Why do we write ± for square roots but not for cube roots?

The ± symbol appears when solving x^2 = p because squaring either a positive or negative number gives a positive result. So x^2 = 16 has two solutions: x = 4 and x = -4. In contrast, cubing preserves the sign: a positive number cubed stays positive, and a negative number cubed stays negative. So x^3 = 8 has only one solution (x = 2) and x^3 = -8 has only one solution (x = -2). The cube root function is one-to-one, meaning each output comes from exactly one input.

What's the difference between sqrt(9) and the solutions to x^2 = 9?

sqrt(9) = 3 (a single number, always the principal root). The solutions to x^2 = 9 are x = 3 and x = -3 (two values). The radical symbol x by definition gives only the non-negative root. But when you solve an equation, you must find all values of x that make it true, which includes both the positive and negative roots.

How do I know if a number is a perfect square?

Check whether it appears in the list 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225. These are 1^2 through 15^2. You can also try to find a whole number that, when multiplied by itself, gives your number. For example, is 144 a perfect square? Check: 12 × 12 = 144, so yes. Is 50 a perfect square? No whole number squared equals 50 (since 7^2 = 49 and 8^2 = 64), so 50 is not a perfect square.

Can a square root be negative?

No. By definition, the square root symbol sqrt(n) always means the principal (non-negative) square root. So sqrt(16) = 4, never -4. However, the **solutions** to x^2 = 16 include both 4 and -4. This is why we use the ± notation when solving: x = (16) = ± 4. The symbol itself is non-negative; the equation yields two roots.

M8MATH-1.3

Square Roots & Cube Roots

Learn to evaluate square roots and cube roots, solve equations with perfect squares and cubes, and understand why √2 is irrational.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Square Roots & Cube Roots, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Square Roots & Cube Roots free

What this lesson covers

Square roots and cube roots are inverse operations—they undo squaring and cubing. When you know that

7^2 = 49

, you automatically know that

\sqrt{49} = 7

. In this lesson, you'll become fluent with perfect squares and cubes up to specific limits, solve equations like

x^2 = 16

and

x^3 = 8

, and learn why some roots, like

\sqrt{2}

, cannot be expressed as a simple fraction. These ideas are fundamental to algebra and set the stage for working with exponents, radicals, and irrational numbers.

Perfect Squares and Square Roots

A perfect square is a number that equals some whole number times itself. For example,

1^2 = 1

2^2 = 4

3^2 = 9

, all the way up to

15^2 = 225

. You need to memorize the perfect squares from 1 to 225. The square root

\sqrt{n}

asks: what positive number, when multiplied by itself, gives

n

?

For instance,

\sqrt{16} = 4

because

4 \times 4 = 16

. When we write

\sqrt{16}

, the radical symbol

\sqrt{\phantom{x}}

always refers to the principal (positive) square root.

However, when you solve an equation like

x^2 = 16

, there are two real solutions:

x = 4

and

x = -4

. Both work because

(4)^2 = 16

and

(-4)^2 = 16

. We can write this as

x = \pm 4

. The symbol

\pm

(plus or minus) reminds us that both the positive and negative roots satisfy the equation. This is a critical distinction: the symbol

\sqrt{16}

equals only 4, but the solutions to

x^2 = 16

are

\pm 4

Perfect Cubes and Cube Roots

A perfect cube is a number that equals some whole number times itself three times. Examples:

1^3 = 1

2^3 = 8

3^3 = 27

, up to

10^3 = 1000

. The cube root

\sqrt[3]{n}

asks: what number, when multiplied by itself three times, gives

n

?

For instance,

\sqrt[3]{27} = 3

because

3 \times 3 \times 3 = 27

. Unlike square roots, cube roots can be negative. For example,

\sqrt[3]{-8} = -2

because

(-2)^3 = -8

. When you solve

x^3 = p

, there is always exactly one real solution:

x = \sqrt[3]{p}

. No

\pm

is needed for cube roots.

Memorize the perfect cubes:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000

(corresponding to

1^3

through

10^3

). Knowing these helps you quickly evaluate cube root expressions without a calculator.

Why √2 Is Irrational

The number

\sqrt{2}

is the solution to

x^2 = 2

. Unlike

\sqrt{4} = 2

, there is no whole number or simple fraction that, when squared, equals exactly 2.

