Surds

Worked example

1. 1

   $\frac{10}{\sqrt{5}}$ — The question. A root on the bottom is not allowed in a final answer, so we must clear it.

2. 2

   $=\frac{10\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$ — Multiply top and bottom by √5. That is the same as multiplying by 1, so the value does not change.

3. 3

   $=\frac{10\sqrt{5}}{5}$ — On the bottom, √5 × √5 = 5, a whole number. The root has gone from the bottom.

4. 4

   $=2\sqrt{5}$ — Now cancel the whole numbers: 10 ÷ 5 = 2. Always cancel like this to give the simplest form.

Worked example

1. 1

   $\frac{14}{3-\sqrt{2}}$ — The question. The bottom has two terms, so we use the formula above.

2. 2

   $=\frac{14(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$ — Multiply top and bottom by the same bracket with the sign swapped. Same on top and bottom, so the value stays.

3. 3

   $\text{bottom}=3^2-(\sqrt{2}{)}^2$ — Do the bottom first. It is the formula with a = 3 and b = 2: the middle terms −3√2 and +3√2 cancel.

4. 4

   $=9-2=7$ — 3² = 9 and (√2)² = 2. The bottom is now a whole number, 7.

5. 5

   $\text{top}=14\times 3+14\times \sqrt{2}=42+14\sqrt{2}$ — Now the top: multiply 14 by both terms in the bracket.

6. 6

   $=\frac{42+14\sqrt{2}}{7}$ — Put the new top over the new bottom, ready to divide.

7. 7

   $=6+2\sqrt{2}$ — Divide both terms by 7: 42 ÷ 7 = 6 and 14 ÷ 7 = 2.

Worked example

1. 1

   $\sqrt{75}+\sqrt{12}$ — The question. The roots are different, so we cannot add them yet.

2. 2

   $\sqrt{75}=\sqrt{25\times 3}$ — 25 is the biggest square that goes into 75, because 25 × 3 = 75.

3. 3

   $=\sqrt{25}\times \sqrt{3}=5\sqrt{3}$ — Split the root into two roots. √25 = 5 comes out as a whole number; √3 stays.

4. 4

   $\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$ — Same for 12: 4 is the biggest square in it, and √4 = 2.

5. 5

   $5\sqrt{3}+2\sqrt{3}=7\sqrt{3}$ — Both are now lots of √3, so add the numbers in front, just like 5x + 2x = 7x.

Worked example

1. 1

   $4\sqrt{6}\times 3\sqrt{2}$ — The question. Two surds multiplied, each with a number in front of its root.

2. 2

   $=(4\times 3)\times (\sqrt{6}\times \sqrt{2})$ — Outside numbers together, roots together. Order does not matter when you multiply.

3. 3

   $=12\sqrt{12}$ — 4 × 3 = 12 and √6 × √2 = √12.

4. 4

   $=12\times 2\sqrt{3}$ — √12 is not in its simplest form: √12 = √(4 × 3) = 2√3.

5. 5

   $=24\sqrt{3}$ — 12 × 2 = 24.

Worked example

1. 1

   $\frac{18\sqrt{20}}{3\sqrt{5}}=k$ — The question: find the whole number k.

2. 2

   $=\frac{18}{3}\times \sqrt{\frac{20}{5}}$ — Outside divides by outside, inside divides by inside.

3. 3

   $=6\sqrt{4}$ — 18 ÷ 3 = 6 and 20 ÷ 5 = 4.

4. 4

   $=6\times 2=12$ — √4 = 2 is a whole number, so k = 12.

Worked example

1. 1

   $(3+\sqrt{5}{)}^2$ — The question. Squared means the bracket times itself.

2. 2

   $=(3+\sqrt{5})(3+\sqrt{5})$ — Write it out twice so no term gets lost.

3. 3

   $=3\times 3+3\times \sqrt{5}+\sqrt{5}\times 3+\sqrt{5}\times \sqrt{5}$ — Every term in the first bracket times every term in the second: four products.

4. 4

   $=9+3\sqrt{5}+3\sqrt{5}+5$ — Work each one out. √5 × √5 = 5, a whole number.

