Sec 3 A-Math Surds worksheet with answers

A free Sec 3 A-Math worksheet on surds: 9 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 A-Math worksheet: Surds

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Solve

x\sqrt{3} - 2x = 6

, giving your answer in the form

a + b\sqrt{3}

, where

a

and

b

are integers.

Answer[2]

Simplify

\sqrt{45} - \sqrt{20} + 2\sqrt{80}

, giving your answer in the form

a\sqrt{b}

Answer[2]

Express

\dfrac{4 + \sqrt{3}}{2 - \sqrt{3}}

in the form

\dfrac{a + b\sqrt{3}}{c}

, where

a

b

and

c

are integers.

Answer[2]

(a)Simplify

\sqrt{80}

Answer (a)[1]

(b)Simplify

\sqrt{150} + \sqrt{24}

, giving your answer in the form

a\sqrt{b}

Answer (b)[2]

(a)Expand and simplify

\sqrt{2}(1 + \sqrt{2})

Answer (a)[2]

(b)Expand and simplify

(1 + \sqrt{3})^2

Answer (b)[2]

(a)Express

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}}

in the form

\dfrac{a + b\sqrt{3}}{c}

, where

a

b

and

c

are integers.

Answer (a)[2]

(b)Express

\dfrac{4 + \sqrt{7}}{2 - \sqrt{7}}

in the form

\dfrac{a + b\sqrt{7}}{c}

, where

a

b

and

c

are integers.

Answer (b)[2]

Section B

(a)Show that squaring both sides of the equation

{\sqrt{4x + 17} = x - 1}

gives

{x^{2} - 6x - 16 = 0}

[2]

(b)Hence solve the equation

{\sqrt{4x + 17} = x - 1}

Answer (b)[3]

(c)Hence find the positive value of

y

for which

{\sqrt{4y^{2} + 17} = y^{2} - 1}

, giving your answer in simplest surd form.

Answer (c)[2]

Do not use a calculator in this question.

(a)Given that

\sqrt{37 + 20\sqrt{3}} = a + b\sqrt{3}

, where

a

and

b

are positive integers, find the values of

a

and

b

Answer (a)[3]

(b)The equation

x^{2} + bx + c = 0

, where

b

and

c

are integers, has a root

x = 5 + 2\sqrt{3}

. By substituting this root and equating the rational and irrational parts, find the value of

b

Answer (b)[3]

(c)Find the value of

c

Answer (c)[1]

(d)Find the other root of the equation.

Answer (d)[1]

In the diagram, triangle

ABC

is right-angled at

B

and

D

is the mid-point of

BC

AB = (4 + \sqrt{2})

cm and the area of triangle

ABC

(6 + 5\sqrt{2})

cm².

(a)Without using a calculator, find the length of

BC

in the form

(p + q\sqrt{2})

cm, where

p

and

q

are integers.

Answer (a)cm[3]

(b)Find

\tan \angle ADB

in the form

h + k\sqrt{2}

, where

h

and

k

are integers.

Answer (b)[3]

(c)Find

AD^{2}

in the form

(u + v\sqrt{2})

cm², where

u

and

v

are integers.

Answer (c)cm²[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

x = -12 - 6\sqrt{3}

[2]

x(\sqrt{3} - 2) = 6

x = \dfrac{6}{\sqrt{3} - 2}

(\sqrt{3} - 2)(\sqrt{3} + 2) = -1

x = \dfrac{12 + 6\sqrt{3}}{-1}

x = -12 - 6\sqrt{3}

Question 2

Answer:

3\sqrt{5} - 2\sqrt{5} + 8\sqrt{5} = 9\sqrt{5}

[2]

\sqrt{45} = 3\sqrt{5}

-\sqrt{20} = -2\sqrt{5}

2\sqrt{80} = 2 \times 4\sqrt{5}

= 8\sqrt{5}

3\sqrt{5} - 2\sqrt{5} + 8\sqrt{5} = 9\sqrt{5}

Question 3

Answer:

= 11 + 6\sqrt{3}

[2]

\dfrac{4 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{(4 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}

\mathstrut 2^2 - (\sqrt{3})^2

\text{Bottom} = 1

\dfrac{4 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{11 + 6\sqrt{3}}{1}