We can prove

\sqrt{2}

is irrational by contradiction. Suppose

\sqrt{2} = \frac{a}{b}

where

a

and

b

are integers with no common factors (a fraction in lowest terms). Squaring both sides:

2 = \frac{a^2}{b^2}

, so

2b^2 = a^2

. This means

a^2

is even, which forces

a

to be even. Let

a = 2k

. Then

2b^2 = (2k)^2 = 4k^2

, so

b^2 = 2k^2

. This means

b^2

is even, so

b

is even. But now both

a

and

b

are even—they share a factor of 2—which contradicts our assumption. Therefore,

\sqrt{2}

cannot be a fraction and is irrational.

On a number line,

\sqrt{2} \approx 1.414...

The decimal never repeats and never terminates.

Solving Equations with Squares and Cubes

To solve

x^2 = p

where

p

is a perfect square up to 225: take the square root of both sides and remember both signs. For example, if

x^2 = 36

, then

x = \pm\sqrt{36} = \pm 6

.

To solve

x^3 = p

where

p

is a perfect cube up to 1000: take the cube root of both sides. For example, if

x^3 = 125

, then

x = \sqrt[3]{125} = 5

. If

x^3 = -125

, then

x = \sqrt[3]{-125} = -5

. There is one real solution, not two.

A common mistake is to forget the negative solution in square-root equations. Another mistake is to confuse

\sqrt{9}

(which is always 3) with the solutions to

x^2 = 9

(which are

x = 3

and

x = -3

). Keep them separate in your mind: the radical symbol gives you the principal root; solving the equation gives you all roots.

Evaluating Roots and Building Fluency

To evaluate a square root like

\sqrt{64}

, ask yourself: "What number times itself equals 64?" The answer is 8, so

\sqrt{64} = 8

. If the number under the radical is not a perfect square, the result is irrational and cannot be simplified to a single number.

For cube roots, ask: "What number times itself three times equals this?" For

\sqrt[3]{216}

, the answer is 6 because

6^3 = 216

.

Build fluency by drilling the perfect squares and cubes. Write them down, say them aloud, and practice until you can recognize 49 and immediately know that

\sqrt{49} = 7

. This automaticity saves time and reduces errors in later algebra. When you encounter an expression like

\sqrt{121} + \sqrt[3]{64}

, you should instantly recognize

\sqrt{121} = 11

and

\sqrt[3]{64} = 4

, giving you

11 + 4 = 15

without reaching for a calculator.

Key terms

Perfect square.: A whole number that is the product of some whole number multiplied by itself, such as 16 or 81.
Square root.: The inverse of squaring; $\sqrt{n}$ is the non-negative number that, when multiplied by itself, equals $n$ . The symbol $\sqrt{\phantom{x}}$ always denotes the principal (positive) root.
Perfect cube.: A whole number that is the product of some whole number multiplied by itself three times, such as 8 or 125.
Cube root.: The inverse of cubing; $\sqrt[3]{n}$ is the real number that, when multiplied by itself three times, equals $n$ . Cube roots are defined for negative numbers as well.
Irrational number.: A real number that cannot be expressed as a ratio of two integers; its decimal representation neither terminates nor repeats. Examples include $\sqrt{2}$ and $\pi$ .
Radical symbol.: The symbol $\sqrt{\phantom{x}}$ used to denote a root; the index (small number) indicates which root (2 for square root, 3 for cube root, etc.).
Principal root.: The non-negative root when there are multiple roots; for example, the principal square root of 9 is 3, not $-3$ , even though both $3^2$ and $(-3)^2$ equal 9.

Worked example

Solve for

x

: (a)

x^2 = 144

and (b)

x^3 = -216

. Then evaluate

\sqrt{169} + \sqrt[3]{512}

Part (a): Solve $x^2 = 144$

Take the square root of both sides:

x = \pm\sqrt{144}

. We need to find what number times itself equals 144. Check:

12 \times 12 = 144

, so

\sqrt{144} = 12

. Therefore,

x = \pm 12

. The two solutions are

x = 12

and

x = -12

.

Part (b): Solve $x^3 = -216$

Take the cube root of both sides:

x = \sqrt[3]{-216}

. We need to find what number times itself three times equals

-216

. Check:

(-6) \times (-6) \times (-6) = -216

, so

\sqrt[3]{-216} = -6

. There is one solution:

x = -6

. (Note: the cube root of a negative number is negative.)

