Adding and Subtracting Surds
Surds can be added or subtracted only when they are like surds, with the same number under the root, just as \(3x+2x=5x\). Often each surd must be simplified first to show this, as in \(\sqrt{40}-\sqrt{10}=2\sqrt{10}-\sqrt{10}\). In NSW Year 10 Advanced maths this extends to surds with pronumerals and fractions (NESA outcome MA5‑IND‑P‑02).
You will simplify sums and differences such as \(3\sqrt{40}-\sqrt{90}+2\sqrt{160}\), collect terms with pronumerals under the root, find a missing coefficient and solve a simple surd equation. Exact perimeters of rectangles, right-angled triangles and square garden beds use these skills, and lengths on the number plane show when three points lie on one straight line. You will also use surds to decide whether three lengths can form a triangle. It is part of Year 10 Advanced Surds.
Every question with a fully worked solution.
- Adding and Subtracting Surds Watch
Theory
To add or subtract surds, simplify every surd first, then collect like surds by adding or subtracting the numbers in front: \(6\sqrt{5}-3\sqrt{5} = 3\sqrt{5}\). This Year 10 Advanced maths page covers expressions with two kinds of surd, fractions, pronumerals, finding an unknown and exact perimeters.
Like surds have the same number under the root, such as \(4\sqrt{7}\) and \(-\sqrt{7}\). They are added and subtracted exactly like like terms in algebra: \(4\sqrt{7}-\sqrt{7} = 3\sqrt{7}\), just as \(4x - x = 3x\).
Unlike surds such as \(\sqrt{2}\) and \(\sqrt{3}\) cannot be combined into one surd. Never add the numbers under the roots: \(\sqrt{2}+\sqrt{3}\) is about \(3.15\), but \(\sqrt{5}\) is about \(2.24\).
Surds that look different are often like surds in disguise. \(\sqrt{24}\), \(\sqrt{54}\) and \(\sqrt{96}\) simplify to \(2\sqrt{6}\), \(3\sqrt{6}\) and \(4\sqrt{6}\), so always simplify every surd first.
The same idea works with pronumerals under the root (when they are positive) and with surds in fractions, where you first find a common denominator.
| Expression | Simplify first | Result |
|---|---|---|
| \(\sqrt{50}+\sqrt{2}\) | \(5\sqrt{2}+\sqrt{2}\) | \(6\sqrt{2}\) |
| \(\sqrt{40}-\sqrt{10}\) | \(2\sqrt{10}-\sqrt{10}\) | \(\sqrt{10}\) |
| \(\sqrt{3}+\sqrt{12}+\sqrt{5}\) | \(\sqrt{3}+2\sqrt{3}+\sqrt{5}\) | \(3\sqrt{3}+\sqrt{5}\) |
| \(\dfrac{\sqrt{7}}{2}+\dfrac{\sqrt{7}}{3}\) | \(\dfrac{3\sqrt{7}+2\sqrt{7}}{6}\) | \(\dfrac{5\sqrt{7}}{6}\) |
How to add and subtract surds
- Simplify every surd using its largest square factor. Write decimals as fractions over \(100\) first.
- Group the like surds. Underline each family in a different way if there are two.
- Collect each family by adding or subtracting the numbers in front. A surd on its own, \(\sqrt{3}\), has a \(1\) in front.
- Fractions: put the terms over a common denominator before collecting.
- Write the answer with one term per family of surds, plus any whole number.
Other question types
- Unknown coefficient: write every surd as a multiple of the same surd, then compare the numbers in front (Example 3).
- Unknown under the root: if every surd contains \(\sqrt{3x}\), collect them, isolate \(\sqrt{3x}\), then square both sides.
- Patterns: numbers like \(6, 24, 54, 96\) are \(6\times 1^2, 6\times 2^2, 6\times 3^2, 6\times 4^2\), so their roots are \(\sqrt{6}, 2\sqrt{6}, 3\sqrt{6}, 4\sqrt{6}\).
- Measurement: sides from areas (\(\text{side} = \sqrt{\text{area}}\)) or Pythagoras' theorem are added for a perimeter. Keep the exact form until a decimal is asked for.
- Triangle test: three lengths make a triangle only if the two shorter ones add to more than the longest.
(i) \(3\sqrt{40}-\sqrt{90}+2\sqrt{160}\)
(ii) \(4\sqrt{3}+2\sqrt{6}-\sqrt{27}+\sqrt{24}\)
(i) The surds do not look alike yet. Simplify each one first.
| \(3\sqrt{40}\) | \(=\) | \(3\times 2\sqrt{10} = 6\sqrt{10}\) |
| \(\sqrt{90}\) | \(=\) | \(3\sqrt{10}\) |
| \(2\sqrt{160}\) | \(=\) | \(2\times 4\sqrt{10} = 8\sqrt{10}\) |
| \(6\sqrt{10}-3\sqrt{10}+8\sqrt{10}\) | \(=\) | \(\mathbf{11\sqrt{10}}\) |
(ii) Two families appear: \(\sqrt{27} = 3\sqrt{3}\) and \(\sqrt{24} = 2\sqrt{6}\). Collect each family separately.
| \(\) | \(\) | \(4\sqrt{3}-3\sqrt{3}+2\sqrt{6}+2\sqrt{6}\) |
| \(\) | \(=\) | \(\mathbf{\sqrt{3}+4\sqrt{6}}\) |
The answer has two terms. \(\sqrt{3}\) and \(\sqrt{6}\) are not like surds, so it cannot be simplified further.