= 11 + 6\sqrt{3}

Question 4

(a)Answer:

\sqrt{80} = 4\sqrt{5}

[1]

80 = 16 \times 5

\sqrt{80} = 4\sqrt{5}

(b)Answer:

5\sqrt{6} + 2\sqrt{6} = 7\sqrt{6}

[2]

\sqrt{150} = 5\sqrt{6}

\sqrt{24} = 2\sqrt{6}

5\sqrt{6} + 2\sqrt{6} = 7\sqrt{6}

Question 5

(a)Answer:

\sqrt{2}(1 + \sqrt{2}) = 2 + \sqrt{2}

[2]

\sqrt{2} \times 1 = \sqrt{2}

\sqrt{2} \times \sqrt{2} = 2

\sqrt{2}(1 + \sqrt{2}) = 2 + \sqrt{2}

(b)Answer:

(1 + \sqrt{3})^2 = 4 + 2\sqrt{3}

[2]

First:

1 \times 1 = 1

Outer:

1 \times \sqrt{3} = \sqrt{3}

Inner:

\sqrt{3} \times 1 = \sqrt{3}

Last:

\sqrt{3} \times \sqrt{3} = 3

(1 + \sqrt{3})^2 = 4 + 2\sqrt{3}

Question 6

(a)Answer:

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{9 + 7\sqrt{3}}{6}

[2]

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{(1 + 2\sqrt{3})(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})}

\mathstrut 3^2 - (\sqrt{3})^2

\text{Bottom} = 6

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{9 + 7\sqrt{3}}{6}

(b)Answer:

= -5 - 2\sqrt{7}

[2]

\dfrac{4 + \sqrt{7}}{2 - \sqrt{7}} = \dfrac{(4 + \sqrt{7})(2 + \sqrt{7})}{(2 - \sqrt{7})(2 + \sqrt{7})}

\mathstrut 2^2 - (\sqrt{7})^2

\text{Bottom} = -3

\dfrac{4 + \sqrt{7}}{2 - \sqrt{7}} = \dfrac{15 + 6\sqrt{7}}{-3}

= -5 - 2\sqrt{7}

Question 7

(a)[2]

4x + 17 = (x - 1)^{2}

4x + 17 = x^{2} - 2x + 1

x^{2} - 6x - 16 = 0

(shown)

(b)Answer:

x = 8

[3]

(x + 2)(x - 8) = 0

x = -2

x = 8

When

x = -2

\sqrt{9} = 3

, but

x - 1 = -3

, so

x = -2

is rejected.When

x = 8

\sqrt{49} = 7

and

x - 1 = 7

x = 8

(c)Answer:

y = 2\sqrt{2}

[2]

This is the equation in part (b) with

y^{2}

in place of

x

.So

y^{2} = 8

only.

y = \sqrt{8}

(as

y > 0

)

= 2\sqrt{2}

Question 8

(a)Answer:

5 + 2\sqrt{3}

[3]

37 + 20\sqrt{3} = (a + b\sqrt{3})^{2}

= a^{2} + 3b^{2} + 2ab\sqrt{3}

Rational parts:

a^{2} + 3b^{2} = 37

Irrational parts:

2ab = 20

, so

ab = 10

Trying positive integer pairs with

ab = 10

a = 5

b = 2

\sqrt{37 + 20\sqrt{3}} = 5 + 2\sqrt{3}

(b)Answer:

b = -10

[3]

(5 + 2\sqrt{3})^{2} + b(5 + 2\sqrt{3}) + c = 0

37 + 20\sqrt{3} + 5b + 2b\sqrt{3} + c = 0

Irrational parts:

20 + 2b = 0

b = -10

(c)Answer:

c = 13

[1]

Rational parts:

37 + 5b + c = 0

37 - 50 + c = 0

c = 13

(d)Answer:

5 - 2\sqrt{3}

[1]

x^{2} - 10x + 13 = 0

x = \dfrac{10 \pm \sqrt{48}}{2}

= 5 \pm 2\sqrt{3}

The other root is

5 - 2\sqrt{3}

Question 9

(a)Answer:

(2 + 2\sqrt{2})

cm[3]

