Unit 6
4 marks · Normal2. Consider the algebraic expressions:
Simplify (2x^3y^2)^1/2 × (4x^2y^6)^1/4. Give your answer in its simplest index form.
List all the members of the set S = { x | x = 2^n, n ∈ N, x ≤ 32 }.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 2 x^{7/4} y^{5/2}
(b) {2, 4, 8, 16, 32}
Common mistakes: Forgetting index laws; Incorrect list of members of set
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Use index/fractional laws correctly on both parts.
A1 (a) · 1 mark · Obtain correct simplified expression (2 x^{7/4} y^{5/2})
M2 (b) · 1 mark · List all required elements using the rule (powers of 2 up to 32).
A2 (b) · 1 mark · All correct values ({2, 4, 8, 16, 32}).
Worked solution
a) (2x^3y^2)^{1/2} = 2^{1/2} x^{3/2} y^{1}, (4x^2y^6)^{1/4} = 4^{1/4} x^{1/2} y^{3/2}.
4^{1/4} = (2^2)^{1/4} = 2^{1/2}.
When multiplied: 2^{1/2} × 2^{1/2} = 2, x^{3/2} × x^{1/2} = x^{2}, y^1 × y^{3/2} = y^{5/2}.
So, the answer: 2 x^{2} y^{5/2} (PLEASE NOTE: MISTAKE! Check indices carefully).
Let’s do it step by step:
First, write all as indices:
(2 x^3 y^2)^{1/2} = 2^{1/2} x^{3/2} y^1
(4 x^2 y^6)^{1/4} = 4^{1/4} x^{2/4} y^{6/4} = 2^{1/2} x^{1/2} y^{3/2}
Multiply together: (2^{1/2} × 2^{1/2}) = 2, (x^{3/2} × x^{1/2}) = x^2, (y^1 × y^{3/2}) = y^{5/2}.
Final answer is: 2 x^2 y^{5/2}.
b) S = {2, 4, 8, 16, 32}, since 2^1, 2^2, ..., 2^5 = 32.