(i) \(\dfrac{\sqrt{54}}{6}-\dfrac{\sqrt{24}}{8}\)
(ii) \(\sqrt{28y^3}-y\sqrt{7y}\), where \(y\gt 0\)
(i) Simplify each term, then use a common denominator.
| \(\dfrac{\sqrt{54}}{6}\) | \(=\) | \(\dfrac{3\sqrt{6}}{6} = \dfrac{\sqrt{6}}{2}\) |
| \(\dfrac{\sqrt{24}}{8}\) | \(=\) | \(\dfrac{2\sqrt{6}}{8} = \dfrac{\sqrt{6}}{4}\) |
| \(\dfrac{\sqrt{6}}{2}-\dfrac{\sqrt{6}}{4}\) | \(=\) | \(\dfrac{2\sqrt{6}-\sqrt{6}}{4} = \mathbf{\dfrac{\sqrt{6}}{4}}\) |
(ii) \(28y^3 = 4y^2\times 7y\), and \(\sqrt{4y^2} = 2y\) because \(y\gt 0\).
| \(\sqrt{28y^3}-y\sqrt{7y}\) | \(=\) | \(2y\sqrt{7y}-y\sqrt{7y}\) |
| \(\) | \(=\) | \(\boldsymbol{y\sqrt{7y}}\) |
Here the like surd is \(\sqrt{7y}\), and the coefficients \(2y\) and \(y\) are collected like algebra.
Write every surd as a multiple of \(\sqrt{2}\).
| \(\sqrt{8}\) | \(=\) | \(2\sqrt{2}\) |
| \(\sqrt{18}\) | \(=\) | \(3\sqrt{2}\) |
| \(\sqrt{50}\) | \(=\) | \(5\sqrt{2}\) |
Substitute and compare the numbers in front of \(\sqrt{2}\):
| \(2k\sqrt{2}-3\sqrt{2}\) | \(=\) | \(5\sqrt{2}\) |
| \(2k-3\) | \(=\) | \(5\) |
| \(k\) | \(=\) | \(\boldsymbol{4}\) |
Check: \(4\sqrt{8} = 8\sqrt{2}\), and \(8\sqrt{2}-3\sqrt{2} = 5\sqrt{2}\).
Pythagoras' theorem first. Squaring a square root gives back the number under it.
| \(h^2\) | \(=\) | \((\sqrt{20})^2+(\sqrt{45})^2 = 20+45 = 65\) |
| \(h\) | \(=\) | \(\sqrt{65}\) |
\(65 = 5\times 13\) has no square factor, so \(\sqrt{65}\) is already simplest. Now add the sides:
| \(P\) | \(=\) | \(2\sqrt{5}+3\sqrt{5}+\sqrt{65}\) |
| \(\) | \(=\) | \(\mathbf{5\sqrt{5}+\sqrt{65}}\text{ m}\) |
| \(\) | \(\approx\) | \(\mathbf{19.2}\text{ m}\) |
\(5\sqrt{5}\) and \(\sqrt{65}\) are unlike surds, so the exact answer keeps two terms.
Common pitfalls
Frequently asked questions
How do you add surds?
Simplify each surd first. Then add the numbers in front of surds that have the same number under the root, keeping that surd the same. For example \(6\sqrt{7}+3\sqrt{7} = 9\sqrt{7}\).
Can you add root 2 and root 3?
Not into a single surd. \(\sqrt{2}\) and \(\sqrt{3}\) are unlike surds, so \(\sqrt{2}+\sqrt{3}\) stays as it is. It is about \(3.15\), while \(\sqrt{5}\) is only about \(2.24\).
What are like surds?
Like surds have the same number under the root once they are simplified, such as \(2\sqrt{7}\) and \(5\sqrt{7}\). \(\sqrt{24}\) and \(\sqrt{54}\) are also like surds, because they simplify to \(2\sqrt{6}\) and \(3\sqrt{6}\).
How do you subtract surds with different numbers under the root?
Simplify each surd. If they become like surds, subtract the numbers in front. If they are still different, the expression cannot be simplified further and the answer keeps both terms.
How do you add surds that are fractions?
Simplify each surd, then write the terms over a common denominator and collect the like surds in the numerator, the same way you add algebraic fractions.