Simplify fully √(112) + √(175).
Worked Solution & Answer
√(112) + √(175) = 4√(7) + 5√(7) = 9√(7) B1
Answer: 9√(7)
Surds — Notes & Worked Examples Show / Hide
Explanation:
A surd is an irrational root that cannot be written as an exact fraction, such as √2, √3, or √5. To simplify a surd, split the number under the root into its largest square factor multiplied by another factor using √(ab) = √a × √b, and only collect like surds when adding or subtracting.
To rationalise the denominator of a fraction, remove the surd from the bottom: for a√b, multiply top and bottom by √b; for ca + √b, multiply top and bottom by the conjugate a − √b so the middle surd terms cancel out via the difference of two squares.
Example 1: Simplify fully √75 + √48 − 2√12
Solution:
Extract the largest square factor from each surd:
√75 = √25 × √3 = 5√3
√48 = √16 × √3 = 4√3
2√12 = 2 × √4 × √3 = 2 × 2√3 = 4√3
5√3 + 4√3 − 4√3 = 5√3
Answer: 5√3
Example 2: Rationalise the denominator of 143 − √2
Solution:
Multiply numerator and denominator by 3 + √2:
= 14(3 + √2)(3 − √2)(3 + √2)
Denominator: 32 − (√2)2 = 9 − 2 = 7
= 14(3 + √2)7 = 2(3 + √2)
Answer: 6 + 2√2
Simplify fully √(80) + √(125).
Worked Solution & Answer
√(80) + √(125) = 4√(5) + 5√(5) = 9√(5) B1
Answer: 9√(5)
Given that a is a prime number, rationalise the denominator of 1√(a) + 1
Give your answer in its simplest form.
Give your answer in its simplest form.
Worked Solution & Answer
Multiply top and bottom by (√a – 1): (√a – 1) / [(√a + 1)(√a – 1)] = (√a – 1) / (a – 1). M1 A1
Answer: √(a) - 1a - 1
a = √(2) and b = √(18)
Match each expression to its value.
One has been done for you.
Match each expression to its value.
One has been done for you.
a = √2 and b = √18. Match each expression (a², a + b, ab, b/a) to its value. [3 marks]
Worked Solution & Answer
b = √18 = 3√2. M1
a + b = √2 + 3√2 = 4√2. M1
ab = √2 × √18 = √36 = 6.
b/a = √18 / √2 = √9 = 3. A1
a + b = √2 + 3√2 = 4√2. M1
ab = √2 × √18 = √36 = 6.
b/a = √18 / √2 = √9 = 3. A1
Answer: a + b → 4√2, ab → 6, b/a → 3
(a) Rationalise the denominator of 1/√7 (1) (b) Simplify fully √80 – √5 (2)
(a) Rationalise the denominator of 1√(7) [1 mark]
(b) Simplify fully √(80) - √(5) [2 marks]
Worked Solution & Answer
(a)
(1 × √7)/(√7 × √7) = √7/7 B1
Answer (a): √(7)7
(b)
√80 – √5 = √(16 × 5) – √5 = 4√5 – √5 = 3√5 M1 A1
Answer (b): 3√(5)
Show that √(180) - 2√(5)5√(5) - 5 can be written in the form a + √(5)b where a and b are integers.
