Estimate Square and Cube Roots, Then Fix Errors | Grade 8 Math | LevelKite

You already know that $\sqrt{49} = 7$ , because $7^2 = 49$ and $7$ is not negative. You also know that $\sqrt[3]{8} = 2$ , because $2^3 = 8$ . Both roots are whole numbers (the numbers $0, 1, 2, 3, \dots$ ) because $49$ is a perfect square, the square of a whole number, and $8$ is a perfect cube, the cube of a whole number.

Most whole numbers are not perfect squares or perfect cubes. Take $\sqrt{20}$ . No whole number has a square of $20$ : $4^2 = 16$ is too small and $5^2 = 25$ is too big. So you estimate, which means you find a simple number that is close to the root.

Estimating takes two moves. First you trap the root between two numbers you know. Then you test the halfway point to see which of those two numbers is closer.

Why does trapping work? A number is nonnegative when it is $0$ or greater. Among nonnegative numbers, a bigger number always has a bigger square: $4 < 5$ , and $16 < 25$ . So the order of two squares tells you the order of the nonnegative numbers they came from.

For a number $n \ge 0$ , the symbol $\sqrt{n}$ means the principal square root of $n$ : the one nonnegative number whose square is $n$ . Because $16 < 20 < 25$ , and $4$ , $\sqrt{20}$ and $5$ are all nonnegative, you get $4 < \sqrt{20} < 5$ . The trapping numbers $4$ and $5$ are called bounds.

The bounds must be nonnegative. Negative numbers break the pattern: $-6 < 4$ , but $(-6)^2 = 36$ is greater than $4^2 = 16$ . So always trap a square root between nonnegative bounds.

Cube roots are friendlier. A real number is any number on the number line: positive, negative or zero. Every real number $n$ has exactly one real cube root, written $\sqrt[3]{n}$ . It is positive when $n$ is positive, negative when $n$ is negative, and $0$ when $n = 0$ .

For all real numbers, a bigger number has a bigger cube. For example, $-3 < -2$ , and $(-3)^3 = -27$ is less than $(-2)^3 = -8$ . So the trap works for cube roots with any real bounds, including negative ones: $8 < 20 < 27$ gives $2 < \sqrt[3]{20} < 3$ , and $-27 < -20 < -8$ gives $-3 < \sqrt[3]{-20} < -2$ .

Benchmarks: squares and cubes of the whole numbers 1 to 10

Whole number $k$	Square $k^2$	Cube $k^3$
$1$	$1$	$1$
$2$	$4$	$8$
$3$	$9$	$27$
$4$	$16$	$64$
$5$	$25$	$125$
$6$	$36$	$216$
$7$	$49$	$343$
$8$	$64$	$512$
$9$	$81$	$729$
$10$	$100$	$1000$

Key concept

Trap the root, then test the midpoint

Trap. To estimate $\sqrt{n}$ for $n \ge 0$ , find nonnegative numbers $a$ and $b$ with $a^2 < n < b^2$ . Then $a < \sqrt{n} < b$ . For $\sqrt[3]{n}$ , where $n$ is any real number, find real numbers $a$ and $b$ with $a^3 < n < b^3$ . Then $a < \sqrt[3]{n} < b$ .

Test the midpoint. The midpoint $m$ is the number halfway between the bounds: $m = \frac{a + b}{2}$ . Compare $n$ with $m^2$ for a square root, or with $m^3$ for a cube root. If $n$ is less, the root is below $m$ , so it is closer to $a$ . If $n$ is greater, the root is above $m$ , so it is closer to $b$ . When the bounds are next to each other, such as $6$ and $7$ , the closer bound is the rounded root. Write it with $\approx$ , which means is approximately equal to.

\text{If } a, b \ge 0 \text{ and } a^2 < n < b^2, \text{ then } a < \sqrt{n} < b

Worked example

Estimate a square root

Estimate $\sqrt{45}$ to the nearest whole number, and then to the nearest tenth.

