Surds

Look for the largest square number that divides into 128.

1. 128=64\times2 and 64 is a square number
2. \sqrt{128}=\sqrt{64}\times\sqrt2
3. Answer: 8\sqrt2

Multiply the numbers under the square roots together first.

1. \sqrt8\times\sqrt{18}=\sqrt{8\times18}=\sqrt{144}
2. Answer: 12

Multiply the numbers under the roots first, then take out the largest square factor.

1. \sqrt6\times\sqrt8=\sqrt{48}
2. 48=16\times3, so \sqrt{48}=\sqrt{16}\times\sqrt3
3. Answer: 4\sqrt3

Write both surds as multiples of \sqrt2 before adding.

1. 2\sqrt{32}=2\times\sqrt{16}\times\sqrt2=8\sqrt2
2. \sqrt{50}=\sqrt{25}\times\sqrt2=5\sqrt2
3. 8\sqrt2+5\sqrt2=13\sqrt2
4. Answer: 13\sqrt2

Write the left-hand side as a single square root, then square both sides.

1. \sqrt{12}\times\sqrt k=\sqrt{12k}
2. So \sqrt{12k}=18, which means 12k=18^2=324
3. k=324\div12
4. Answer: k=27

Expand the brackets fully: the two middle terms cancel each other out.

1. (3+\sqrt7)(3-\sqrt7)=9-3\sqrt7+3\sqrt7-\sqrt7\times\sqrt7
2. The \sqrt7 terms cancel
3. \sqrt7\times\sqrt7=7, so this is 9-7
4. Answer: 2

Expand the brackets as you would with algebra, remembering that \sqrt3\times\sqrt3=3.

1. (2+\sqrt3)(5-\sqrt3)=10-2\sqrt3+5\sqrt3-\sqrt3\times\sqrt3
2. \sqrt3\times\sqrt3=3
3. =10-3+3\sqrt3
4. Answer: 7+3\sqrt3

Write the square as (\sqrt5-2)(\sqrt5-2) and expand: there are four terms before you simplify.

1. (\sqrt5-2)^2=(\sqrt5-2)(\sqrt5-2)
2. =5-2\sqrt5-2\sqrt5+4
3. Answer: 9-4\sqrt5

Multiply the top and bottom by \sqrt{15}, then simplify the surd on top before cancelling.

1. Multiply top and bottom by \sqrt{15}: \dfrac{10\sqrt3\times\sqrt{15}}{15}
2. \sqrt3\times\sqrt{15}=\sqrt{45}=3\sqrt5
3. So the fraction is \dfrac{30\sqrt5}{15}
4. Cancel: 30\div15=2
5. Answer: 2\sqrt5

Write each of the three surds as a multiple of \sqrt5 first.

1. \sqrt{125}=5\sqrt5, \sqrt{20}=2\sqrt5 and \sqrt{500}=10\sqrt5
2. So the fraction is \dfrac{5\sqrt5+2\sqrt5}{10\sqrt5}=\dfrac{7\sqrt5}{10\sqrt5}
3. Cancel the \sqrt5
4. Answer: \dfrac{7}{10}

Use Pythagoras' theorem: squaring a square root just removes it.

1. Hypotenuse^2=(\sqrt{12})^2+(\sqrt{24})^2
2. =12+24=36
3. Hypotenuse =\sqrt{36}
4. Answer: 6 cm

Simplify \sqrt{96} and \sqrt{24} as multiples of \sqrt6 before dividing.

1. \sqrt{96}=\sqrt{16}\times\sqrt6=4\sqrt6 and \sqrt{24}=\sqrt4\times\sqrt6=2\sqrt6
2. So x\sqrt6=4\sqrt6-2\sqrt6=2\sqrt6
3. Divide both sides by \sqrt6
4. Answer: x=2

Multiply the top and the bottom by 3+\sqrt5, changing the sign between the terms.

1. Multiply top and bottom by 3+\sqrt5
2. Bottom: (3-\sqrt5)(3+\sqrt5)=9-5=4
3. Top: 8(3+\sqrt5)=24+8\sqrt5
4. \dfrac{24+8\sqrt5}{4}=6+2\sqrt5
5. Answer: 6+2\sqrt5

Multiply the top and the bottom by \sqrt3-1, then expand the top carefully: \sqrt3\times\sqrt3=3.

