Algebra & Functions

The quadratic equation 2x² + kx + 8 = 0 has equal roots, where k > 0. Find the value of k.

• Ak = 4
• Bk = 8✓
• Ck = 2
• Dk = 16

Answer: Equal roots require the discriminant b² − 4ac = 0, so k² − 4(2)(8) = k² − 64 = 0, giving k² = 64 and the positive root k = 8.

Simplify fully (3 + √5)(3 − √5).

• A9 − √5
• B14
• C6
• D4✓

Answer: This is a difference of two squares: (3)² − (√5)² = 9 − 5 = 4.

Given f(x) = x² − 6x + 11, express f(x) in completed-square form a(x + b)² + c.

• A(x − 6)² + 11
• B(x + 3)² + 2
• C(x − 3)² + 2✓
• D(x − 3)² − 2

Answer: Completing the square: x² − 6x + 11 = (x − 3)² − 9 + 11 = (x − 3)² + 2.

Evaluate 27^(2/3) without a calculator.

• A6
• B81
• C18
• D9✓

Answer: 27^(2/3) = (27^(1/3))² = 3² = 9, since the cube root of 27 is 3.

In the binomial expansion of (1 + x)⁵, find the coefficient of x².

• A5
• B10✓
• C25
• D20

Answer: The coefficient of x² is the binomial coefficient C(5,2) = 10.

The equation x² + kx + 9 = 0 has two distinct real roots. Find the complete set of values of k.

• A−6 < k < 6
• Bk > 6 only
• Ck < −6 or k > 6✓
• Dk < −3 or k > 3

Answer: Two distinct real roots require discriminant b² − 4ac > 0, so k² − 4(1)(9) = k² − 36 > 0, giving k² > 36 and hence k < −6 or k > 6.

Express 2x² + 12x + 5 in the form a(x + p)² + q.

• A2(x + 3)² − 13✓
• B2(x + 3)² + 5
• C2(x + 6)² − 13
• D(x + 6)² − 31

Answer: Factor 2 from the x-terms: 2(x² + 6x) + 5 = 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13.

Solve the simultaneous equations y = x + 1 and x² + y² = 25. Which pair gives both solution points?

• A(4, 3) and (−3, −4)
• B(3, 4) and (−4, −3)✓
• C(3, 4) only
• D(0, 1) and (5, 6)

Answer: Substitute y = x + 1: x² + (x + 1)² = 25 → 2x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → (x + 4)(x − 3) = 0. So x = 3 (y = 4) or x = −4 (y = −3), giving (3, 4) and (−4, −3).

Solve the inequality x² − x − 6 < 0.

• Ax < −2 or x > 3
• B−3 < x < 2
• C−2 < x < 3✓
• Dx < −3 or x > 2

Answer: Factorise: x² − x − 6 = (x − 3)(x + 2). The parabola opens upwards and is below the x-axis between its roots, so (x − 3)(x + 2) < 0 gives −2 < x < 3.

Express (5x + 1)/((x + 1)(x − 2)) as partial fractions A/(x + 1) + B/(x − 2). Find A and B.

• AA = 11/3, B = 4/3
• BA = 4/3, B = 11/3✓
• CA = −4/3, B = 11/3
• DA = 4, B = 11

Answer: Write 5x + 1 = A(x − 2) + B(x + 1). Let x = 2: 11 = 3B, so B = 11/3. Let x = −1: −4 = −3A, so A = 4/3.

Solve the modulus equation |2x − 3| = 5.

• Ax = 4 or x = 1
• Bx = 4 only
• Cx = −4 or x = 1
• Dx = 4 or x = −1✓

Answer: Either 2x − 3 = 5, giving 2x = 8 and x = 4, or 2x − 3 = −5, giving 2x = −2 and x = −1. Both satisfy the equation.

The functions are f(x) = 2x + 1 and g(x) = x². Find the value of the composite fg(3), that is f(g(3)).

• A49
• B19✓
• C37
• D13

Answer: fg(3) means apply g first: g(3) = 3² = 9, then f(9) = 2(9) + 1 = 19. (Option A is gf(3) = (2·3 + 1)² = 49, a common order mix-up.)