Roots

A common error is to think “square root” means “divide by 2,” giving \( 36 \div 2 = 18 \). But the square root of 36 asks: “What number multiplied by itself gives 36?” Since \( 6 \times 6 = 36 \), we know \( \sqrt{36} = 6 \).

The “root means divide by the index” misconception breaks down quickly with a simple check: if \( \sqrt{36} \) were 18, then \( 18 \times 18 \) would need to equal 36 — but \( 18 \times 18 = 324 \). Square rooting undoes squaring; it does not halve.

Many students believe you cannot take any root of a negative number. This is true for square roots, but cube roots are different. We need a number that, multiplied by itself three times, gives −8. Since \( (-2) \times (-2) \times (-2) = 4 \times (-2) = -8 \), we have \( \sqrt[3]{-8} = -2 \).

The key insight is that a negative number multiplied an odd number of times stays negative. This means cube roots of negative numbers work perfectly well.

It is tempting to think you can “combine” square roots like fractions. But \( \sqrt{9} = 3 \) and \( \sqrt{16} = 4 \), so \( \sqrt{9} + \sqrt{16} = 3 + 4 = 7 \). Meanwhile, \( \sqrt{25} = 5 \). Since \( 7 \neq 5 \), the two sides are not equal.

The “distributing the root over addition” misconception fails because square rooting is not a linear operation. Notice that \( \sqrt{9 + 16} = \sqrt{25} = 5 \), which is less than \( \sqrt{9} + \sqrt{16} = 7 \).

Following the steps: \( (-3)^2 = (-3) \times (-3) = 9 \), and then \( \sqrt{9} = 3 \) (not −3). So squaring −3 and then square rooting gives positive 3. The square root function always returns the non-negative root.

This reveals the “squaring and square rooting are perfect inverses” misconception. Squaring destroys information about the sign — both \( 3^2 \) and \( (-3)^2 \) give 9 — so the square root cannot “know” whether the original number was positive or negative.

We just need one counter-example. If we choose \( x = -5 \), then \( x^2 = 25 \). The square root of 25 is 5. So we started with -5 and ended with 5. Therefore, \( \sqrt{x^2} \neq x \) when \( x \) is negative.

Mathematically, \( \sqrt{x^2} = |x| \) (the absolute value of \( x \)). This is a crucial distinction in higher-level algebra!

Example: 8 (because \( \sqrt[3]{8} = 2 \), since \( 2 \times 2 \times 2 = 8 \))

Another: 27 (because \( \sqrt[3]{27} = 3 \), since \( 3 \times 3 \times 3 = 27 \))

Creative: 1000 (because \( \sqrt[3]{1000} = 10 \), since \( 10 \times 10 \times 10 = 1000 \) — students often don’t think beyond small cubes)

Trap: 9 — a student might offer this thinking “\( \sqrt[3]{9} = 3 \) because \( 3 \times 3 = 9 \),” but that’s the square root, not the cube root. The cube root requires three equal factors: \( 3 \times 3 \times 3 = 27 \), not 9.

Example: 16 (because \( \sqrt{16} = 4 \), since \( 4 \times 4 = 16 \))

Another: 25 (because \( \sqrt{25} = 5 \), since \( 5 \times 5 = 25 \))

Creative: 81 (because \( \sqrt{81} = 9 \) — students sometimes forget that \( 9 \times 9 = 81 \) is a two-digit perfect square)

Trap: 32 — a student might reason “since \( \sqrt{16} = 4 \), and 32 is double 16, then \( \sqrt{32} \) must be double 4, which is 8.” But \( 8 \times 8 = 64 \neq 32 \). This exploits the “square root scales linearly” misconception.

Example: 64 (because \( 64 = 8^2 \) and \( 64 = 4^3 \))

Another: 1 (because \( 1 = 1^2 \) and \( 1 = 1^3 \))

Creative: 729 (because \( 729 = 27^2 \) and \( 729 = 9^3 \) — students rarely think of examples this large)

Trap: 8 — a student might offer 8 thinking “it’s a perfect cube (\( 2^3 = 8 \)) and I think it’s also a perfect square.” But \( \sqrt{8} \) is not a whole number.