\dfrac{1}{2}(4 + \sqrt{2}) \times BC = 6 + 5\sqrt{2}

BC = \dfrac{2(6 + 5\sqrt{2})}{4 + \sqrt{2}}

= \dfrac{2(6 + 5\sqrt{2})(4 - \sqrt{2})}{16 - 2}

= \dfrac{2(14 + 14\sqrt{2})}{14}

= 2 + 2\sqrt{2}

(b)Answer:

-2 + 3\sqrt{2}

[3]

BD = \dfrac{1}{2}BC

= 1 + \sqrt{2}

\tan \angle ADB = \dfrac{AB}{BD}

= \dfrac{4 + \sqrt{2}}{1 + \sqrt{2}}

= \dfrac{(4 + \sqrt{2})(1 - \sqrt{2})}{1 - 2}

= \dfrac{2 - 3\sqrt{2}}{-1}

= -2 + 3\sqrt{2}

(c)Answer:

(21 + 10\sqrt{2})

cm²[2]

AD^{2} = AB^{2} + BD^{2}

= (4 + \sqrt{2})^{2} + (1 + \sqrt{2})^{2}

= (18 + 8\sqrt{2}) + (3 + 2\sqrt{2})

= 21 + 10\sqrt{2}

Practice questions by skill

Sec 3 A-Math Surds worksheet with answers

A free Sec 3 A-Math worksheet on surds: 9 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Surds notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)Show that squaring both sides of the equation

{\sqrt{4x + 17} = x - 1}

gives

{x^{2} - 6x - 16 = 0}

[2]

(b)Hence solve the equation

{\sqrt{4x + 17} = x - 1}

Answer (b)[3]

(c)Hence find the positive value of

y

for which

{\sqrt{4y^{2} + 17} = y^{2} - 1}

, giving your answer in simplest surd form.

Answer (c)[2]

Answers and worked solutions

Question 1

Answer:

x = -12 - 6\sqrt{3}

[2]

x(\sqrt{3} - 2) = 6

x = \dfrac{6}{\sqrt{3} - 2}

(\sqrt{3} - 2)(\sqrt{3} + 2) = -1

x = \dfrac{12 + 6\sqrt{3}}{-1}

x = -12 - 6\sqrt{3}

Question 2

Answer:

3\sqrt{5} - 2\sqrt{5} + 8\sqrt{5} = 9\sqrt{5}

[2]

\sqrt{45} = 3\sqrt{5}

-\sqrt{20} = -2\sqrt{5}

2\sqrt{80} = 2 \times 4\sqrt{5}

= 8\sqrt{5}

3\sqrt{5} - 2\sqrt{5} + 8\sqrt{5} = 9\sqrt{5}

Question 3

Answer:

= 11 + 6\sqrt{3}

[2]

\dfrac{4 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{(4 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}

\mathstrut 2^2 - (\sqrt{3})^2

\text{Bottom} = 1

\dfrac{4 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{11 + 6\sqrt{3}}{1}

= 11 + 6\sqrt{3}

Question 4

(a)Answer:

\sqrt{80} = 4\sqrt{5}

[1]

80 = 16 \times 5

\sqrt{80} = 4\sqrt{5}

(b)Answer:

5\sqrt{6} + 2\sqrt{6} = 7\sqrt{6}

[2]

\sqrt{150} = 5\sqrt{6}

\sqrt{24} = 2\sqrt{6}

5\sqrt{6} + 2\sqrt{6} = 7\sqrt{6}

Question 5

(a)Answer:

\sqrt{2}(1 + \sqrt{2}) = 2 + \sqrt{2}

[2]

\sqrt{2} \times 1 = \sqrt{2}

\sqrt{2} \times \sqrt{2} = 2

\sqrt{2}(1 + \sqrt{2}) = 2 + \sqrt{2}

(b)Answer:

(1 + \sqrt{3})^2 = 4 + 2\sqrt{3}

[2]

First:

1 \times 1 = 1

Outer:

1 \times \sqrt{3} = \sqrt{3}

Inner:

\sqrt{3} \times 1 = \sqrt{3}

Last:

\sqrt{3} \times \sqrt{3} = 3

(1 + \sqrt{3})^2 = 4 + 2\sqrt{3}

Question 6

(a)Answer:

