Square Root Of 40: Value, Simplified Form And Methods

What Is A Square Root?

A square root of a number $n$ is a value $r$ with $r^2 = n$. The symbol $\sqrt{n}$ always means the non-negative one, called the principal square root.

Every positive number has two square roots, one positive and one negative. For 40, they are $2\sqrt{10}$ and $-2\sqrt{10}$, and $\sqrt{40}$ names only the positive one.

Where Does The Square Root Of 40 Show Up, And Why Does It Matter?

The square root of 40 is the length of the diagonal of a rectangle that measures 2 by 6. By the Pythagorean theorem, that diagonal is $\sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$.

It is also the side of any square with an area of 40 square units. A square tabletop of 40 square feet is $\sqrt{40} \approx 6.32$ feet along each edge, which is the number a carpenter or a tile-cutter actually needs.

The same distance appears on a coordinate grid. The points $(0, 0)$ and $(2, 6)$ are exactly $\sqrt{40} = 2\sqrt{10}$ units apart, twice the distance from $(0, 0)$ to $(1, 3)$.

Quick Reference Table

Here are the square roots from 36 to 49, the two perfect squares on either side of 40. The exact column shows the simplest radical form.

Only the numbers with a perfect-square factor (40, 44, 45, 48) simplify. The rest, such as $\sqrt{37}$ and $\sqrt{41}$, are already in simplest form.

Is 40 A Perfect Square?

No, 40 is not a perfect square. It sits between $6^2 = 36$ and $7^2 = 49$, and no whole number squared lands on 40.

The repeated-subtraction test shows it directly. Subtract the odd numbers 1, 3, 5, 7, 9, 11 in turn from 40:

The remainder 4 is smaller than the next odd number, 13, so the subtraction never reaches 0. For a perfect square it always would; the six steps taken also tell you the whole-number part of $\sqrt{40}$ is 6.

The prime factors say the same thing. $40 = 2^3 \times 5$, and a perfect square needs every prime exponent to be even, while 3 and 1 are both odd.

Is The Square Root Of 40 Rational Or Irrational?

The square root of 40 is irrational. It cannot be written as a fraction $\frac{p}{q}$ of whole numbers, and its decimal never ends or repeats.

Here is the proof sketch. Suppose $\sqrt{40} = \frac{p}{q}$ with whole numbers $p$ and $q$; squaring gives $p^2 = 40q^2$.

Now count the factors of 5 on each side. In $p^2$ the number of 5s is even (every prime appears twice as often in a square), but $40q^2 = 2^3 \times 5 \times q^2$ has an odd number of 5s.

A whole number cannot have both an even and an odd count of the same prime, so no such fraction exists. The same argument shows $\sqrt{10}$ is an irrational number, which fits, since $\sqrt{40} = 2\sqrt{10}$.

How Do You Simplify The Square Root Of 40?

To simplify $\sqrt{40}$, pull out the largest perfect square that divides 40. That square is 4, because $40 = 4 \times 10$ and 10 has no square factor left.

$$\sqrt{40} = \sqrt{4 \times 10}$$

$$\sqrt{40} = \sqrt{4} \times \sqrt{10}$$

$$\sqrt{40} = 2\sqrt{10}$$

So the square root of 40 in simplest radical form, also called surd form, is $2\sqrt{10}$. It cannot go further, because $10 = 2 \times 5$ contains no repeated prime.

This also links two pages you may already know. Since $\sqrt{40} = 2\sqrt{10}$, the decimal is just twice the square root of 10: $2 \times 3.16227766 = 6.32455532$.

How Do You Find The Square Root Of 40?

Simplifying gives the exact form, but the decimal needs a method. Here are four, from the one textbooks ask you to write out to the quickest checks.

Method 1: Prime Factorization

Break 40 into primes with a factor tree:

$$40 = 2 \times 20 = 2 \times 2 \times 10 = 2 \times 2 \times 2 \times 5$$

Pair equal primes. One pair of 2s comes out of the root as a single 2, and the unpaired 2 and 5 stay inside:

$$\sqrt{40} = \sqrt{(2 \times 2) \times 2 \times 5}$$

$$\sqrt{40} = 2\sqrt{10}$$

Method 2: Long Division

Long division gives the decimal one digit at a time. Write 40 as $40.00,00,00,00,00$ and group the digits in pairs from the decimal point.

