Simplifying Surds

We need to find the largest perfect square factor of 72. Since \( 72 = 36 \times 2 \), we get \( \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2} \). We can verify: \( (6\sqrt{2})^2 = 36 \times 2 = 72 \) ✓.

A common incomplete simplification is \( \sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18} \). This isn’t wrong, but it isn’t fully simplified because 18 still contains the square factor 9. The “not using the largest square factor” misconception leads to extra work — always look for the largest square factor to simplify in one step.

The left-hand side is \( \sqrt{25} = 5 \). The right-hand side is \( \sqrt{9} + \sqrt{16} = 3 + 4 = 7 \). Since \( 5 \neq 7 \), the two expressions are not equal. The square root does not “distribute” over addition.

This addresses the “splitting surds over addition” misconception — students who treat \( \sqrt{a + b} \) as \( \sqrt{a} + \sqrt{b} \) by analogy with multiplication, where \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \) genuinely does work. The product rule works, but there is no sum rule for surds.

Walking the edges (3 + 4 = 7) is not equal to walking the diagonal (5).

The square root of 36 asks “what number multiplied by itself gives 36?” Since \( 6 \times 6 = 36 \), we have \( \sqrt{36} = 6 \), not 18. The answer 18 comes from halving 36 — dividing by 2, not finding the square root.

This targets the “square root means halve” misconception. Students confuse \( \sqrt{} \) with \( \div 2 \), which happens to give the correct answer for \( \sqrt{4} \) (since \( 4 \div 2 = 2 = \sqrt{4} \)) but fails everywhere else. A good test: halving 9 gives 4.5, but \( \sqrt{9} = 3 \).

Using the multiplication rule for surds: \( \sqrt{3} \times \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36} = 6 \). So the product is 6, a whole number — even though both individual factors are irrational.

Alternatively, simplify \( \sqrt{12} = 2\sqrt{3} \) first, giving \( \sqrt{3} \times 2\sqrt{3} = 2 \times (\sqrt{3})^2 = 2 \times 3 = 6 \). Students often assume that multiplying surds always gives another surd. This question shows that when the product under the root is a perfect square, the result is rational.

To rationalise the denominator, we multiply the numerator and denominator by \( \sqrt{3} \). This gives \( \frac{6 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{6\sqrt{3}}{3} \). Simplifying the fraction \( \frac{6}{3} = 2 \), we are left with \( 2\sqrt{3} \).

Alternatively, think of 6 as \( 2 \times 3 \), and since \( 3 = \sqrt{3} \times \sqrt{3} \), we have \( \frac{2 \times \sqrt{3} \times \sqrt{3}}{\sqrt{3}} = 2\sqrt{3} \).

Example: \( \sqrt{8} = 2\sqrt{2} \) (since \( 8 = 4 \times 2 \))

Another: \( \sqrt{50} = 5\sqrt{2} \) (since \( 50 = 25 \times 2 \))

Creative: \( \sqrt{200} = 10\sqrt{2} \) — the largest square factor is 100, which students might miss, going via \( \sqrt{4 \times 50} = 2\sqrt{50} \) and then needing a second step.

Trap: \( \sqrt{15} \) — a student might try to write this as \( \sqrt{3} \times \sqrt{5} \) and think they’ve “simplified” it, but 15 has no perfect square factor other than 1, so \( \sqrt{15} \) is already in its simplest surd form.

Example: \( \sqrt{18} = 3\sqrt{2} \) (since \( 18 = 9 \times 2 \))

Another: \( \sqrt{27} = 3\sqrt{3} \) (since \( 27 = 9 \times 3 \))

Creative: \( \sqrt{45} = 3\sqrt{5} \) (since \( 45 = 9 \times 5 \)) — students tend to gravitate toward familiar surds like \( \sqrt{18} \) and \( \sqrt{27} \), but any \( 9k \) works where \( k \) has no square factor.

Trap: \( \sqrt{36} = 6 \) — a student might argue the coefficient is “3 × 2”, but \( \sqrt{36} \) is a rational number (6), not a surd with coefficient 3. The question asks for a surd in the form \( 3\sqrt{n} \) where \( n \) is not a perfect square.

