Part I Review

1: D. Combine real parts (6+2=8) and imaginary parts (1−5=−4): 8 − 4i. Choice A drops the negative sign on the imaginary part while combining.

2: F. 50 is not a perfect square and its square root does not terminate or repeat, so √50 is irrational. Choice G is a perfect square (√64=8), which is rational, not irrational.

3: D. 30 = 4(7) + 2, so i³⁰ = (i⁴)⁷·i² = 1·(−1) = −1. Choice A reports i⁰ instead of completing the reduction to i²=−1.

4: F. Same base, add the exponents: 3/4 + 1/4 = 1, so the result is x. Choice G multiplies the exponents instead of adding them, computing 3/4 × 1/4 = 3/16.

5: B. (7)(3) − (2)(1) = 21 − 2 = 19. Choice F adds the two products instead of subtracting, computing (7)(3) + (2)(1) = 23.

6: G. (4−i)(3+2i) = 12 + 8i − 3i − 2i² = 12 + 5i + 2 = 14 + 5i. Choice F drops the sign on the cross terms during distribution, giving 14 − 5i instead of 14 + 5i.

7: A. 72 = 36·2, and √36 = 6, leaving 6√2. Choice B misidentifies the perfect-square factor of 72, using an incorrect factorization.

8: H. Squaring gives x+7 = x²−10x+25, so x²−11x+18=0, giving x=9 or x=2. Checking x=9: √16=4, and 9−5=4, matches. Checking x=2: √9=3, but 2−5=−3; 3≠−3, rejected. Only x=9 is valid. Choice G reports both candidate solutions without checking that x=2 is extraneous.

9: C. Multiply by the conjugate (1+i)/(1+i): (5+3i)(1+i) ÷ (1−i)(1+i) = (5+5i+3i+3i²) ÷ 2 = (2+8i) ÷ 2 = 1 + 4i. Choice A swaps the real and imaginary parts of the final result.

10: F. √((−8)²+15²) = √289 = 17. Choice G adds the absolute values of the components (8+15=23) instead of applying the magnitude formula.

11: B. Multiplying a 2×3 matrix by a 3×2 matrix takes the outer dimensions: 2×2. Choice A uses the outer dimensions in the wrong order, swapping which matrix's rows and which matrix's columns form the result.

12: G. x^(−3/4) = 1/8 means x^(3/4) = 8, so x = 8^(4/3) = (³√8)⁴ = 2⁴ = 16. Choice F reports the value of x^(3/4) instead of completing the solution for x.

13: A. Multiply by the conjugate (4−√2)/(4−√2): 3(4−√2) ÷ (16−2) = (12−3√2) ÷ 14. Choice H uses 16 instead of 14 for the denominator, mishandling the conjugate multiplication (4+√2)(4−√2)=16−2=14.

14: J. 3A = [[9,−3],[6,15]]; 2B = [[2,8],[−4,0]]; 3A−2B = [[7,−11],[10,15]]. Choice H reverses the sign of the entry in row 2, column 2.

15: B. Expanding gives (x−2y) + (2x+y)i = −1+8i, so x−2y=−1 and 2x+y=8. Solving simultaneously: x=3, y=2, so x+y=5. Choice A reports x − y instead of the requested sum x + y.

16: G. The determinant is 2(4) − x(1) = 8−x. Setting 8−x=10 gives x=−2. Choice F drops the negative sign while solving 8−x=10.

17: A. 2u = (6, −4); 2u−v = (6−1, −4−(−4)) = (5, 0); magnitude = √(25+0) = 5. Choice G results from a sign or order error while computing 2u − v before applying the magnitude formula.

18: H. (2i)⁵ = 2⁵·i⁵ = 32·i⁴·i = 32·1·i = 32i. Choice F mishandles the power of i within the expansion, producing the sign-reversed value.

19: C. Division by zero is undefined; there is no number that, multiplied by 0, gives 7. Choice A confuses this with 0 ÷ 7, which is defined and equals 0.

20: J. √48 = 4√3, √27 = 3√3, √12 = 2√3; 4√3 + 3√3 − 2√3 = 5√3. Choice F drops one of the three terms before combining the coefficients.