Rationalising Denominators and Surd Equations - Worksheets, Questions and Revision

20 original exam-style questions - 9 pages of questions with a full mark scheme - free printable PDF.

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GCSE · Algebra - Surds

N2 Rationalising Denominators and Surd Equations

AQA 8365 · Calculator allowed · about 105 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Simplify each surd, giving your answer in the form a √b where b has no square factors.

(a)Simplify √20.(1)

(b)Simplify √63.(1)

(c)Simplify √8 x √18, giving your answer as an integer.(2)

(Total for Question 1 is 4 marks)

Rationalise the denominator in each expression, simplifying fully.

(a)1 / √3(2)

(b)5 / √5(2)

(Total for Question 2 is 6 marks)

Expand and simplify each expression.

(a)(2 + √3)(2 - √3)(2)

(b)(√5 + 1)²(2)

(c)√7 (3 - √7)(2)

(Total for Question 3 is 6 marks)

Rationalise the denominator in each expression, simplifying fully.

(a)4 / (3 + √2)(3)

(b)6 / (√5 - 2)(3)

(Total for Question 4 is 6 marks)

Solve each equation.

(a)√x = 7(1)

(b)√x + 3 = 5(2)

(c)√2x - 1 = 3(2)

(Total for Question 5 is 5 marks)

Circle the letter next to the correct answer for each question. No working is required.

(a)Which of these is equal to 1/√7 after rationalising the denominator?(1)

A) √7
B) √7/7
C) 7sqrt(7)
D) 1/7

(b)Which of these is the simplified form of √45?(1)

A) 9sqrt(5)
B) 3sqrt(5)
C) 5sqrt(3)
D) 15sqrt(3)

(c)Which value of x satisfies √x = 6?(1)

A) 3
B) 12
C) 36
D) 6

(Total for Question 6 is 3 marks)

Solve each equation. Check that your answer satisfies the original equation.

(a)√3x + 4 - 2 = 3(3)

(b)√5x + 1 + 2 = 7(3)

(Total for Question 7 is 6 marks)

Show that each statement is true. You must show all your working.

(a)Show that √3 + √12 = 3sqrt(3).(2)

(b)Show that (5 + √2)² = 27 + 10sqrt(2).(4)

(Total for Question 8 is 6 marks)

A rectangle has length (5 + √2) cm and width (3 + √2) cm.

(a)Find the perimeter of the rectangle, giving your answer in the form a + b √2.(2)

(b)Find the area of the rectangle, giving your answer in the form p + q √2.(3)

(Total for Question 9 is 5 marks)

Simplify each expression fully.

(a)(√18 + √2) / √2(3)

(b)Rationalise and simplify (√3 + 1) / (√3 - 1).(3)

(Total for Question 10 is 6 marks)

Solve √3x + 10 = x, showing why any extraneous solution must be rejected.

(Total for Question 11 is 5 marks)

Show that 1 / (√5 - 2) = √5 + 2.

(Total for Question 12 is 4 marks)

A is the point (0, 0) and B is the point (2, 2sqrt(5)).

(a)Find the length AB, giving your answer in the form k √6.(3)

(b)Find the exact y-coordinate of the midpoint M of AB.(2)

(c)Explain why AB² is a rational number even though the y-coordinate of B is irrational.(2)

(Total for Question 13 is 7 marks)

Solve each equation.

(a)√3x - 2 = √x + 8(2)

(b)√5x + 1 = 2sqrt(x + 1)(3)

(Total for Question 14 is 5 marks)

Simplify fully: (3 + √2)/(2 - √2) + (3 - √2)/(2 + √2).

(Total for Question 15 is 5 marks)

Solve √x + 4 = x - 2, showing clearly why any extraneous solution is rejected.

(Total for Question 16 is 6 marks)

Prove that 1 / (√n+1 - √n) = √n+1 + √n for all positive integers n.

(Total for Question 17 is 5 marks)

A right-angled triangle has two shorter sides of length 1 cm and √2 cm.

Diagram NOT accurately drawn

(a)Show that the hypotenuse has length √3 cm.(2)

(b)Theta is the angle between the side of length 1 cm and the hypotenuse. Find cos(θ), rationalising your answer.(3)

(c)Find the exact area of the triangle.(1)

(Total for Question 18 is 6 marks)

Solve (x + 3) / √x = √x + 4, where x > 0.

(Total for Question 19 is 7 marks)

Prove that 1/(√n+1 + √n) + 1/(√n+1 - √n) = 2sqrt(n+1) for all positive integers n.

