E1.18 - Simplifying and combining surds | Cambridge IGCSE Mathematics

Some square roots cannot be written as terminating or recurring decimals. In this Extended lesson, you will keep them exact, simplify them by extracting square factors, combine matching surds and multiply surds without losing accuracy.

Extracting square factors

Can $\sqrt{72}$ be written more simply without replacing it by a rounded decimal? A square factor gives the answer: $72=36\times2$ , so its square root contains a factor of $6$ .

Surd

A surd is an irrational root written in exact form, such as $\sqrt2$ or $\sqrt7$ . Not every root is a surd: $\sqrt{49}=7$ is rational, and $\sqrt0=0$ . The square-root symbol means the non-negative square root; $\sqrt{49}$ is $7$ , not $\pm7$ .

Recall that squaring and taking the non-negative square root undo one another for non-negative numbers. Since $6^2=36$ , a factor of $36$ under a root becomes a factor of $6$ outside it.

Splitting a product under a square root

\sqrt{ab}=\sqrt a\sqrt b\qquad(a\ge0,\ b\ge0)

Both sides are non-negative and their squares equal $ab$ , which explains the rule. Negative numbers have no real square roots, so these conditions matter.

Not in the List of formulas: learn it.

Choose the largest square factor you can see. If you start with a smaller one, the method still works, but you may need another simplification.

Worked example 1: taking out a square factor

Question. Simplify $\sqrt{72}$ .

Plan. Split $72$ into a square factor and the remaining factor.

\begin{aligned} \sqrt{72}&=\sqrt{36\times2} &&\text{choose the square factor 36}\\ &=\sqrt{36}\sqrt2 &&\text{split the product}\\ &=6\sqrt2 &&\text{take the square root of 36} \end{aligned}

Answer. $6\sqrt2$ .

Check. $(6\sqrt2)^2=36\times2=72$ , and the answer is positive, as required.

Change only the square factor from $36$ to $49$ . The remaining surd stays the same.

Worked example 2: a different square factor

Question. Simplify $\sqrt{98}$ .

Plan. Use $98=49\times2$ .

\begin{aligned} \sqrt{98}&=\sqrt{49\times2} &&\text{expose a square factor}\\ &=7\sqrt2 &&\text{take out the factor 7} \end{aligned}

Answer. $7\sqrt2$ .

Check. Its square is $49\times2=98$ . Also, $9<\sqrt{98}<10$ , consistent with $7\sqrt2$ .

Watch out

$\sqrt{36\times2}=36\sqrt2$ is wrong. The factor outside must be the square root of $36$ , so it is $6\sqrt2$ . In simplest form, the number left under the square root has no square factor greater than $1$ .

Combining like surds

Three lots of $\sqrt5$ plus two lots of $\sqrt5$ make five lots of $\sqrt5$ . The root is a fixed number; you are counting equal-sized quantities. Thus $3\sqrt5+2\sqrt5=5\sqrt5$ , just as three metres plus two metres make five metres.

Like surds

Like surds have the same root part after simplification. Their numerical coefficients can be added or subtracted. For example, $2\sqrt3$ and $7\sqrt3$ are like surds; $2\sqrt3$ and $2\sqrt5$ are not.

Simplify first: $\sqrt{12}$ and $\sqrt{27}$ look different, but both contain the root part $\sqrt3$ . This is how you decide which terms can combine.

Worked example 3: reveal matching roots

Question. Simplify $\sqrt{50}+\sqrt{18}$ .

Plan. Extract square factors, then add the coefficients of $\sqrt2$ .

\begin{aligned} \sqrt{50}+\sqrt{18}&=5\sqrt2+3\sqrt2 &&\text{use 25 and 9 as square factors}\\ &=(5+3)\sqrt2 &&\text{count equal root parts}\\ &=8\sqrt2 &&\text{add the coefficients} \end{aligned}

Answer. $8\sqrt2$ .

