Square Root of 1500 - How to Simplify √1500?

What Is A Square Root?

The square root of a number $n$ is the value $r$ for which $r^2 = n$. For 1500, that is the number which, squared, returns 1500.

No integer fits. $38^2 = 1444$ is too small and $39^2 = 1521$ is too large, so $\sqrt{1500}$ lies between 38 and 39, near 38.73. The symbol $\sqrt{1500}$ denotes the principal (positive) root; the equation $x^2 = 1500$ has both $+\sqrt{1500}$ and $-\sqrt{1500}$ as solutions. For the operation in general, Wolfram MathWorld covers the square root.

Where Does √1500 Appear?

$\sqrt{1500}$ is the side length of a square whose area is 1500 square units, so a square plot of 1500 square metres measures about 38.73 metres on each edge. The value sits just under a clean 39, since $39^2 = 1521$. Whenever a quantity of 1500 needs a square footprint, or a $\sqrt{15}$ pattern gets scaled by ten, $10\sqrt{15}$ is what appears. The same digit-by-digit process also handles nearby roots such as √1000.

Quick Reference Table

How Do You Simplify √1500? (Prime Factorization)

Can the square root of 1500 be simplified? Yes. Factor 1500 into primes, take out each matched pair, and leave the rest under the radical.

$$1500 = 2^2 \times 3 \times 5^3$$

$$\sqrt{1500} = \sqrt{2^2 \times 5^2 \times 3 \times 5}$$

$$\sqrt{1500} = 2 \times 5 \times \sqrt{3 \times 5}$$

$$\sqrt{1500} = 10\sqrt{15}$$

The pair $2^2$ gives a 2, and two of the three 5s pair up to give a 5, so $2 \times 5 = 10$ comes out front. One 5 and the lone 3 have no partners, so they stay inside as $3 \times 5 = 15$. That fixes $\sqrt{1500} = 10\sqrt{15}$, and the same routine drives all simplifying radical expressions.

The shortcut is to spot the largest perfect-square factor. Since $1500 = 100 \times 15$ and $100 = 10^2$:

$$\sqrt{1500} = \sqrt{100 \times 15}$$

$$\sqrt{1500} = \sqrt{100} \times \sqrt{15}$$

$$\sqrt{1500} = 10\sqrt{15}$$

Estimating up front, one of these square root tricks, tells you the answer should be near 38.7 before you begin.

How Do You Find √1500 By Long Division?

Long division produces the decimal $38.73\ldots$ one digit at a time. Pair the digits outward from the decimal point and resolve each place in turn.

Step 1: Group the digits: $\overline{15},\overline{00},.,\overline{00},\overline{00}$

Step 2: The largest square below 15 is $9 = 3^2$, so the first digit is 3, remainder $15 - 9 = 6$.

Step 3: Bring down 00 to make 600. Double the quotient (3) to get 6, then find $d$ with $(60 + d) \times d \le 600$. Here $68 \times 8 = 544$, so $d = 8$.

Step 4: The quotient is now 38, remainder $600 - 544 = 56$. Bring down 00 to make 5600.

Step 5: Double 38 to get 76, then find $d$ with $(760 + d) \times d \le 5600$. Here $767 \times 7 = 5369$, so $d = 7$, and the quotient reads 38.7.

Step 6: One more place gives 2, so $\sqrt{1500} \approx 38.72$, which rounds to 38.73.

The digits never settle into a pattern, the signature of an irrational value.

Is The Square Root Of 1500 Rational Or Irrational?

$\sqrt{1500}$ is irrational because 1500 is not a perfect square. A whole number has a rational square root only when it is itself a perfect square, and 1500 falls between $38^2$ and $39^2$.

The prime-factor test makes it exact. Write $1500 = 2^2 \times 3^1 \times 5^3$. A perfect square needs an even exponent on every prime (the reason numbers like 100, 1444, and 1521 have whole roots, as the perfect squares list shows). Here both the 3 and one factor of 5 carry odd exponents, so the root cannot be rational. That is the same reason the leftover $\sqrt{15}$ is irrational; Wolfram MathWorld sets out the idea under irrational number.

Keep $10\sqrt{15}$ as the exact form in algebra, and reach for $38.730$ only when a decimal is required.

Examples Of Square Root Of 1500

Example 1

Simplify $\sqrt{1500}$ to simplest radical form.

$$\sqrt{1500} = \sqrt{100 \times 15}$$

$$\sqrt{1500} = 10\sqrt{15}$$

Final answer: $10\sqrt{15}$.