1. Multiply top and bottom by \sqrt3-1
2. Bottom: (\sqrt3+1)(\sqrt3-1)=3-1=2
3. Top: 6\sqrt3(\sqrt3-1)=18-6\sqrt3
4. \dfrac{18-6\sqrt3}{2}=9-3\sqrt3
5. Answer: 9-3\sqrt3

Multiply the top and the bottom by \sqrt2-1; the bottom becomes 1, so the answer is just the expanded top.

1. Multiply top and bottom by \sqrt2-1
2. Bottom: (\sqrt2+1)(\sqrt2-1)=2-1=1
3. Top: (5+\sqrt2)(\sqrt2-1)=5\sqrt2-5+2-\sqrt2
4. =4\sqrt2-3
5. Answer: -3+4\sqrt2

Multiply the top and the bottom by \sqrt7+\sqrt5, and remember that \sqrt7\times\sqrt5=\sqrt{35}.

1. Multiply top and bottom by \sqrt7+\sqrt5
2. Bottom: (\sqrt7-\sqrt5)(\sqrt7+\sqrt5)=7-5=2
3. Top: (3\sqrt7+\sqrt5)(\sqrt7+\sqrt5)=21+3\sqrt{35}+\sqrt{35}+5
4. =26+4\sqrt{35}
5. \dfrac{26+4\sqrt{35}}{2}=13+2\sqrt{35}
6. Answer: 13+2\sqrt{35}

Expand the left-hand side and collect the terms with \sqrt5 separately from those without. Answer part (a) before attempting part (b).

Part (a)

1. Expand: (p+2\sqrt5)(3-\sqrt5)=3p-p\sqrt5+6\sqrt5-10
2. =(3p-10)+(6-p)\sqrt5
3. The \sqrt5 terms must match: 6-p=-1
4. Answer: p=7

Part (b)

1. The terms without \sqrt5 must match: q=3p-10
2. With p=7: q=21-10
3. Answer: q=11

Treat \sqrt5 like any other number: subtract the coordinates to find the horizontal and vertical distances, then use Pythagoras' theorem.

Part (a)

1. Difference in x: 3\sqrt5-\sqrt5=2\sqrt5
2. Difference in y: 6-2=4
3. AB^2=(2\sqrt5)^2+4^2=20+16=36
4. Answer: AB=6

Part (b)

1. The midpoint is halfway between the two points in each direction
2. x: \dfrac{\sqrt5+3\sqrt5}{2}=\dfrac{4\sqrt5}{2}=2\sqrt5
3. y: \dfrac{2+6}{2}=4
4. Answer: M=(2\sqrt5,\ 4)

Square (a+b\sqrt2) and compare the result with 11+6\sqrt2: the parts with \sqrt2 must match and so must the parts without. Answer part (a) before attempting part (b).

Part (a)

1. Area =(a+b\sqrt2)^2=a^2+2ab\sqrt2+2b^2
2. Compare with 11+6\sqrt2: 2ab=6, so ab=3
3. a and b are positive integers, so a=3,\ b=1 or a=1,\ b=3
4. a^2+2b^2=11 works for a=3,\ b=1 (9+2=11) but not for a=1,\ b=3 (1+18=19)
5. Answer: 3+\sqrt2 cm

Part (b)

1. Diagonal =\sqrt2\times side (Pythagoras in a square)
2. =\sqrt2(3+\sqrt2)=3\sqrt2+2
3. Answer: 2+3\sqrt2 cm

Height = volume ÷ base area; to divide by 4+3\sqrt2, multiply top and bottom by 4-3\sqrt2, and watch the sign of the denominator.

Part (a)

1. Area =(2+\sqrt2)(1+\sqrt2)
2. =2+2\sqrt2+\sqrt2+\sqrt2\times\sqrt2
3. \sqrt2\times\sqrt2=2, so =2+3\sqrt2+2
4. Answer: 4+3\sqrt2 cm^2

Part (b)

1. Height = volume \div base area =\dfrac{18+13\sqrt2}{4+3\sqrt2}
2. Multiply top and bottom by the conjugate 4-3\sqrt2 (change the sign between the terms)
3. Bottom: (4+3\sqrt2)(4-3\sqrt2)=16-18=-2
4. Top: (18+13\sqrt2)(4-3\sqrt2)=72-54\sqrt2+52\sqrt2-78=-6-2\sqrt2
5. \dfrac{-6-2\sqrt2}{-2}=3+\sqrt2
6. Answer: 3+\sqrt2 cm