Part (c): Evaluate $\sqrt{169} + \sqrt[3]{512}$

First, evaluate

\sqrt{169}

. What number squared equals 169? Check:

13 \times 13 = 169

, so

\sqrt{169} = 13

.

Next, evaluate

\sqrt[3]{512}

. What number cubed equals 512? Check:

8 \times 8 \times 8 = 512

, so

\sqrt[3]{512} = 8

.

Add them:

\sqrt{169} + \sqrt[3]{512} = 13 + 8 = 21

Practice questions

Which of the following is the complete solution set to

x^2 = 49

$x = 7$
$x = -7$
$x = 7$ or $x = -7$
$x = \pm\sqrt{49} = \pm 7$ are both true, so $x = 7$ or $x = -7$

Answer: $x = 7$ or $x = -7$

When solving

x^2 = 49

, both

7^2 = 49

and

(-7)^2 = 49

are true. The symbol

\sqrt{49} = 7

refers only to the principal (positive) root, but the equation

x^2 = 49

has two real solutions:

x = 7

and

x = -7

. We write this as

x = \pm 7

. A common mistake is to forget the negative solution.

\sqrt{50}

rational or irrational? Explain your reasoning.

Answer: $\sqrt{50}$ is irrational.

The number 50 is not a perfect square. Since

7^2 = 49

and

8^2 = 64

, we know that

\sqrt{50}

lies between 7 and 8 and cannot equal any whole number. Moreover,

\sqrt{50}

cannot be expressed as a fraction of two integers. We can simplify

\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}

, and since

\sqrt{2}

is irrational (as shown in the lesson),

\sqrt{50}

is also irrational. Its decimal expansion goes on forever without repeating.

Evaluate: (a)

\sqrt[3]{343}

and (b)

\sqrt[3]{-125}

. Explain the difference between these two.

Answer: (a) $\sqrt[3]{343} = 7$ ; (b) $\sqrt[3]{-125} = -5$ .

For (a), we ask: what number cubed equals 343? Since

7^3 = 7 \times 7 \times 7 = 343

, we have

\sqrt[3]{343} = 7

. For (b), we ask: what number cubed equals

-125

? Since

(-5)^3 = (-5) \times (-5) \times (-5) = -125

, we have

\sqrt[3]{-125} = -5

. The key difference from square roots is that cube roots are defined for negative numbers and the result is also negative. There is no

\pm

in cube-root equations because cubing a positive number gives a positive result, and cubing a negative number gives a negative result—each output comes from exactly one input.

FAQ

Why do we write $\pm$ for square roots but not for cube roots?: The $\pm$ symbol appears when solving $x^2 = p$ because squaring either a positive or negative number gives a positive result. So $x^2 = 16$ has two solutions: $x = 4$ and $x = -4$ . In contrast, cubing preserves the sign: a positive number cubed stays positive, and a negative number cubed stays negative. So $x^3 = 8$ has only one solution ( $x = 2$ ) and $x^3 = -8$ has only one solution ( $x = -2$ ). The cube root function is one-to-one, meaning each output comes from exactly one input.
What's the difference between $\sqrt{9}$ and the solutions to $x^2 = 9$ ?: $\sqrt{9} = 3$ (a single number, always the principal root). The solutions to $x^2 = 9$ are $x = 3$ and $x = -3$ (two values). The radical symbol $\sqrt{\phantom{x}}$ by definition gives only the non-negative root. But when you solve an equation, you must find all values of $x$ that make it true, which includes both the positive and negative roots.
How do I know if a number is a perfect square?: Check whether it appears in the list $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225$ . These are $1^2$ through $15^2$ . You can also try to find a whole number that, when multiplied by itself, gives your number. For example, is 144 a perfect square? Check: $12 \times 12 = 144$ , so yes. Is 50 a perfect square? No whole number squared equals 50 (since $7^2 = 49$ and $8^2 = 64$ ), so 50 is not a perfect square.
Can a square root be negative?: No. By definition, the square root symbol $\sqrt{n}$ always means the principal (non-negative) square root. So $\sqrt{16} = 4$ , never $-4$ . However, the solutions to $x^2 = 16$ include both $4$ and $-4$ . This is why we use the $\pm$ notation when solving: $x = \pm\sqrt{16} = \pm 4$ . The symbol itself is non-negative; the equation yields two roots.

Learn this with a teacher, not a page

The Crimsora tutor teaches Square Roots & Cube Roots live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free