Example: −64 (because \( \sqrt[3]{-64} = -4 \), since \( (-4) \times (-4) \times (-4) = -64 \))

Another: −27 (because \( \sqrt[3]{-27} = -3 \), since \( (-3) \times (-3) \times (-3) = -27 \))

Creative: −1 (because \( \sqrt[3]{-1} = -1 \), since \( (-1) \times (-1) \times (-1) = -1 \) — the simplest case, often overlooked)

Trap: −9 — a student might claim “\( \sqrt[3]{-9} = -3 \) because \( -3 \times 3 = -9 \).” But cube root requires three equal factors: \( (-3) \times (-3) \times (-3) = -27 \), not −9.

This is true for perfect squares: for example, \( \sqrt{16} = 4 \) and \( \sqrt{49} = 7 \). But most whole numbers are not perfect squares — \( \sqrt{5} \), \( \sqrt{10} \), and \( \sqrt{20} \), for instance, are not whole numbers.

Between any two consecutive perfect squares (say 9 and 16), there are several whole numbers (10, 11, 12, 13, 14, 15) whose square roots are not whole numbers.

This is always true. If a number is negative, its cube root must also be negative, because a negative number cubed gives a negative result: \( (-n) \times (-n) \times (-n) = -(n^3) \).

Students often fall for the “negative numbers don’t have roots” misconception, confusing this with square roots. The difference is that cubing preserves the sign (negative stays negative), while squaring destroys it.

(With a caveat!) In the set of Real Numbers (the numbers we use every day), this is never true. A positive number squared is positive, and a negative number squared is positive. No real number squared results in a negative.

Note for advanced students: Mathematicians invented “Imaginary Numbers” to handle this, where \( \sqrt{-1} \) is defined as \( i \). But for standard calculations in the Real Number system, we say this has no solution.

For positive numbers greater than 1, this is true: \( \sqrt[3]{27} = 3 \), and \( 3 \lt 27 \). But for 0 and 1, the cube root equals the number. And for negative numbers, the cube root is actually greater than the number: \( \sqrt[3]{-27} = -3 \), and \( -3 \gt -27 \) on the number line.

We know that \( 7^2 = 49 \) and \( 8^2 = 64 \). Since 50 is just 1 away from 49, but 14 away from 64, its square root will be very close to 7 (approx 7.07), making it much closer to 7 than to 8.

The student has the “root means divide by the index” misconception. But the cube root of 125 asks: “What number, multiplied by itself three times, gives 125?” Since \( 5 \times 5 \times 5 = 125 \), the answer is \( \sqrt[3]{125} = 5 \).

A quick disproof: if \( \sqrt[3]{125} \) were 41.666…, then \( 41.666\ldots \times 41.666\ldots \times 41.666\ldots \) would need to equal 125, but this gives a number over 72,000.

The student gets the correct answer (5), but their reasoning contains the “± square root” misconception. The \( \sqrt{} \) symbol specifically denotes the Principal Square Root, so \( \sqrt{25} = 5 \) — it does not equal ±5.

It is true that 25 has two square roots (5 and −5), but the function \( \sqrt{x} \) is defined to return only the positive one. Confusing solving \( x^2 = 25 \) (two answers) with finding \( \sqrt{25} \) (one answer) is a critical error to fix.

The student has the “confusing −√n with √(−n)” misconception. The expression \( -\sqrt{16} \) means “the negative of the square root of 16,” not “the square root of negative 16.”

Reading it carefully: first find \( \sqrt{16} = 4 \), then apply the minus sign to get −4. So \( -\sqrt{16} = -4 \). The minus sign is outside the root, so it is perfectly valid.

The student has the “distributing the root over addition” misconception — treating \( \sqrt{a + b} \) as if it equals \( \sqrt{a} + \sqrt{b} \). The correct approach: first add inside the root, giving \( \sqrt{36 + 64} = \sqrt{100} = 10 \).

Square roots do NOT distribute over addition. A clear disproof: \( \sqrt{9 + 16} = \sqrt{25} = 5 \), while \( \sqrt{9} + \sqrt{16} = 3 + 4 = 7 \). The answers are different.