Check. Subtract $\sqrt{18}=3\sqrt2$ from the result: $8\sqrt2-3\sqrt2=5\sqrt2=\sqrt{50}$ .

Keep the roots and change only addition to subtraction. The coefficients now subtract.

Worked example 4: the same roots, a subtraction

Question. Simplify $\sqrt{50}-\sqrt{18}$ .

Plan. Use the same simplified forms and retain the subtraction sign.

\begin{aligned} \sqrt{50}-\sqrt{18}&=5\sqrt2-3\sqrt2 &&\text{simplify each root}\\ &=(5-3)\sqrt2 &&\text{subtract matching quantities}\\ &=2\sqrt2 &&\text{collect the coefficients} \end{aligned}

Answer. $2\sqrt2$ .

Check. Add back $3\sqrt2$ : $2\sqrt2+3\sqrt2=5\sqrt2$ , recovering the first term.

Some expressions cannot combine further. The exact answer $\sqrt2+\sqrt3$ stays as it is; there is no common root part. Likewise $2\sqrt3+5-\sqrt3=\sqrt3+5$ : combine the surds and leave the rational term separate.

Watch out

$\sqrt a+\sqrt b=\sqrt{a+b}$ is not a rule. For instance, $\sqrt9+\sqrt{16}=3+4=7$ , whereas $\sqrt{9+16}=5$ . Splitting a product works; splitting a sum this way does not.

Products of surds and exact answers

Multiplication uses a different decision from addition: the root parts do not need to match. Multiply the numerical factors together, multiply the roots using the product rule, then simplify.

For a non-negative number, $\sqrt a\times\sqrt a=a$ : that is the defining property of its square root. A product of irrational numbers can therefore be rational. For example, $\sqrt2\times\sqrt2=2$ .

Worked example 5: a product that remains a surd

Question. Work out $2\sqrt3\times\sqrt6$ , giving your answer in simplest exact form.

Plan. Multiply the root parts, then extract a square factor.

\begin{aligned} 2\sqrt3\times\sqrt6&=2\sqrt{18} &&\text{multiply the numbers under the roots}\\ &=2\times3\sqrt2 &&\text{use }18=9\times2\\ &=6\sqrt2 &&\text{multiply the coefficients} \end{aligned}

Answer. $6\sqrt2$ .

Check. The original positive product has square $4\times3\times6=72$ ; the answer has square $36\times2=72$ too.

Change only the second root, from $\sqrt6$ to $\sqrt{12}$ . The product under the root is now a square.

Worked example 6: a rational product

Question. Work out $2\sqrt3\times\sqrt{12}$ .

Plan. Combine the roots before taking the square root.

\begin{aligned} 2\sqrt3\times\sqrt{12}&=2\sqrt{36} &&\text{multiply 3 by 12}\\ &=2\times6 &&\text{36 is a square}\\ &=12 &&\text{finish the multiplication} \end{aligned}

Answer. $12$ .

Check. Alternatively, $\sqrt{12}=2\sqrt3$ , so the product is $2\times2\times3=12$ .

An exact answer preserves the full value, such as $6\sqrt2$ . A decimal such as $8.49$ is a rounded approximation. Use decimals for an estimate if helpful, but retain roots throughout working when an exact answer is requested.

Exam tip

“Simplify” requires the square factors to be extracted and any like surds combined. A decimal approximation does not replace an exact simplified surd.

Quick Recap

Extract square factors: their square roots become coefficients outside the root.
Simplify first, then combine coefficients of matching root parts.
Multiply non-negative root parts using $\sqrt a\sqrt b=\sqrt{ab}$ ; multiplication can produce a rational answer.
The square-root symbol gives the non-negative root, and exact answers retain surds where needed.

The opening expression simplifies exactly to $\sqrt{72}=6\sqrt2$ , without rounding any digits.

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E1.18 - Simplifying and combining surds

Extracting square factors

Combining like surds

Products of surds and exact answers

Quick Recap

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