfect squares

Since $2^2 = 4$ and $3^2 = 9$, start at $2.2$: $2.2^2 = 4.84$, a little low. Try $2.23$: $2.23^2 = 4.9729$. Try $2.236$: $2.236^2 = 4.999696$. Each guess closes in on $5$, settling at $2.236$ for everyday work — the same logic as the average (Babylonian) method.

What are the most common mistakes with √5?

Mistake 1: Trying to simplify a prime radicand

Where it slips in: A student applies the "pull out a square factor" rule before checking whether $5$ has one.

Don't do this: Writing $\sqrt{5} = \sqrt{4} + \sqrt{1} = 2 + 1 = 3$, or splitting the radicand to "simplify."

The correct way: $5$ is prime — no square factor — so $\sqrt{5}$ is already simplest.

Mistake 2: Splitting the root over addition

Where it slips in: When $\sqrt{5}$ shows up as $\sqrt{1 + 4}$ inside the diagonal calculation.

Don't do this: $\sqrt{1 + 4} = \sqrt{1} + \sqrt{4} = 1 + 2 = 3$.

The correct way: $\sqrt{1 + 4} = \sqrt{5} \approx 2.236$. Square roots do not distribute over addition.

Mistake 3: Misplacing √5 in the golden-ratio formula

Where it slips in: Recalling $\varphi$ but forgetting the division by $2$.

Don't do this: $\varphi = 1 + \sqrt{5} = 3.236$.

The correct way: $\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.618$. The whole numerator is halved, not just the $1$.

Examples of Square Root of 5

Example 1

Simplify $\sqrt{20}$ using $\sqrt{5}$.

$\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} \approx 4.4721$. The $4$ comes out as $2$; the $5$ stays under the root.

Example 2 (Wrong path first)

Find the diagonal of a $1 \times 2$ rectangle.

Wrong attempt. A student adds the sides: diagonal $= 1 + 2 = 3$.

Why it breaks. The diagonal of a rectangle is always shorter than going along two sides; a length of $3$ would equal the full perimeter path, which is impossible for a straight diagonal.

Correct. Use Pythagoras: diagonal $= \sqrt{1^2 + 2^2} = \sqrt{5} \approx 2.236$.

Example 3

Compute the golden ratio from $\sqrt{5}$.

$\varphi = \dfrac{1 + \sqrt{5}}{2} = \dfrac{1 + 2.236}{2} = \dfrac{3.236}{2} \approx 1.618$.

Example 4

Rationalise $\dfrac{2}{\sqrt{5}}$.

Multiply top and bottom by $\sqrt{5}$: $\dfrac{2}{\sqrt{5}} = \dfrac{2\sqrt{5}}{5} \approx 0.8944$.

Example 5

Evaluate $\sqrt{45} + \sqrt{5}$.

$\sqrt{45} = 3\sqrt{5}$, so $\sqrt{45} + \sqrt{5} = 3\sqrt{5} + \sqrt{5} = 4\sqrt{5} \approx 8.944$. Like terms in $\sqrt{5}$ add the way $3x + x = 4x$ does.

Conclusion

• The square root of 5 is approximately $2.236$ — irrational, non-terminating, non-repeating.

• $5$ is prime, so $\sqrt{5}$ cannot be simplified.

• Long division and estimation both reach the value by hand.

• $\sqrt{5}$ is the diagonal of a $1 \times 2$ rectangle and the engine inside the golden ratio.

• Square roots do not distribute over addition: $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$.

A practical next step

1. Find $\sqrt{6}$ to three decimal places by long division and check by squaring.

2. Show that $\sqrt{20}$ simplifies to $2\sqrt{5}$ while $\sqrt{5}$ does not simplify.

3. Use $\varphi = \frac{1 + \sqrt{5}}{2}$ to verify that $\varphi^2 = \varphi + 1$.

Want a live Bhanzu trainer to walk through more square-root problems? Book a free demo class — online globally.