5. 5

   $=14+6\sqrt{5}$ — Collect: 9 + 5 = 14, and 3√5 + 3√5 = 6√5.

Worked example

1. 1

   $(4-\sqrt{3})(2+5\sqrt{3})$ — The question: write it as $a+b\sqrt{3}$.

2. 2

   $=8+20\sqrt{3}-2\sqrt{3}-5\sqrt{3}\times \sqrt{3}$ — Four products: 4 × 2, 4 × 5√3, −√3 × 2 and −√3 × 5√3. Keep each sign with its term.

3. 3

   $(-5)\times (\sqrt{3}\times \sqrt{3})=(-5)\times 3=-15$ — The hard term alone. Numbers outside multiply, roots multiply, and √3 × √3 = 3.

4. 4

   $=8+20\sqrt{3}-2\sqrt{3}-15$ — Put −15 back in its place.

5. 5

   $=-7+18\sqrt{3}$ — Whole numbers: 8 − 15 = −7. Roots: 20√3 − 2√3 = 18√3.

6. 6

   $a=-7,b=18$ — Match the parts: the whole number is a, the number in front of √3 is b.

Worked example

1. 1

   $\frac{10}{2-\sqrt{6}}$ — The question. Two terms on the bottom, so swap the sign.

2. 2

   $=\frac{10(2+\sqrt{6})}{(2-\sqrt{6})(2+\sqrt{6})}$ — Multiply top and bottom by (2 + √6), the bottom with its sign swapped, so the roots on the bottom will cancel.

3. 3

   $\text{bottom}=2^2-(\sqrt{6}{)}^2=4-6$ — The formula with a = 2 and b = 6.

4. 4

   $=-2$ — 6 is bigger than 4, so the bottom is −2. It is a whole number, which is what we wanted.

5. 5

   $\text{top}=20+10\sqrt{6}$ — 10 × 2 = 20 and 10 × √6 = 10√6.

6. 6

   $=\frac{20+10\sqrt{6}}{-2}$ — Put the new top over the new bottom, ready to divide.

7. 7

   $=-10-5\sqrt{6}$ — Divide each term by −2: 20 ÷ (−2) = −10 and 10√6 ÷ (−2) = −5√6. Both signs flip.

Worked example

1. 1

   $\frac{4+\sqrt{2}}{3-\sqrt{2}}$ — The question: write it as $a+\sqrt{b}$.

2. 2

   $=\frac{(4+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$ — Multiply top and bottom by the bottom bracket with its sign swapped.

3. 3

   $\text{bottom}=3^2-(\sqrt{2}{)}^2=9-2=7$ — The formula with a = 3 and b = 2.

4. 4

   $\text{top}=12+4\sqrt{2}+3\sqrt{2}+2$ — Four products: 4 × 3, 4 × √2, √2 × 3 and √2 × √2 = 2.

5. 5

   $=14+7\sqrt{2}$ — 12 + 2 = 14 and 4√2 + 3√2 = 7√2.

6. 6

   $=\frac{14+7\sqrt{2}}{7}=2+\sqrt{2}$ — Divide both terms by 7: 14 ÷ 7 = 2 and 7√2 ÷ 7 = √2.

7. 7

   $a=2,b=2$ — 2 + √2 matches $a+\sqrt{b}$.

Worked example

1. 1

   $\text{width}=\frac{\text{area}}{\text{length}}=\frac{12}{\sqrt{5}+1}$ — The area is 12 and the length is √5 + 1. Width = area ÷ length.

2. 2

   $=\frac{12(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}$ — Two terms on the bottom, so multiply top and bottom by the bracket with the sign swapped.

3. 3

   $\text{bottom}=(\sqrt{5}{)}^2-1^2=5-1=4$ — The middle terms cancel, leaving 5 − 1 = 4.

4. 4

   $\text{top}=12\sqrt{5}-12$ — Multiply 12 by both terms: 12 × √5 and 12 × (−1).

5. 5

   $=\frac{12\sqrt{5}-12}{4}=3\sqrt{5}-3$ — Divide both terms by 4: 12 ÷ 4 = 3.