Example: \( \sqrt{2} + \sqrt{2} = 2\sqrt{2} \)

Another: \( 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5} \)

Creative: \( \sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} \) — surds that don’t look alike at first but simplify to like surds.

Think of \( \sqrt{5} \) like the letter \( x \). Just as \( 3x + 2x = 5x \), we know that \( 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5} \). You can only add them if the ‘letter’ (the radicand) is exactly the same.

Trap: \( \sqrt{2} + \sqrt{3} = \sqrt{5} \) — this is false. A student applying the “sum rule” misconception might add the radicands, but \( \sqrt{2} + \sqrt{3} \approx 1.41 + 1.73 = 3.14 \), while \( \sqrt{5} \approx 2.24 \). You can only add surds with the same radicand (like terms).

Example: \( \sqrt{72} \) (since \( 72 = 36 \times 2 \), so \( \sqrt{72} = 6\sqrt{2} \))

Another: \( 2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2} \)

Creative: \( \sqrt{8} \times \sqrt{9} = 2\sqrt{2} \times 3 = 6\sqrt{2} \) — using multiplication of surds rather than simplification of a single surd.

Trap: \( \sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3} \) — a student might think “6 and √2 are both there so it must be \( 6\sqrt{2} \)”, but actually \( \sqrt{6} \times \sqrt{2} = \sqrt{12} \), not \( 6\sqrt{2} \). This exposes confusion between the coefficient and the radicand.

Example: \( (2 + \sqrt{3})(2 – \sqrt{3}) = 4 – 3 = 1 \)

Another: \( (\sqrt{5} + 1)(\sqrt{5} – 1) = 5 – 1 = 4 \)

Creative: \( (\sqrt{8} + \sqrt{2})(\sqrt{8} – \sqrt{2}) = 8 – 2 = 6 \)

Trap: \( (2 + \sqrt{3})(2 + \sqrt{3}) \) — this expands to \( 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \), which is irrational. It must be the difference of two squares to eliminate the middle surd terms.

It depends on whether \( n \) has a perfect square factor greater than 1. True case: \( \sqrt{12} = 2\sqrt{3} \) — here \( 12 = 4 \times 3 \), so the square factor 4 allows simplification. False case: \( \sqrt{7} \) cannot be simplified because 7 is prime and has no perfect square factor other than 1.

Students who think all surds can be simplified will try to “break down” primes like \( \sqrt{7} \). Conversely, students who think no surds simplify will miss opportunities like \( \sqrt{12} \). A surd \( \sqrt{n} \) simplifies if and only if \( n \) has a factor that is a perfect square greater than 1.

This is the product rule for surds, and it is always true for positive values of \( a \) and \( b \). For example, \( \sqrt{3} \times \sqrt{5} = \sqrt{15} \), and \( \sqrt{4} \times \sqrt{9} = \sqrt{36} = 6 = 2 \times 3 \) ✓.

Students may doubt this rule because it feels “too easy” or because they confuse it with the (incorrect) addition rule. The product rule is one of the foundational laws for simplifying surds — for instance, \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \) uses this rule in reverse.

This is never true for positive values of \( a \) and \( b \). Consider any example: let \( a = 4 \), \( b = 9 \). Then \( \sqrt{4 + 9} = \sqrt{13} \approx 3.61 \), but \( \sqrt{4} + \sqrt{9} = 2 + 3 = 5 \). We can prove it algebraically: if it were true, squaring both sides would give \( a + b = a + 2\sqrt{ab} + b \), which simplifies to \( 2\sqrt{ab} = 0 \), meaning \( ab = 0 \). Since \( a \) and \( b \) are positive, this is impossible.

This is the “splitting surds over addition” misconception — one of the most persistent errors in surd work. Students overgeneralise the (correct) product rule \( \sqrt{ab} = \sqrt{a}\sqrt{b} \) and assume a similar rule exists for addition.

Squaring \( a\sqrt{b} \) means \( a\sqrt{b} \times a\sqrt{b} = a \times a \times \sqrt{b} \times \sqrt{b} = a^2 \times b \). For example, \( (3\sqrt{2})^2 = 9 \times 2 = 18 \), and \( (5\sqrt{7})^2 = 25 \times 7 = 175 \).