(Total for Question 20 is 6 marks)

Mark scheme · N2 Rationalising Denominators and Surd Equations

Question 1

(a) B1 2sqrt(5) oe
(a) Answer: 2sqrt(5)
(b) B1 3sqrt(7) oe
(b) Answer: 3sqrt(7)
(c) M1 combines to √8 x 18 = √144 oe (e.g. 2sqrt(2) x 3sqrt(2))
(c) A1 12 cao
(c) Answer: 12

Question 2

(a) M1 multiplies numerator and denominator by √3
(a) A1 √3/3 oe cao
(a) Answer: √3/3
(b) M1 multiplies numerator and denominator by √5
(b) A1 √5 cao
(b) Answer: √5
(c) M1 multiplies numerator and denominator by √6
(c) A1 2sqrt(6) cao
(c) Answer: 2sqrt(6)

Question 3

(a) M1 attempts full expansion, e.g. 4 - 2sqrt(3) + 2sqrt(3) - 3
(a) A1 1 cao
(a) Answer: 1
(b) M1 attempts full expansion 5 + √5 + √5 + 1 oe
(b) A1 6 + 2sqrt(5) oe cao
(b) Answer: 6 + 2sqrt(5)
(c) M1 multiplies out to 3sqrt(7) - 7 oe (unsimplified acceptable)
(c) A1 3sqrt(7) - 7 oe cao
(c) Answer: 3sqrt(7) - 7

Question 4

(a) M1 multiplies numerator and denominator by the conjugate (3 - √2)
(a) A1 denominator simplifies to 7
(a) A1 (12 - 4sqrt(2))/7 oe cao
(a) Answer: (12 - 4sqrt(2))/7
(b) M1 multiplies numerator and denominator by the conjugate (√5 + 2)
(b) A1 denominator simplifies to 1
(b) A1 6sqrt(5) + 12 oe cao
(b) Answer: 6sqrt(5) + 12

Question 5

(a) B1 x = 49 cao
(a) Answer: x = 49
(b) M1 squares both sides: x + 3 = 25
(b) A1 x = 22 cao
(b) Answer: x = 22
(c) M1 squares both sides: 2x - 1 = 9
(c) A1 x = 5 cao
(c) Answer: x = 5

Question 6

(a) B1 B
(a) Answer: B
(b) B1 B
(b) Answer: B
(c) B1 C
(c) Answer: C

Question 7

(a) M1 rearranges to √3x+4 = 5
(a) M1 squares both sides to 3x + 4 = 25 (dependent on previous M1)
(a) A1 x = 7 cao
(a) Answer: x = 7
(b) M1 rearranges to √5x+1 = 5
(b) M1 squares both sides to 5x + 1 = 25 (dependent on previous M1)
(b) A1 x = 4.8 oe (24/5) cao
(b) Answer: x = 4.8 (24/5)

Question 8

(a) M1 identifies √12 = 2sqrt(3)
(a) A1 √3 + 2sqrt(3) = 3sqrt(3) cso
(a) Answer: 3sqrt(3) (shown)
(b) M1 attempts full expansion (5+sqrt2)(5+sqrt2)
(b) M1 obtains 25 + 5sqrt(2) + 5sqrt(2) + 2 oe (unsimplified)
(b) A1 collects rational terms 25 + 2 = 27 and surd terms 5sqrt(2)+5sqrt(2) = 10sqrt(2)
(b) A1 27 + 10sqrt(2) cso, no errors seen
(b) Answer: 27 + 10sqrt(2) (shown)

Question 9

(a) M1 P = 2[(5+sqrt2)+(3+sqrt2)] oe
(a) A1 16 + 4sqrt(2) cao
(a) Answer: 16 + 4sqrt(2) cm
(b) M1 attempts full expansion (5+sqrt2)(3+sqrt2)
(b) M1 simplifies √2 x √2 = 2 within the expansion
(b) A1 17 + 8sqrt(2) cao
(b) Answer: 17 + 8sqrt(2) cm²

Question 10

(a) M1 √18 = 3sqrt(2)
(a) M1 combines numerator to 4sqrt(2)
(a) A1 4 cao
(a) Answer: 4
(b) M1 multiplies numerator and denominator by (√3+1)
(b) M1 denominator = 2 and numerator = 4 + 2sqrt(3) obtained
(b) A1 2 + √3 cao
(b) Answer: 2 + √3

