8th Grade · 8.EE.A.2

Square roots and cube roots in 8th grade

Square roots and cube roots undo squaring and cubing, so the square root of 49 is 7 and the cube root of 8 is 2. In 8th grade, your child uses the root symbols to solve equations such as $x^2 = 49$ and learns the small perfect squares and cubes by heart. Your child also meets $\sqrt{2}$ , a number that no fraction can equal.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form x² = p and x³ = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational.

What it means, in plain words

Squaring a number means multiplying it by itself, so $6^2 = 36$ . The square root of 36 is the number that gives 36 when multiplied by itself, so $\sqrt{36} = 6$ . Cubing means using a number three times as a factor, so $4^3 = 64$ , and the cube root goes back the other way: $\sqrt[3]{64} = 4$ . A perfect square is a whole number squared (1, 4, 9, 16, 25 and so on). A perfect cube is a whole number cubed (1, 8, 27, 64, 125 and so on).

In 8th grade, square roots and cube roots become the answers to equations. The equation $x^2 = 25$ has two solutions, 5 and $-5$ , because $(-5) \times (-5) = 25$ as well. Children write this as $x = \pm 5$ , read "plus or minus 5". The equation $x^3 = 27$ has only one solution, $x = 3$ , because $(-3)^3 = -27$ . When the number is not a perfect square, the root symbol is the answer: $x^2 = 2$ gives $x = \pm\sqrt{2}$ .

Children also learn that $\sqrt{2}$ is irrational. That means no fraction of two whole numbers equals it, and its decimal, 1.41421..., never ends and never repeats. A calculator shows only the first few digits, so 1.414 is close to $\sqrt{2}$ but not equal to it.

Your child should be able to:

Find square roots of small perfect squares, such as $\sqrt{81} = 9$ , and cube roots of small perfect cubes, such as $\sqrt[3]{125} = 5$
Solve an equation such as $x^2 = 16$ and give both answers, $x = \pm 4$
Solve an equation such as $x^3 = 8$ with a cube root
Explain that $\sqrt{2}$ is irrational and that 1.414 is only close to it

Example

Solve $x^2 = 81$ .

Look for every number that gives 81 when multiplied by itself.
$9 \times 9 = 81$ and $(-9) \times (-9) = 81$
So $x = \pm 9$ . The two solutions are 9 and negative 9.

Example

Solve $x^3 = 64$ .

Look for a number that gives 64 when used three times as a factor.
$4 \times 4 \times 4 = 16 \times 4 = 64$
$-4$ does not work, because $(-4)^3 = -64$ .
So $x = \sqrt[3]{64} = 4$ . The only solution is 4.

Where children usually get stuck

Your child solves $x^2 = 81$ and writes only $x = 9$ .
The symbol $\sqrt{81}$ means only the positive root, 9. The equation asks for every number that squares to 81, and $-9$ is one of them.
What helps. After each answer, ask your child to check it by squaring. Then ask what else squares to 81. A negative times a negative is positive, so $-9$ works too.
Your child writes $\sqrt{16} = 8$ or $\sqrt[3]{27} = 9$ .
The square root gets mixed up with halving, and the cube root with dividing by 3. The small 2 and 3 in squaring and cubing make division feel like the way back.
What helps. Ask your child to multiply the answer back. $8 \times 8 = 64$ , not 16, and $9 \times 9 \times 9 = 729$ , not 27. The check takes a few seconds and shows the mix-up right away.
Your child types $\sqrt{2}$ into a calculator and writes $\sqrt{2} = 1.414$ .
The calculator screen looks exact. Nothing on it says that the digits go on forever without a pattern.
What helps. Ask your child to square 1.414 on the calculator. The result is 1.999396, not 2, so 1.414 is close but not equal. For a rounded value, the sign $\approx$ ("is about") is the honest one: $\sqrt{2} \approx 1.414$ .

Try four practice questions

Real 8th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Which values of x make the equation true? Select all that apply.
$\displaystyle x^2 = 4$
A
$\displaystyle -2$
B
$\displaystyle 2$
C
$\displaystyle 4$
D
$\displaystyle -4$
E
$\displaystyle 3$
Show the answer and how to solve it
The answers are $-2$ and $2$ because both square to 4, while 4, $-4$ , and 3 do not.
1. 1Look at the equation. You need every value of $x$ whose square is 4.
2. 2The positive root works: $2^2 = 4$ , so $x = 2$ is a solution.
3. 3The negative root also works because squaring a negative gives a positive, so $x = -2$ is a solution too. The other choices do not work because their squares are not 4.
Common mistake. A common mistake is selecting only $2$ or choosing $4$ itself. Remember that $x$ must be a number whose square equals 4, and both $2$ and $-2$ satisfy that.
The idea underneath. For $x^{2} = p$ with $p > 0$ , the two solutions are the numbers the same distance from 0 because squaring removes the sign.
2On level
Which integer equals this square root?
$\displaystyle \sqrt{36}$
A
6
B
7
C
18
D
36
Show the answer and how to solve it
The answer is $6$ because $6 \times 6 = 36$ , so $\sqrt{36} = 6$ .
1. 1Look at the expression shown. A square root asks for the positive number whose square equals the radicand, 36.
2. 2Check which integer multiplied by itself makes 36.
3. 3So the square root of 36 is 6. Choose 6.
Common mistake. A common mistake is choosing 36, the number shown under the radical sign. The square root asks for a smaller number that times itself equals 36, which is 6, not 36 itself.
The idea underneath. Square roots undo squaring, and $\sqrt{n}$ names the positive number whose square is $n$ .
3On level
Solve the equation below. Write the full solution in ± form.
$\displaystyle x^2 = 2$
Loading...
Show the answer and how to solve it
Answer. x = \pm\sqrt{2}
The answer is $x = \pm\sqrt{2}$ because both $\sqrt{2}$ and $-\sqrt{2}$ square to $2$ .
1. 1Look at the equation. It says $x$ squared equals 2.
2. 2Take the square root of both sides. Since 2 is not a perfect square, keep the radical.
3. 3Write the two solutions together with ±.
Common mistake. A common mistake is writing $x = \pm 2$ instead of $x = \pm\sqrt{2}$ . The square root of 2 is not 2 — taking the square root and dividing by 2 are different operations.
The idea underneath. Squaring removes the sign, so undoing $x^{2} = p$ with $p > 0$ gives both a positive and a negative square root.
4Stretch
Look at the square-root statements. Select the line with the mistake.
$\displaystyle \begin{array}{l}\text{A. }\sqrt{16} = 4\\\text{B. }\sqrt{121} = -11\\\text{C. }\sqrt{49} = 7\\\text{D. }\sqrt{81} = 9\end{array}$
A
A
B
B
C
C
D
D
Show the answer and how to solve it
The incorrect line is B because $\sqrt{121} = 11$ , not $-11$ ; the radical symbol always denotes the principal nonnegative root.
1. 1A square root gives the principal nonnegative number whose square equals the radicand. Check each line by squaring its right-hand side.
2. 2Lines A, C, and D are all correct.
3. 3Line B claims $\sqrt{121} = -11$ . Squaring 11 gives 121, so the correct root is 11. The $\sqrt{\phantom{0}}$ symbol always denotes the principal nonnegative root, so the answer cannot be negative.
4. 4The incorrect line is B.
Common mistake. A common mistake is writing $\sqrt{121} = -11$ because $(-11)^2 = 121$ is true. However, the $\sqrt{\phantom{0}}$ symbol always means the principal nonnegative square root, so the answer must be 11.
The idea underneath. You can test any square-root statement by squaring the number on the right; if it does not return the radicand, that line is wrong.