Trap the root between whole numbers. From the table, $6^2 = 36$ and $7^2 = 49$ , and $45$ lies between them. The bounds $6$ and $7$ are nonnegative, so the roots are in the same order as the squares.
$36 < 45 < 49 \quad\text{so}\quad 6 < \sqrt{45} < 7$
Test the midpoint. Halfway between $6$ and $7$ is $6.5$ . Square it so that you can compare it with $45$ . Multiply in parts: $6.5 \times 6 = 39$ and $6.5 \times 0.5 = 3.25$ , and $39 + 3.25 = 42.25$ .
$6.5^2 = 42.25$
Compare. $45$ is greater than $42.25$ , and $\sqrt{45}$ and $6.5$ are both nonnegative, so $\sqrt{45}$ is greater than $6.5$ . The root is in the upper half, closer to $7$ than to $6$ .
$45 > 42.25 \quad\text{so}\quad \sqrt{45} > 6.5$
Now trap the root between tenths. It is between $6.5$ and $7$ , so square the tenths in that range until a square passes $45$ : $6.6^2 = 43.56$ , $6.7^2 = 44.89$ and $6.8^2 = 46.24$ . The bounds $6.7$ and $6.8$ are nonnegative, so the trap holds.
$44.89 < 45 < 46.24 \quad\text{so}\quad 6.7 < \sqrt{45} < 6.8$
Test the new midpoint, $6.75$ . Its square is $6.75 \times 6.75 = 45.5625$ . Because $45$ is less than $45.5625$ , the root is less than $6.75$ , so it is closer to $6.7$ than to $6.8$ .
$45 < 45.5625 \quad\text{so}\quad \sqrt{45} < 6.75$

Answer

To the nearest whole number, $\sqrt{45} \approx 7$ . To the nearest tenth, $\sqrt{45} \approx 6.7$ .

Worked example

Estimate the cube root of a negative number

Estimate $\sqrt[3]{-50}$ to the nearest integer.

A negative number has a negative cube root, so use cubes of negative integers. From the table, $3^3 = 27$ and $4^3 = 64$ , so $(-3)^3 = -27$ and $(-4)^3 = -64$ . The number $-50$ lies between $-64$ and $-27$ .
$-64 < -50 < -27$
Cubing keeps the order of all real numbers, so the cube roots are in the same order as the cubes.
$-4 < \sqrt[3]{-50} < -3$
Test the midpoint of $-4$ and $-3$ , which is $-3.5$ . Cube it in two steps: $(-3.5)^2 = 12.25$ , and then $12.25 \times (-3.5) = -42.875$ .
$(-3.5)^3 = -42.875$
Compare. $-50$ is less than $-42.875$ , so $\sqrt[3]{-50}$ is less than $-3.5$ . Less than $-3.5$ means to its left on the number line, which is the half closer to $-4$ .
$-50 < -42.875 \quad\text{so}\quad \sqrt[3]{-50} < -3.5$

Answer

To the nearest integer, $\sqrt[3]{-50} \approx -4$ .

Do not round by the closest perfect cube. To estimate $\sqrt[3]{17}$ , a tempting shortcut says: $17$ is $9$ away from $8$ and $10$ away from $27$ , so it is closer to $8$ , and the root rounds to $2$ . That is wrong. The midpoint of $2$ and $3$ is $2.5$ , and $2.5^3 = 15.625$ is less than $17$ . So the root is above $2.5$ , and $\sqrt[3]{17} \approx 3$ .

The shortcut is not reliable because it tests the wrong halfway point. The halfway root $2.5$ has the cube $15.625$ , but halfway between $8$ and $27$ is $17.5$ . Every number between $15.625$ and $17.5$ is closer to $8$ , yet its cube root is above $2.5$ . Squares have the same mismatch: $4.5^2 = 20.25$ , but halfway between $16$ and $25$ is $20.5$ . So always compare with the square or cube of the midpoint.

You can use the same ideas to find errors in work with roots. Do not look only at the final answer. Test each line against what the symbols mean by asking three questions.

Does it undo? Square a claimed square root, or cube a claimed cube root. For an exact answer, written with $=$ , you must get back exactly the number under the root sign.
Is the sign right? For $n \ge 0$ , $\sqrt{n}$ is the one nonnegative root. The equation $x^2 = n$ , where $x$ is an unknown real number, is different. It has two real solutions, $\sqrt{n}$ and $-\sqrt{n}$ (written together as $\pm\sqrt{n}$ ), when $n > 0$ ; one solution, $0$ , when $n = 0$ ; and no real solution when $n < 0$ . Every real number has exactly one real cube root: positive for a positive number, negative for a negative number, and $0$ for $0$ .
Is the estimate justified? The bounds must trap the root itself, and for a square root they must be nonnegative. The rounding must come from the midpoint test. An estimate needs $\approx$ , not $=$ .