Question 11

M1 squares both sides: 3x + 10 = x²
M1 rearranges to x² - 3x - 10 = 0
M1 factorises correctly: (x-5)(x+2) = 0, or uses the quadratic formula correctly
A1 x = 5 and x = -2 both found
A1 rejects x = -2 with valid reason (sqrt gives a non-negative value, but x = -2 is negative, so it cannot satisfy the original equation); final answer x = 5
Answer: x = 5 (x = -2 rejected)

Question 12

M1 multiplies numerator and denominator by the conjugate (√5+2)
M1 expands denominator correctly: (√5-2)(√5+2) = 5 - 4
A1 denominator simplifies to 1
A1 √5 + 2 cso, printed result obtained with no errors
Answer: √5 + 2 (shown)

Question 13

(a) M1 correct distance formula set up: √(2-0)² + (2sqrt5-0)²
(a) M1 simplifies to √4 + 20 = √24
(a) A1 2sqrt(6) cao
(a) Answer: 2sqrt(6)
(b) M1 uses midpoint formula (0 + 2sqrt(5))/2
(b) A1 √5 cao
(b) Answer: √5
(c) B1 (2sqrt(5))² = 4 x 5 = 20, a rational number
(c) B1 correct conclusion: AB² = 2² + (2sqrt5)² = 4 + 20 = 24, the sum of two rational numbers, so AB² is rational
(c) Answer: AB² = 24, which is rational

Question 14

(a) M1 squares both sides to 3x - 2 = x + 8
(a) A1 x = 5 cao
(a) Answer: x = 5
(b) M1 squares both sides correctly, including squaring the 2, to give 5x + 1 = 4(x+1)
(b) M1 expands and rearranges: 5x + 1 = 4x + 4
(b) A1 x = 3 cao
(b) Answer: x = 3

Question 15

M1 identifies common denominator (2-sqrt2)(2+sqrt2) = 2
M1 expands first numerator (3+sqrt2)(2+sqrt2) = 8 + 5sqrt(2)
M1 expands second numerator (3-sqrt2)(2-sqrt2) = 8 - 5sqrt(2)
A1 combines numerators: surd terms cancel to give 16
A1 8 cao
Answer: 8

Question 16

M1 squares both sides: x + 4 = (x-2)²
M1 expands (x-2)² = x² - 4x + 4
M1 rearranges to x² - 5x = 0
M1 factorises correctly: x(x-5) = 0
A1 identifies both roots x = 0 and x = 5
A1 rejects x = 0 with valid reason (requires x - 2 ≥ 0, so x ≥ 2; at x=0, RHS = -2 which is negative and cannot equal a square root); final answer x = 5
Answer: x = 5 (x = 0 rejected)

Question 17

M1 multiplies numerator and denominator by the conjugate (√n+1 + √n)
M1 correctly expands denominator (√n+1-√n)(√n+1+√n) = (n+1) - n
A1 denominator simplifies to 1
A1 reaches √n+1 + √n cso, printed result obtained
B1 recognises the simplification holds regardless of the value of n, so the identity is true for all positive integers n
Answer: √n+1 + √n (proven for all positive integers n)

Question 18

(a) M1 uses Pythagoras: 1² + (√2)² = 1 + 2 = 3
(a) A1 hypotenuse = √3 cso
(a) Answer: √3 cm (shown)
(b) M1 cos(θ) = adjacent/hypotenuse = 1/√3
(b) M1 rationalises by multiplying by √3/√3
(b) A1 √3/3 cao
(b) Answer: √3/3
(c) B1 √2/2 oe cao
(c) Answer: √2/2 cm²

Question 19

B1 states or uses x > 0 so that √x is defined and non-zero
M1 multiplies both sides by √x
M1 correctly obtains x + 3 = x + 4sqrt(x)
M1 simplifies to 3 = 4sqrt(x)
A1 √x = 3/4, squares to give x = 9/16
A1 verifies x = 9/16 satisfies the original equation; final answer x = 9/16 cao
Answer: x = 9/16

Question 20

M1 combines the two fractions over a common denominator (√n+1+√n)(√n+1-√n)
M1 correctly obtains numerator (√n+1-√n) + (√n+1+√n) = 2sqrt(n+1)
M1 correctly expands denominator to (n+1) - n
A1 denominator simplifies to 1
A1 reaches 2sqrt(n+1) cso, printed result obtained with no errors
B1 recognises the simplification is independent of n, so the identity holds for all positive integers n
Answer: 2sqrt(n+1) (proven for all positive integers n)

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