Worked Solution & Answer
√180 = 6√5 M3
⇒ numerator = 4√5. Multiply top and bottom by (5√5 + 5): (4√5)(5√5 + 5) / (125 – 25) = (100 + 20√5) / 100 = 1 + √5/5 C1
⇒ numerator = 4√5. Multiply top and bottom by (5√5 + 5): (4√5)(5√5 + 5) / (125 – 25) = (100 + 20√5) / 100 = 1 + √5/5 C1
Answer: 1 + √(5)5 (a = 1, b = 5)
(a) Show that √363 / √3 simplifies to an integer. [2 marks]
20 (b) Rationalise the denominator and simplify 20 / √5. [2 marks]
20 (b) Rationalise the denominator and simplify 20 / √5. [2 marks]
(a) Show that √(363)√(3) simplifies to an integer. [2 marks]
(b) Rationalise the denominator and simplify 20√(5) [2 marks]
Answer
Worked Solution & Answer
(a)
√363 / √3 = √(363 ÷ 3) = √121 = 11 M1 C1
Answer (a): √(363/3) = √121 = 11
(b)
(20 / √5) × (√5 / √5) = 20√5 / 5 = 4√5 M1 A1
Answer (b): Answer 4√(5)
(a) Rationalise the denominator of 22/√11 in simplest form. (2) (b) Show that √3 / (2√3 – 1) can be written in the form (a + √3)/b. (3)
(a) Rationalise the denominator of 22√(11)
Give your answer in its simplest form. [2 marks]
Give your answer in its simplest form. [2 marks]
(b) Show that √(3)2√(3) - 1 can be written in the form a + √(3)b where a and b are integers. [3 marks]
Worked Solution & Answer
(a)
22√11 / 11 = 2√11 M1 A1
Answer (a): 2√(11)
(b)
Multiply top and bottom by (2√3 + 1): √3(2√3 + 1) / (12 – 1) = (6 + √3) / 11 M2 C1
Answer (b): 6 + √(3)11
Work out
(a) Rationalise the denominator of 35√(7). (2)
(b) Write √(27) - 12 - √(3) in the form a + b√(3). Work out a and b. (4)
(a) Rationalise the denominator of 35√(7). (2)
(b) Write √(27) - 12 - √(3) in the form a + b√(3). Work out a and b. (4)
(a) Rationalise the denominator of 35√(7)
Give your answer in its simplest form. [2 marks]
Give your answer in its simplest form. [2 marks]
(b) Work out the value of a and the value of b. [4 marks]
Worked Solution & Answer
(a)
(35 × √7) ÷ (√7 × √7) = 35√7 ÷ 7 = 5√7 M1 A1
Answer (a): 5√(7)
(b)
√27 = 3√3. Multiply top and bottom by 2 + √3: M2
Denominator = (2 – √3)(2 + √3) = 4 – 3 = 1. M1
Numerator = (3√3 – 1)(2 + √3) = 6√3 + 9 – 2 – √3 = 7 + 5√3 ⇒ a = 7, b = 5 A1
Denominator = (2 – √3)(2 + √3) = 4 – 3 = 1. M1
Numerator = (3√3 – 1)(2 + √3) = 6√3 + 9 – 2 – √3 = 7 + 5√3 ⇒ a = 7, b = 5 A1
Answer (b): a = 7, b = 5
Work out
(a) Simplify.
√(3) × √(15)
(b) Rationalise the denominator and simplify.
40√(15)
(c) Work out.
27<span class="opg-frac"><span class="opg-num">4</span><span class="opg-den">3</span></span>
(a) Simplify.
√(3) × √(15)
(b) Rationalise the denominator and simplify.
40√(15)
(c) Work out.
27<span class="opg-frac"><span class="opg-num">4</span><span class="opg-den">3</span></span>
(a) Simplify.
√(3) × √(15) [2 marks]
√(3) × √(15) [2 marks]
(b) Rationalise the denominator and simplify.
40√(15) [3 marks]
40√(15) [3 marks]
(c) Work out.
27<span class="opg-frac"><span class="opg-num">4</span><span class="opg-den">3</span></span> [2 marks]
27<span class="opg-frac"><span class="opg-num">4</span><span class="opg-den">3</span></span> [2 marks]
Worked Solution & Answer
(a)
√3 × √15 = √45 = √(9 × 5) = 3√5. M1 A1
Answer (a): 3√(5)
(b)
40 / √15 = (40 × √15) / 15 = (8√15) / 3. M2 A1
Answer (b): 8√(15)3
(c)
27 {}4/3) = (∛27)⁴ = 3⁴ = 81. M1 A1
Answer (c): 81