The rule at each step: double the root so far, write it as a trial divisor with a blank on the end, and find the largest digit $d$ so that (trial divisor with $d$) times $d$ fits under the current number.

Each digit is the largest that fits. In step 2, for example, $124 \times 4 = 496$ is already more than 400, so the digit is 3.

The root to five decimal places is $6.32455$, which rounds to $\sqrt{40} \approx 6.3246$. The remainder never reaches 0, which is the long-division picture of an irrational root.

Method 3: Estimation Between Perfect Squares

Since $36 < 40 < 49$, the root lies between 6 and 7. A first estimate spreads the gap evenly: 40 is 4 of the 13 steps from 36 to 49, so $\sqrt{40} \approx 6 + \frac{4}{13} \approx 6.31$.

That estimate is about 0.017 too low, because the square-root curve bends. Try squaring nearby decimals to tighten it:

$$6.32^2 = 39.9424$$

$$6.33^2 = 40.0689$$

So $\sqrt{40}$ is between 6.32 and 6.33, and much closer to 6.32.

Method 4: Newton (Babylonian) Iteration

Average a guess with 40 divided by that guess, and repeat:

$$x_{\text{new}} = \frac{1}{2}\left(x + \frac{40}{x}\right)$$

Start from $x = 6$:

$$x_1 = \frac{1}{2}\left(6 + \frac{40}{6}\right) = \frac{19}{3} \approx 6.3333$$

$$x_2 \approx 6.3245614$$

$$x_3 \approx 6.3245553$$

Three rounds match the calculator to seven decimal places. On a calculator, entering $\sqrt{40}$ shows $6.32455532$, the same digits.

What Are The Positive And Negative Square Roots Of 40?

The equation $x^2 = 40$ has two solutions, $x = 2\sqrt{10}$ and $x = -2\sqrt{10}$, written together as $x = \pm 2\sqrt{10}$. Both work, since squaring a negative number gives a positive result.

The symbol $\sqrt{40}$ on its own means only the positive root, about 6.3246. Write the $\pm$ when you are solving an equation, and leave it off when you are evaluating $\sqrt{40}$.

Examples Of Square Root Of 40

Example 1

Simplify $3\sqrt{40}$.

$$3\sqrt{40} = 3 \times 2\sqrt{10}$$

$$3\sqrt{40} = 6\sqrt{10} \approx 18.9737$$

Final answer: $6\sqrt{10}$

Example 2

Find $\sqrt{40}$ by splitting 40 into two perfect squares.

A tempting move: $40 = 36 + 4$, so $\sqrt{40} = \sqrt{36} + \sqrt{4} = 6 + 2 = 8$.

Check it by squaring: $8^2 = 64$, not 40. The answer is also bigger than $\sqrt{49} = 7$, which is impossible for a number less than 49.

Square roots split over multiplication, not addition, so write 40 as a product instead:

$$\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}$$

Final answer: $\sqrt{40} = 2\sqrt{10} \approx 6.3246$

Example 3

Find $\sqrt{40} \times \sqrt{10}$ and $\sqrt{8} \times \sqrt{40}$.

Multiply under one root, then simplify:

$$\sqrt{40} \times \sqrt{10} = \sqrt{400} = 20$$

$$\sqrt{8} \times \sqrt{40} = \sqrt{320} = \sqrt{64 \times 5} = 8\sqrt{5} \approx 17.8885$$

Final answer: $20$ and $8\sqrt{5}$

Example 4

Simplify $\sqrt{-40}$.

No real number squares to $-40$, so the answer uses the imaginary unit $i$, where $i^2 = -1$:

$$\sqrt{-40} = \sqrt{40} \times \sqrt{-1}$$

$$\sqrt{-40} = 2\sqrt{10},i \approx 6.3246,i$$

Final answer: $2\sqrt{10},i$, which is not a real number.

Example 5

A square tabletop has an area of 40 square feet. How long is each side, and how much LED strip runs around the edge?

$$\text{side} = \sqrt{40} = 2\sqrt{10} \approx 6.32,\text{ft}$$

$$\text{perimeter} = 4 \times 2\sqrt{10} = 8\sqrt{10} \approx 25.30,\text{ft}$$

Final answer: each side is about 6.32 ft, and the strip needs about 25.30 ft.