Students often make errors here, such as only squaring the coefficient (getting \( (3\sqrt{2})^2 = 9\sqrt{2} \)) or only squaring the surd (getting \( (3\sqrt{2})^2 = 3 \times 2 = 6 \)). The correct approach is to square every factor: both the coefficient and the surd part.

Rationalising the denominator gives \( \frac{a\sqrt{a}}{\sqrt{a}\sqrt{a}} = \frac{a\sqrt{a}}{a} = \sqrt{a} \). Alternatively, remember that \( a = \sqrt{a} \times \sqrt{a} \), so dividing \( a \) by \( \sqrt{a} \) leaves \( \sqrt{a} \).

The student has correctly multiplied the coefficients (\( 3 \times 4 = 12 \)) but has ignored one of the surd parts instead of multiplying them together. This is the “dropping a surd when multiplying” misconception — the student treats the surds like labels rather than quantities that must also be multiplied.

The correct method: multiply the coefficients (\( 3 \times 4 = 12 \)) AND multiply the surds (\( \sqrt{2} \times \sqrt{5} = \sqrt{10} \)), giving \( 12\sqrt{10} \). A decimal check confirms: \( 3\sqrt{2} \times 4\sqrt{5} \approx 4.24 \times 8.94 \approx 37.9 \), and \( 12\sqrt{10} \approx 12 \times 3.16 \approx 37.9 \) ✓, while \( 12\sqrt{5} \approx 26.8 \) ✗.

If the student was right, then \( 3\sqrt{2} \times 4\sqrt{5} \) would be the exact same as \( 3 \times 4\sqrt{5} \). Does it make sense that multiplying by \( \sqrt{2} \) does nothing?

The student has the correct answer — \( \sqrt{3} \times \sqrt{3} = 3 \) — but for the wrong reason. Square root signs do not “cancel” in the way the student describes. The correct reasoning is: by definition, \( \sqrt{3} \) is the positive number whose square is 3, so \( \sqrt{3} \times \sqrt{3} = (\sqrt{3})^2 = 3 \). Alternatively, using the product rule: \( \sqrt{3} \times \sqrt{3} = \sqrt{9} = 3 \).

The student’s “cancelling” reasoning would incorrectly predict that \( \sqrt{3} + \sqrt{3} = 3 \) (the roots “cancel” by addition too), when in fact \( \sqrt{3} + \sqrt{3} = 2\sqrt{3} \approx 3.46 \). Correct reasoning matters even when the answer happens to be right.

The student has correctly applied the product rule — \( \sqrt{48} = \sqrt{4 \times 12} = 2\sqrt{12} \) — but has not fully simplified. This is the “not using the largest square factor” misconception. The number under the square root, 12, still contains the square factor 4: \( 12 = 4 \times 3 \).

The fully simplified answer is \( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \). Using the largest square factor (16) in one step avoids the need for multiple rounds of simplification. From the student’s answer: \( 2\sqrt{12} = 2 \times 2\sqrt{3} = 4\sqrt{3} \), confirming the correct final form.

The student has only squared the surd part, not the coefficient — this is the “forgetting to square the coefficient” misconception. The expression \( (2\sqrt{3})^2 \) means \( 2\sqrt{3} \times 2\sqrt{3} \). We must square both the 2 and the \( \sqrt{3} \): \( (2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12 \).

The correct answer is 12, not 6. The student’s error is treating \( (2\sqrt{3})^2 \) as \( 2 \times (\sqrt{3})^2 \) instead of \( (2)^2 \times (\sqrt{3})^2 \). A quick check: \( 2\sqrt{3} \approx 3.46 \), and \( 3.46^2 \approx 12 \), confirming the answer is 12.

The mistake is dividing a surd by a rational integer directly. You cannot divide the radicand (20) by a number outside the root (2). To divide them, they must both be surds.

We can rewrite 2 as \( \sqrt{4} \). Then \( \frac{\sqrt{20}}{\sqrt{4}} = \sqrt{5} \). Alternatively, simplify the numerator first: \( \sqrt{20} = 2\sqrt{5} \), so \( \frac{2\sqrt{5}}{2} = \sqrt{5} \).