Revision notes for CIE IGCSE Maths Surds. Open the guide for explanations and worked examples. Written against the CIE IGCSE Maths (0580) specification, so the content matches what's examinable rather than general Maths background.
Revision notes for CIE IGCSE Maths Surds. Open the guide for explanations and worked examples. Written against the CIE IGCSE Maths (0580) specification, so the content matches what's examinable rather than general Maths background.
Before surds, you need to be confident with square numbers: 1, 4, 9, 16, 25, 36, 49, and so on.
A square root asks: “What number squares to give this?” For example, 36=6\sqrt{36}=636=6 because 62=366^2=3662=36.
Key vocabulary
So 25\sqrt{25}25 is not a surd because it equals 5, but 5\sqrt{5}5 is a surd.
Recognising surds
Decide which of 64\sqrt{64}64, 18\sqrt{18}18, and 10\sqrt{10}10 are surds.
First check for exact square roots:
64=8\sqrt{64}=864=8Since 64\sqrt{64}64 is a whole number, it is not a surd.
The numbers 18 and 10 are not perfect squares, so 18\sqrt{18}18 and 10\sqrt{10}10 are surds.
To simplify a surd, look for the largest perfect-square factor inside the square root.
Pull out square factors
For non-negative values, you can split a product inside a square root: ab=ab\sqrt{ab}=\sqrt{a}\sqrt{b}ab=ab. This lets you take square factors out of the root.
For example, if a number contains a factor of 4, 9, 16, 25, 36, and so on, that factor can usually help you simplify.
Writing a surd in simplest form
Write 72\sqrt{72}72 in the form k2k\sqrt{2}k2, where kkk is an integer.
Find a square factor of 72 that leaves 2:
72=36×272=36 \times 272=36×2Split the square root and simplify the square part:
72=36×2=362=62\sqrt{72}=\sqrt{36 \times 2}=\sqrt{36}\sqrt{2}=6\sqrt{2}72=36×2=362=62Therefore the answer is 626\sqrt{2}62.
If there is already a number multiplying the surd, simplify the root first, then multiply the outside numbers.
Simplifying with a coefficient
Write 4454\sqrt{45}445 in the form k5k\sqrt{5}k5, where kkk is an integer.
Simplify the surd part first:
45=9×5=35\sqrt{45}=\sqrt{9 \times 5}=3\sqrt{5}45=9×5=35Now multiply by the coefficient 4:
445=4×35=1254\sqrt{45}=4 \times 3\sqrt{5}=12\sqrt{5}445=4×35=125So k=12k=12k=12.
Do not split sums inside roots
You may split multiplication, but not addition. For example, 9+16\sqrt{9+16}9+16 is not the same as 9+16\sqrt{9}+\sqrt{16}9+16.
Like surds
Like surds have the same root part, such as 373\sqrt{7}37 and 575\sqrt{7}57. You can add or subtract like surds, just like collecting like terms in algebra.
For example, 23+73=932\sqrt{3}+7\sqrt{3}=9\sqrt{3}23+73=93, but 23+752\sqrt{3}+7\sqrt{5}23+75 cannot be collected.
Collecting like surds
Simplify 348+2753\sqrt{48}+2\sqrt{75}348+275.
Simplify each surd separately:
48=16×3=43\sqrt{48}=\sqrt{16 \times 3}=4\sqrt{3}48=16×3=43Also simplify 75\sqrt{75}75:
75=25×3=53\sqrt{75}=\sqrt{25 \times 3}=5\sqrt{3}75=25×3=53Substitute these back in:
348+275=3(43)+2(53)3\sqrt{48}+2\sqrt{75}=3(4\sqrt{3})+2(5\sqrt{3})348+275=3(43)+2(53)Collect the like surds:
123+103=22312\sqrt{3}+10\sqrt{3}=22\sqrt{3}123+103=223Expand brackets with surds in the same way as algebra brackets: multiply every term in the first bracket by every term in the second bracket.
Remember that aa=a\sqrt{a}\sqrt{a}=aaa=a. For example, 55=5\sqrt{5}\sqrt{5}=555=5.
Expanding two brackets
Expand and simplify (2+7)(3−7)(2+\sqrt{7})(3-\sqrt{7})(2+7)(3−7).
Multiply out the four products:
(2+7)(3−7)=6−27+37−(7)2(2+\sqrt{7})(3-\sqrt{7})=6-2\sqrt{7}+3\sqrt{7}-(\sqrt{7})^2(2+7)(3−7)=6−27+37−(7)2Simplify the square root squared:
6−27+37−76-2\sqrt{7}+3\sqrt{7}-76−27+37−7Collect the ordinary numbers and the like surds:
−1+7-1+\sqrt{7}−1+7Conjugates
Two expressions such as a+ba+\sqrt{b}a+b and a−ba-\sqrt{b}a−b are called conjugates. Their product has no surd part because the middle terms cancel.
This is just the difference of two squares:
(a+b)(a−b)=a2−b(a+\sqrt{b})(a-\sqrt{b})=a^2-b(a+b)(a−b)=a2−bUsing conjugates
Expand and simplify (5+6)(5−6)(5+\sqrt{6})(5-\sqrt{6})(5+6)(5−6).
Recognise the brackets as conjugates.
Use the difference of two squares:
(5+6)(5−6)=52−(6)2(5+\sqrt{6})(5-\sqrt{6})=5^2-(\sqrt{6})^2(5+6)(5−6)=52−(6)2Simplify:
25−6=1925-6=1925−6=19When you square a bracket, write it twice. This helps you avoid missing the middle term.