Keeping $8\sqrt{10}$ until the last line matters here. Rounding the side to 6.32 first and then multiplying by 4 gives 25.28 ft, about half a centimetre short of the true length.

Example 6

Rationalize $\frac{1}{\sqrt{40}}$ and give its decimal value.

Simplify the root, then multiply top and bottom by $\sqrt{10}$:

$$\frac{1}{\sqrt{40}} = \frac{1}{2\sqrt{10}}$$

$$\frac{1}{2\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{10}}{20}$$

$$\frac{\sqrt{10}}{20} \approx 0.1581$$

Final answer: $\frac{\sqrt{10}}{20} \approx 0.1581$

The habit that runs through Examples 3, 5 and 6 is simplifying $\sqrt{40}$ to $2\sqrt{10}$ before doing anything else. Students who do that first rarely lose a factor later, because every later step works with a smaller, cleaner number.

Where Do Students Go Wrong With The Square Root Of 40?

Mistake 1: Writing $2\sqrt{5}$ Instead Of $2\sqrt{10}$

Where it slips in: Simplifying, when the 4 has been pulled out and the leftover factor is written down too fast.

Don't do this: Writing $\sqrt{40} = 2\sqrt{5}$. That is the square root of 20, since $2\sqrt{5} = \sqrt{4 \times 5}$.

The correct way: After taking out 4, the factor left inside is $40 \div 4 = 10$. Squaring back catches it at once: $(2\sqrt{10})^2 = 4 \times 10 = 40$, while $(2\sqrt{5})^2 = 20$.

This error is common enough that it even shows up in auto-generated answers to this exact search. The rusher who skips the square-back check is the one it catches.

Mistake 2: Splitting A Sum Under The Root

Where it slips in: Seeing 40 as $36 + 4$ and trying to use the two perfect squares.

Don't do this: Writing $\sqrt{36 + 4} = 6 + 2 = 8$.

The correct way: Only products split: $\sqrt{ab} = \sqrt{a}\sqrt{b}$, never $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$. The first instinct to split a sum comes from how naturally brackets expand in other places, and the size check ($8 > 7 = \sqrt{49}$) is the quickest way to stop it.

Mistake 3: Rounding Too Early

Where it slips in: Multi-step problems such as perimeters, areas or distances that contain $\sqrt{40}$.

Don't do this: Replacing $\sqrt{40}$ with $6.3$ on the first line and carrying it through.

The correct way: Keep $2\sqrt{10}$ until the final step, then round once. The memorizer who knows "root 40 is about 6.3" will get $4 \times 6.3 = 25.2$ ft for the tabletop edge, about 3 cm short of the true $8\sqrt{10} \approx 25.30$ ft.

Where Can You Get Help With Square Roots And Algebra?

Simplifying radicals like $\sqrt{40}$ is the first step into surds, quadratic equations and the distance formula. To practise them with a teacher, look at an algebra tutor, help with algebra, a middle school math tutor or math classes online.

Conclusion

• The square root of 40 is $2\sqrt{10}$, approximately 6.3246.

• 40 is not a perfect square, so $\sqrt{40}$ is irrational and its decimal never ends.

• Simplify by pulling out the square factor 4: $\sqrt{40} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}$.

• Long division, estimation between 36 and 49, and Newton iteration all confirm $6.3246$.

• The common mistakes are writing $2\sqrt{5}$, splitting $\sqrt{36 + 4}$, and rounding before the last step.

Practice These Before Moving On

1. Simplify $5\sqrt{40}$.

2. Find $\sqrt{40} \times \sqrt{90}$.

3. Simplify $\sqrt{40} + \sqrt{90}$.

4. Solve $x^2 = 160$ and give both roots in simplest radical form.

5. Between which two whole numbers does $\sqrt{40}$ lie, and which is it closer to?

Answers:

1. $5 \times 2\sqrt{10} = 10\sqrt{10} \approx 31.6228$

2. $\sqrt{3600} = 60$

3. $2\sqrt{10} + 3\sqrt{10} = 5\sqrt{10} \approx 15.8114$

4. $x = \pm\sqrt{160} = \pm 4\sqrt{10} \approx \pm 12.6491$

5. Between 6 and 7, closer to 6, since $6.5^2 = 42.25$ is already more than 40

If question 3 stalls you, go back to the simplifying section and write each root as a multiple of $\sqrt{10}$ first.

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