Operations with Surds

Many students assume the multiplication rule — \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) — also applies to addition, writing \( \sqrt{a} + \sqrt{b} = \sqrt{a + b} \). This is the “addition under the root” misconception.

We can visualise this contradiction with a right-angled triangle. If the legs are \( \sqrt{2} \) and \( \sqrt{3} \), Pythagoras’ Theorem dictates the hypotenuse is exactly \( \sqrt{5} \).

Since the shortest distance between two points is a straight line, walking along the two legs (\( \sqrt{2} + \sqrt{3} \)) must be longer than walking straight along the hypotenuse (\( \sqrt{5} \)). Clearly, \( \sqrt{2} + \sqrt{3} \neq \sqrt{5} \).

To simplify \( \sqrt{50} \), look for the largest perfect square factor of 50. Since \( 50 = 25 \times 2 \) and \( 25 = 5^2 \), we get \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \). This demonstrates the simplification of surds — the key is finding perfect square factors, not just any factors. Students who struggle here often don’t systematically check for square number factors such as 4, 9, 16, 25.

We can verify by squaring: \( (5\sqrt{2})^2 = 25 \times 2 = 50 \), so \( 5\sqrt{2} = \sqrt{50} \). Both expressions have the same value (\( \approx 7.071 \)).

Students often expand \( (\sqrt{3} + 1)^2 \) by squaring each term separately: \( (\sqrt{3})^2 + 1^2 = 3 + 1 = 4 \). This is the “missing middle term” misconception. An area model beautifully demonstrates why this is false by showing all four expanded terms:

The correct expansion is \( (\sqrt{3})^2 + 2(\sqrt{3})(1) + 1^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \). The total area is not just the 3 and the 1, but also the two \( \sqrt{3} \) rectangles.

Multiply both numerator and denominator of \( \frac{1}{\sqrt{2}} \) by \( \sqrt{2} \): \( \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{2} \). This is multiplying by \( \frac{\sqrt{2}}{\sqrt{2}} = 1 \), which doesn’t change the value — it only changes the form. This addresses the “rationalising changes the value” misconception, where students believe rewriting a fraction with a rational denominator produces a different number.

We can verify both are equal numerically: \( \frac{1}{\sqrt{2}} = 1 \div 1.414\ldots \approx 0.707 \), and \( \frac{\sqrt{2}}{2} = 1.414\ldots \div 2 \approx 0.707 \). Rationalising the denominator expresses the same value in a more standard form.

In algebra, \( x \) represents an unknown value. \( 5x – 2x \) means “5 lots of \( x \) minus 2 lots of \( x \)”, which logically leaves 3 lots of \( x \). A surd like \( \sqrt{7} \) is simply a specific (irrational) number. Just like \( x \), we can treat it as a single unit.

“5 lots of \( \sqrt{7} \) minus 2 lots of \( \sqrt{7} \) leaves 3 lots of \( \sqrt{7} \)”. We can make this algebraic connection completely explicit by factorising out the common term: \( 5\sqrt{7} – 2\sqrt{7} = \sqrt{7}(5 – 2) = \sqrt{7}(3) = 3\sqrt{7} \). Connecting surd operations directly to algebra builds immense confidence.

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

Another: \( 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2} \)

Creative: \( \sqrt{8} + \sqrt{2} = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2} \) — these don’t look like like surds at first, but simplifying \( \sqrt{8} = 2\sqrt{2} \) reveals they can be collected.

Trap: \( \sqrt{2} + \sqrt{3} \) — a student might write this as \( \sqrt{5} \), applying the false rule \( \sqrt{a} + \sqrt{b} = \sqrt{a + b} \). In reality, \( \sqrt{2} \) and \( \sqrt{3} \) are unlike surds and cannot be combined into a single term, just as you cannot add \( 2x + 3y \) into a single term.

Example: \( \sqrt{3} \times \sqrt{3} = 3 \)

Another: \( \sqrt{2} \times \sqrt{8} = \sqrt{16} = 4 \)

Creative: \( \sqrt{6} \times \sqrt{24} = \sqrt{144} = 12 \) — less obvious because neither radicand is a perfect square, yet the product is.

Trap: \( \sqrt{2} \times \sqrt{3} = \sqrt{6} \) — a student might assume that multiplying any two surds gives a whole number. In fact, \( \sqrt{6} \approx 2.449 \), which is irrational. The product is only rational when the combined radicand is a perfect square.

Example: \( \sqrt{5} + \sqrt{5} + \sqrt{5} = 3\sqrt{5} \)

Another: \( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \)

Creative: \( \frac{6\sqrt{10}}{2\sqrt{2}} = 3 \times \frac{\sqrt{10}}{\sqrt{2}} = 3\sqrt{5} \) — using division of surds to produce the target expression.

Trap: \( \sqrt{15} \) — a student might think \( \sqrt{15} = \sqrt{3 \times 5} = 3\sqrt{5} \), confusing \( \sqrt{3 \times 5} \) with \( 3 \times \sqrt{5} \). In fact, \( \sqrt{15} \approx 3.873 \) while \( 3\sqrt{5} \approx 6.708 \). The notation \( 3\sqrt{5} \) means 3 multiplied by \( \sqrt{5} \), not \( \sqrt{3 \times 5} \).

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

Another: \( (\sqrt{3} + \sqrt{2})(\sqrt{3} \; – \; \sqrt{2}) = 3 \; – \; 2 = 1 \)

Creative: \( (2\sqrt{7} + 3)(2\sqrt{7} \; – \; 3) = (2\sqrt{7})^2 \; – \; 3^2 = 28 \; – \; 9 = 19 \) — conjugates with coefficients, producing a less obvious whole number.