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{9 + 7\sqrt{3}}{6}

[2]

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{(1 + 2\sqrt{3})(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})}

\mathstrut 3^2 - (\sqrt{3})^2

\text{Bottom} = 6

\dfrac{1 + 2\sqrt{3}}{3 - \sqrt{3}} = \dfrac{9 + 7\sqrt{3}}{6}

(b)Answer:

= -5 - 2\sqrt{7}

[2]

\dfrac{4 + \sqrt{7}}{2 - \sqrt{7}} = \dfrac{(4 + \sqrt{7})(2 + \sqrt{7})}{(2 - \sqrt{7})(2 + \sqrt{7})}

\mathstrut 2^2 - (\sqrt{7})^2

\text{Bottom} = -3

\dfrac{4 + \sqrt{7}}{2 - \sqrt{7}} = \dfrac{15 + 6\sqrt{7}}{-3}

= -5 - 2\sqrt{7}

Question 7

(a)[2]

4x + 17 = (x - 1)^{2}

4x + 17 = x^{2} - 2x + 1

x^{2} - 6x - 16 = 0

(shown)

(b)Answer:

x = 8

[3]

(x + 2)(x - 8) = 0

x = -2

x = 8

When

x = -2

\sqrt{9} = 3

, but

x - 1 = -3

, so

x = -2

is rejected.When

x = 8

\sqrt{49} = 7

and

x - 1 = 7

x = 8

(c)Answer:

y = 2\sqrt{2}

[2]

This is the equation in part (b) with

y^{2}

in place of

x

.So

y^{2} = 8

only.

y = \sqrt{8}

(as

y > 0

)

= 2\sqrt{2}

Question 8

(a)Answer:

5 + 2\sqrt{3}

[3]

37 + 20\sqrt{3} = (a + b\sqrt{3})^{2}

= a^{2} + 3b^{2} + 2ab\sqrt{3}

Rational parts:

a^{2} + 3b^{2} = 37

Irrational parts:

2ab = 20

, so

ab = 10

Trying positive integer pairs with

ab = 10

a = 5

b = 2

\sqrt{37 + 20\sqrt{3}} = 5 + 2\sqrt{3}

(b)Answer:

b = -10

[3]

(5 + 2\sqrt{3})^{2} + b(5 + 2\sqrt{3}) + c = 0

37 + 20\sqrt{3} + 5b + 2b\sqrt{3} + c = 0

Irrational parts:

20 + 2b = 0

b = -10

(c)Answer:

c = 13

[1]

Rational parts:

37 + 5b + c = 0

37 - 50 + c = 0

c = 13

(d)Answer:

5 - 2\sqrt{3}

[1]

x^{2} - 10x + 13 = 0

x = \dfrac{10 \pm \sqrt{48}}{2}

= 5 \pm 2\sqrt{3}

The other root is

5 - 2\sqrt{3}

Question 9

(a)Answer:

(2 + 2\sqrt{2})

cm[3]

\dfrac{1}{2}(4 + \sqrt{2}) \times BC = 6 + 5\sqrt{2}

BC = \dfrac{2(6 + 5\sqrt{2})}{4 + \sqrt{2}}

= \dfrac{2(6 + 5\sqrt{2})(4 - \sqrt{2})}{16 - 2}

= \dfrac{2(14 + 14\sqrt{2})}{14}

= 2 + 2\sqrt{2}

(b)Answer:

-2 + 3\sqrt{2}

[3]

BD = \dfrac{1}{2}BC

= 1 + \sqrt{2}

\tan \angle ADB = \dfrac{AB}{BD}

= \dfrac{4 + \sqrt{2}}{1 + \sqrt{2}}

= \dfrac{(4 + \sqrt{2})(1 - \sqrt{2})}{1 - 2}

= \dfrac{2 - 3\sqrt{2}}{-1}

= -2 + 3\sqrt{2}

(c)Answer:

(21 + 10\sqrt{2})

cm²[2]

AD^{2} = AB^{2} + BD^{2}

= (4 + \sqrt{2})^{2} + (1 + \sqrt{2})^{2}

= (18 + 8\sqrt{2}) + (3 + 2\sqrt{2})

= 21 + 10\sqrt{2}