Squaring a surd bracket
Write (3−5)2(3-\sqrt{5})^2(3−5)2 in the form a+b5a+b\sqrt{5}a+b5.
Write the squared bracket as two identical brackets:
(3−5)2=(3−5)(3−5)(3-\sqrt{5})^2=(3-\sqrt{5})(3-\sqrt{5})(3−5)2=(3−5)(3−5)Expand:
9−35−35+59-3\sqrt{5}-3\sqrt{5}+59−35−35+5Collect terms:
14−6514-6\sqrt{5}14−65Missing the middle term
(3−5)2(3-\sqrt{5})^2(3−5)2 is not just 32+(5)23^2+(\sqrt{5})^232+(5)2. The two middle terms are essential.
Rationalising the denominator
To rationalise the denominator means rewriting a fraction so there is no surd in the denominator.
A fraction is not considered fully simplified if the denominator still contains a surd.
If the denominator is just one square root, multiply the numerator and denominator by that square root.
Rationalising a single surd denominator
Rationalise 123\frac{12}{\sqrt{3}}312.
Multiply top and bottom by 3\sqrt{3}3:
123×33=1233\frac{12}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{12\sqrt{3}}{3}312×33=3123Simplify the fraction:
1233=43\frac{12\sqrt{3}}{3}=4\sqrt{3}3123=43If the numerator has more than one term, multiply the whole numerator by the surd.
Rationalising with a bracket on top
Simplify fully 4+62\frac{4+\sqrt{6}}{\sqrt{2}}24+6.
Multiply the numerator and denominator by 2\sqrt{2}2:
4+62×22=42+122\frac{4+\sqrt{6}}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}+\sqrt{12}}{2}24+6×22=242+12Simplify 12\sqrt{12}12:
12=23\sqrt{12}=2\sqrt{3}12=23Divide both terms in the numerator by 2:
42+232=22+3\frac{4\sqrt{2}+2\sqrt{3}}{2}=2\sqrt{2}+\sqrt{3}242+23=22+3If the denominator is something like 3+23+\sqrt{2}3+2, multiply by its conjugate 3−23-\sqrt{2}3−2.
Choosing the multiplier
For a one-term denominator, multiply by the same surd. For a two-term denominator, multiply by the conjugate.
Rationalising a two-term denominator
Show that 10+23+2\frac{10+\sqrt{2}}{3+\sqrt{2}}3+210+2 simplifies to 4−24-\sqrt{2}4−2.
Multiply top and bottom by the conjugate of the denominator:
10+23+2×3−23−2\frac{10+\sqrt{2}}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}3+210+2×3−23−2The denominator becomes a difference of two squares:
(3+2)(3−2)=9−2=7(3+\sqrt{2})(3-\sqrt{2})=9-2=7(3+2)(3−2)=9−2=7Expand the numerator:
(10+2)(3−2)=30−102+32−2=28−72(10+\sqrt{2})(3-\sqrt{2})=30-10\sqrt{2}+3\sqrt{2}-2=28-7\sqrt{2}(10+2)(3−2)=30−102+32−2=28−72Divide by 7:
28−727=4−2\frac{28-7\sqrt{2}}{7}=4-\sqrt{2}728−72=4−2Only changing the denominator
You must multiply the numerator and denominator by the same expression. Otherwise you change the value of the fraction.
Sometimes the denominator contains a fraction, such as 12+1\frac{1}{\sqrt{2}}+121+1. First combine the denominator, then rationalise if needed.
Simplifying a fraction inside a fraction
Simplify 312+1\frac{3}{\frac{1}{\sqrt{2}}+1}21+13.
Write the denominator as one fraction:
12+1=1+22\frac{1}{\sqrt{2}}+1=\frac{1+\sqrt{2}}{\sqrt{2}}21+1=21+2Divide by this fraction by multiplying by its reciprocal:
31+22=321+2\frac{3}{\frac{1+\sqrt{2}}{\sqrt{2}}}=\frac{3\sqrt{2}}{1+\sqrt{2}}21+23=1+232Rationalise using the conjugate 1−21-\sqrt{2}1−2:
321+2×1−21−2=32−6−1=6−32\frac{3\sqrt{2}}{1+\sqrt{2}}\times\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{3\sqrt{2}-6}{-1}=6-3\sqrt{2}1+232×1−21−2=−132−6=6−32The same rules work when letters are involved. For example, xx=x\sqrt{x}\sqrt{x}=xxx=x, as long as x≥0x \ge 0x≥0.
Letters under roots
In IGCSE surd algebra, assume quantities under square roots are non-negative unless the question says otherwise.
Algebraic conjugates
Simplify (x+y)(x−y)(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})(x+y)(x−y).
Recognise the conjugate pair:
(x+y)(x−y)=(x)2−(y)2(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=(\sqrt{x})^2-(\sqrt{y})^2(x+y)(x−y)=(x)2−(y)2Simplify each squared root:
x−yx-yx−ySquaring an algebraic surd bracket
Expand and simplify (3p+q)2(3p+\sqrt{q})^2(3p+q)2.
Write the bracket twice:
(3p+q)2=(3p+q)(3p+q)(3p+\sqrt{q})^2=(3p+\sqrt{q})(3p+\sqrt{q})(3p+q)2=(3p+q)(3p+q)Expand all four terms:
9p2+3pq+3pq+q9p^2+3p\sqrt{q}+3p\sqrt{q}+q9p2+3pq+3pq+qCollect the middle terms:
9p2+6pq+q9p^2+6p\sqrt{q}+q9p2+6pq+qIn the exam
Check yourself
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