Example 2

A student simplifies $\sqrt{1500} = \sqrt{25 \times 60} = 5\sqrt{60}$ and calls it done. Is it fully simplified?

The tempting move is to grab a perfect square you recognise, and 25 divides 1500 cleanly.

$$\sqrt{1500} = \sqrt{25 \times 60} = 5\sqrt{60}$$

The arithmetic checks out, but $\sqrt{60}$ still hides the perfect square 4, since $60 = 4 \times 15$.

$$5\sqrt{60} = 5 \times 2\sqrt{15} = 10\sqrt{15}$$

The fix is to keep extracting perfect squares until none remain inside. Stopping at $5\sqrt{60}$ is the classic "almost there" error.

Final answer: $10\sqrt{15}$.

Example 3

Evaluate $\sqrt{1500} + \sqrt{375}$.

$$\sqrt{1500} = 10\sqrt{15}$$

$$\sqrt{375} = \sqrt{25 \times 15} = 5\sqrt{15}$$

$$10\sqrt{15} + 5\sqrt{15} = 15\sqrt{15}$$

Final answer: $15\sqrt{15} \approx 58.09$.

Example 4

Simplify $\dfrac{\sqrt{1500}}{\sqrt{15}}$.

$$\frac{\sqrt{1500}}{\sqrt{15}} = \sqrt{\frac{1500}{15}}$$

$$\sqrt{\frac{1500}{15}} = \sqrt{100}$$

$$\sqrt{100} = 10$$

Final answer: $10$. Cancelling the shared factor of 15 leaves a clean whole number.

Example 5

A square field covers 1500 square metres. What is the length of one side, to two decimal places?

$$\text{side} = \sqrt{1500}$$

$$\text{side} = 10\sqrt{15}$$

$$\text{side} \approx 38.73 \text{ m}$$

Final answer: about $38.73$ m.

Common Mistakes

Mistake 1: Leaving a perfect square trapped inside the radical

Where it slips in: Extracting only the first perfect square you notice, such as $\sqrt{1500} = 5\sqrt{60}$.

Don't do this: Reporting $5\sqrt{60}$ when $\sqrt{60}$ still contains the perfect square 4.

The correct way: Factor down to $10\sqrt{15}$, where the number under the root has no perfect-square factor above 1. The learner who stops at the first easy factor almost always leaves this one half-done.

Mistake 2: Splitting the root across addition

Where it slips in: Trying to break $\sqrt{100 \times 15}$ into separate roots the wrong way.

Don't do this: Writing $\sqrt{100 \times 15} = \sqrt{100} + \sqrt{15}$.

The correct way: A root distributes over multiplication, not addition: $\sqrt{100 \times 15} = \sqrt{100} \times \sqrt{15} = 10\sqrt{15}$. Mixing up multiplying and adding under the radical is a habit worth catching early.

Mistake 3: Rounding to 38.73 before the final step

Where it slips in: Carrying $\sqrt{1500}$ through several multiplications in a longer problem.

Don't do this: Swapping in $38.73$ at the start and reusing that rounded figure at every stage.

The correct way: Hold the exact $10\sqrt{15}$ until the end. Early rounding compounds, and this is precisely the class of error behind the 1991 Patriot missile failure at Dhahran, where a small repeated rounding drift in the system clock let a target slip through.

Conclusion

• The square root of 1500 is $10\sqrt{15}$, roughly $38.730$.

• 1500 is not a perfect square, so $\sqrt{1500}$ is irrational.

• Prime factorization ($1500 = 2^2 \times 3 \times 5^3$) and the largest-perfect-square shortcut ($1500 = 100 \times 15$) both give $10\sqrt{15}$.

• Keep the exact radical through a multi-step problem and round only at the finish.

To push radical work further with a teacher, try an algebra tutor or structured online math classes. Working toward board or entrance exams? A high school math tutor can fold this into surds practice, or you can book a free demo class.

Read More

• Square Root of 1200 - a neighbouring four-digit root that simplifies to $20\sqrt{3}$.

• Square Root of 2000 - simplify a nearby value the same way.

• Square Root of 5 - one of the primes hiding inside $\sqrt{15}$.

• Squares and Square Roots - the groundwork behind perfect squares and roots.

• Square Root 1 to 30 - the full reference table of roots in this range.