Trap: \( (\sqrt{5} + \sqrt{3})(\sqrt{5} + \sqrt{3}) = (\sqrt{5} + \sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15} \) — a student might think squaring the expression is the same as multiplying conjugates. But \( (\sqrt{5} + \sqrt{3})^2 \) is not the same as \( (\sqrt{5} + \sqrt{3})(\sqrt{5} \; – \; \sqrt{3}) \). The signs must differ for the cross-term to cancel via the difference of two squares.

This depends on whether the product of the radicands is a perfect square. True case: \( \sqrt{2} \times \sqrt{3} = \sqrt{6} \), which is irrational since 6 is not a perfect square. False case: \( \sqrt{2} \times \sqrt{18} = \sqrt{36} = 6 \), which is rational. The “surd products are always irrational” misconception ignores the possibility that the radicands multiply to give a perfect square.

Two surds that are “different” can still produce a rational product. The key is whether the combined radicand \( a \times b \) is a perfect square.

True case: Start with 5. Square it: 25. Take the square root: \( \sqrt{25} = 5 \) ✓. False case: Start with −3. Square it: 9. Take the square root: \( \sqrt{9} = 3 \), not −3 ✗. The process \( \sqrt{a^2} \) gives \( |a| \), the absolute value of \( a \), which only equals \( a \) when \( a \) is non-negative.

This exposes the “inverse operations always undo” misconception — students often forget that the square root function, by definition, returns only the non-negative root. The statement is true for all non-negative numbers and false for all negative numbers.

By definition, \( \sqrt{a} \) is the non-negative number whose square is \( a \). So \( \sqrt{a} \times \sqrt{a} = (\sqrt{a})^2 = a \), which is rational (since \( a \) is a rational number in all standard GCSE contexts). Examples: \( \sqrt{3} \times \sqrt{3} = 3 \), \( \sqrt{7} \times \sqrt{7} = 7 \), \( \sqrt{11} \times \sqrt{11} = 11 \).

Students sometimes work this out via the multiplication rule — \( \sqrt{3} \times \sqrt{3} = \sqrt{9} = 3 \) — which is correct but obscures the simpler reasoning: squaring undoes the square root directly. There is no counterexample: multiplying any surd by itself always returns the original radicand.

Simplifying a surd means finding a perfect square factor of the radicand (other than 1) and extracting its root. A prime number \( p \) has only two factors: 1 and \( p \) itself. Since \( p \) is not a perfect square (no prime is), there is no perfect square factor other than 1 to extract. Therefore \( \sqrt{p} \) is already in its simplest form for any prime \( p \).

Examples: \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \), \( \sqrt{7} \), \( \sqrt{11} \), \( \sqrt{13} \) — none can be simplified. The “all surds can be simplified” misconception leads students to attempt false simplifications such as writing \( \sqrt{7} = \sqrt{1 \times 7} = 1\sqrt{7} \) (which is pointless) or incorrectly factoring primes.

The student has applied the “addition under the root” misconception, incorrectly assuming \( \sqrt{a} + \sqrt{b} = \sqrt{a + b} \). There is no such rule — you cannot add the radicands of separate surds. The correct approach is to first simplify \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \), then collect like surds: \( \sqrt{3} + 2\sqrt{3} = 3\sqrt{3} \).

Checking: \( \sqrt{15} \approx 3.87 \) while \( 3\sqrt{3} \approx 5.20 \) — these are clearly different values. The multiplication rule \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) does work, but students must not confuse it with addition.

The student forgot to apply the power to the coefficient. When squaring a product, you must square every part inside the bracket: \( (ab)^2 = a^2b^2 \). The “cancelling” logic only works on the surd part, not the whole expression.

The correct calculation is \( 3^2 \times (\sqrt{2})^2 = 9 \times 2 = 18 \). Forgetting to square the integer coefficient is an incredibly common structural error in surds.

The student has made the “extracting a non-square factor” error. They split \( 18 = 3 \times 6 \) and moved the 3 outside the root, as if \( \sqrt{3 \times 6} = 3\sqrt{6} \). But \( \sqrt{3 \times 6} = \sqrt{3} \times \sqrt{6} \), not \( 3\sqrt{6} \). To take a number outside the square root, it must be a perfect square: you extract its square root, not the number itself.

The correct simplification uses the largest perfect square factor: \( 18 = 9 \times 2 = 3^2 \times 2 \), so \( \sqrt{18} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} \). Checking: \( 3\sqrt{6} \approx 7.35 \) while \( 3\sqrt{2} \approx 4.24 \) — the student’s answer is significantly larger than the correct value.

The student has made the “multiply only the denominator” error when rationalising. They correctly identified that \( \sqrt{5} \times \sqrt{5} = 5 \), but only multiplied the denominator — this changes the value of the fraction. To keep the value unchanged, both numerator and denominator must be multiplied by the same amount: \( \frac{1 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{5}}{5} \).

Checking: \( \frac{1}{5} = 0.2 \), while the original \( \frac{1}{\sqrt{5}} \approx 0.447 \). The student has roughly halved the value. The correct answer \( \frac{\sqrt{5}}{5} \approx 0.447 \) matches the original, confirming the value is preserved.

Multiplying by \( \sqrt{3} \) does not eliminate the surd from a binomial (two-term) denominator. The student gets \( 2\sqrt{3} + 3 \) on the bottom, which is still irrational!

To rationalise a two-term denominator, you must multiply the numerator and denominator by its conjugate, \( 2 – \sqrt{3} \). This uses the difference of two squares to eliminate the surds entirely: \( (2 + \sqrt{3})(2 – \sqrt{3}) = 4 – 3 = 1 \). So the correct answer is \( 3(2 – \sqrt{3}) \) or \( 6 – 3\sqrt{3} \).