Find the error, then correct it

Work shown	What went wrong	Corrected work
$\sqrt{64} = 32$	Halved $64$ instead of taking a root. Undo check: $32^2 = 1024$ , not $64$ .	$\sqrt{64} = 8$ , because $8^2 = 64$ and $8 \ge 0$ .
$\sqrt{49} = \pm 7$	Sign: $\sqrt{49}$ names only the nonnegative root.	$\sqrt{49} = 7$ . The equation $x^2 = 49$ is what has two solutions, $7$ and $-7$ .
$x^2 = -25$ , so $x = -5$	Undo check: $(-5)^2 = 25$ , not $-25$ . No real number has a negative square.	$x^2 = -25$ has no real solution.
$\sqrt[3]{-64}$ is not a real number	Sign: a negative number has no real square root, but it does have a real cube root.	$\sqrt[3]{-64} = -4$ , because $(-4)^3 = -64$ .
$\sqrt{40}$ is between $36$ and $49$	Estimate: $36$ and $49$ trap $40$ , not its root. The root is trapped by $\sqrt{36} = 6$ and $\sqrt{49} = 7$ .	$36 < 40 < 49$ , so $6 < \sqrt{40} < 7$ .

Worked example

Find and correct errors in a solution

Kai solved $x^2 = 30$ for a real number $x$ and rounded to the nearest integer. Find and correct the errors in Kai's work. Line 1: $x = \sqrt{30}$ . Line 2: $25 < 30 < 36$ , so $5 < \sqrt{30} < 6$ . Line 3: The midpoint of $5$ and $6$ is $5.5$ , and $5.5^2 = 30.25$ . Line 4: $30.25$ is very close to $30$ , so $\sqrt{30} = 5.5$ , which rounds up to $6$ . So $x = 6$ .

Check Line 1 with the sign question. In $x^2 = 30$ , the number $30$ is greater than $0$ , so the equation has two real solutions. Kai wrote only the nonnegative one. This is the first error.
$x = \sqrt{30} \quad\text{or}\quad x = -\sqrt{30}$
Check Line 2. $5^2 = 25$ and $6^2 = 36$ , the bounds $5$ and $6$ are nonnegative, and $30$ is between $25$ and $36$ . Line 2 is correct.
$25 < 30 < 36 \quad\text{so}\quad 5 < \sqrt{30} < 6$
Check Line 3 by redoing the arithmetic. $5.5 \times 5 = 27.5$ and $5.5 \times 0.5 = 2.75$ , and $27.5 + 2.75 = 30.25$ . Line 3 is correct.
$5.5^2 = 30.25$
Check Line 4 with the undo question. $5.5^2$ is $30.25$ , not $30$ , so $\sqrt{30}$ does not equal $5.5$ . The midpoint test needs a comparison instead: $30$ is less than $30.25$ , so $\sqrt{30}$ is less than $5.5$ and is closer to $5$ than to $6$ . This is the second error.
$30 < 30.25 \quad\text{so}\quad \sqrt{30} < 5.5$
Correct the work. The nonnegative solution rounds to $5$ . The other solution is its opposite, so it lies between $-5.5$ and $-5$ and rounds to $-5$ . Write $\approx$ , because these are estimates.
$x = \sqrt{30} \approx 5 \quad\text{or}\quad x = -\sqrt{30} \approx -5$

Answer

Kai made two errors. Line 1 left out the negative solution, and Line 4 treated $\sqrt{30}$ as exactly $5.5$ and rounded the wrong way. The corrected solutions are $x = \sqrt{30} \approx 5$ and $x = -\sqrt{30} \approx -5$ .

Takeaway. To estimate a root, trap it between two numbers whose squares or cubes you know. Then compare the number under the root sign with the square or cube of the midpoint to see which bound is closer. Square roots need nonnegative bounds. Cube roots work with any real bounds, and a negative number has a negative cube root. Write $\approx$ for every estimate. To find an error in work with roots, test each line: undo the root, check the sign, and check that the estimate came from the midpoint test.