Quadratics & Algebraic Fractions

Solve x² + 6x + 2 = 0 by completing the square, giving your answer in surd form.

• Ax = -3 ± √7✓
• Bx = -3 ± √11
• Cx = 3 ± √7
• Dx = -3 ± √5

Answer: Completing the square on x² + 6x + 2 gives (x + 3)² − 7 = 0, since half of 6 is 3 and 3² − 2 = 7. Rearranging gives (x + 3)² = 7, so x = −3 ± √7. A sign error on the 3 or a slip combining the constants produces the other results.

Solve x² - 4x - 3 = 0 by completing the square, leaving your answer in surd form.

• Ax = 2 ± √11
• Bx = -2 ± √7
• Cx = 2 ± √5
• Dx = 2 ± √7✓

Answer: Completing the square on x² − 4x − 3 gives (x − 2)² − 7 = 0, because half of −4 is −2 and (−2)² − 3 = 7. Rearranging gives (x − 2)² = 7, so x = 2 ± √7. Losing the sign on the 2, or mishandling the constant term, produces the other values.

A student completes the square on 2x² + 8x + 3. Which expression is equivalent?

• A2(x + 4)² − 5
• B2(x + 2)² − 5✓
• C(x + 2)² − 5
• D2(x + 2)² − 8

Answer: Factoring out the 2 gives 2(x² + 4x) + 3, and completing the square inside the bracket uses half of 4, which is 2, so x² + 4x = (x + 2)² − 4. Multiplying back out and adding the 3 gives 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5. Forgetting to multiply the −4 by the outer 2, or dropping that factor entirely, produces the other forms.

Solve 3x² - 12x + 5 = 0 by completing the square, giving your answers to 2 decimal places.

• Ax = 3.53 or x = 0.47✓
• Bx = 4.53 or x = -0.53
• Cx = 3.61 or x = 0.39
• Dx = 3.15 or x = 0.85

Answer: Dividing 3x² − 12x + 5 = 0 by 3 gives x² − 4x + 5/3 = 0, and completing the square gives (x − 2)² = 4 − 5/3 = 7/3. Taking the square root gives x − 2 = ±1.53 (2 d.p.), so x = 3.53 or x = 0.47. Forgetting to divide the constant by 3, or a rounding slip on the square root, produces the other pairs.

Solve 2x² + 3x - 5 = 0 using the quadratic formula.

• Ax = -1 or x = 2.5
• Bx = 2 or x = -2.5
• Cx = 1 or x = -5
• Dx = 1 or x = -2.5✓

Answer: For 2x² + 3x − 5 = 0, the discriminant is 3² − 4(2)(−5) = 49, and √49 = 7. Substituting into x = (−3 ± 7)/4 gives x = 1 or x = −2.5. A sign slip on b, or an error evaluating the discriminant, produces the other pairs.

Solve 3x² - 5x - 2 = 0 using the quadratic formula, giving exact values.

• Ax = 4 or x = -1/3
• Bx = 2 or x = -1/2
• Cx = -2 or x = 1/3
• Dx = 2 or x = -1/3✓

Answer: For 3x² − 5x − 2 = 0, the discriminant is (−5)² − 4(3)(−2) = 49, and √49 = 7. Substituting into x = (5 ± 7)/6 gives x = 2 or x = −1/3. A sign error on the roots, or a slip simplifying the fraction, produces the other pairs.

Use the quadratic formula to solve x² - 3x - 7 = 0, giving your answers to 2 decimal places.

• Ax = 6.08 or x = -3.08
• Bx = 3.54 or x = -0.54
• Cx = 4.54 or x = 1.54
• Dx = 4.54 or x = -1.54✓

Answer: For x² − 3x − 7 = 0, the discriminant is (−3)² − 4(1)(−7) = 37, and √37 ≈ 6.08. Substituting into x = (3 ± 6.08)/2 gives x ≈ 4.54 or x ≈ −1.54. Misreading the discriminant, dropping a sign, or forgetting to divide by 2 produces the other pairs.

A ball's height in metres is modelled by h = -5t² + 20t + 1, where t is time in seconds. Using the quadratic formula, find the positive value of t when h = 0, to 2 decimal places.

• At = 3.95 seconds
• Bt = 4.05 seconds✓
• Ct = 4.10 seconds
• Dt = 4.00 seconds

Answer: Rearranging gives 5t² − 20t − 1 = 0, so the discriminant is (−20)² − 4(5)(−1) = 420 and √420 ≈ 20.49. Substituting into t = (20 ± 20.49)/10 and taking the positive root gives t ≈ 4.05 seconds. A small rounding slip on the square root produces the other close values.

Factorise 2x² + 7x + 3.

• A(x + 1)(2x + 7)
• B(2x + 1)(x + 3)✓
• C(2x + 1)(x + 1)
• D(2x + 3)(x + 1)

Answer: Factorising 2x² + 7x + 3 requires two brackets whose outer and inner products sum to 7x and whose constants multiply to 3, which (2x + 1)(x + 3) satisfies since 6x + x = 7x. Swapping which bracket holds the 1 and the 3, or pairing the wrong factors of 3, produces expressions with a different middle or constant term.

Factorise 3x² - 5x - 2.

• A(3x + 2)(x - 1)
• B(3x - 1)(x + 2)
• C(3x + 1)(x - 2)✓
• D(3x - 2)(x + 1)

Answer: Factorising 3x² − 5x − 2 needs factors of −2 placed so the cross terms combine to −5x, which (3x + 1)(x − 2) satisfies since −6x + x = −5x. Swapping the signs of the 1 and 2, or which bracket holds them, produces expressions with the wrong middle term.

Factorise 6x² + 13x + 6.

• A(2x + 2)(3x + 3)
• B(2x + 3)(3x + 2)✓
• C(2x + 6)(3x + 1)
• D(6x + 3)(x + 2)

Answer: Factorising 6x² + 13x + 6 needs factor pairs of 6 and 6 whose cross products sum to 13x, which (2x + 3)(3x + 2) satisfies since 4x + 9x = 13x. Mismatching which factors pair together produces a different, incorrect middle term.

Factorise 4x² - 4x - 3.

• A(2x + 3)(2x - 1)
• B(4x + 3)(x - 1)
• C(2x - 3)(2x + 1)✓
• D(4x - 3)(x + 1)

Answer: Factorising 4x² − 4x − 3 needs factors of 4 and −3 whose cross terms combine to −4x, which (2x − 3)(2x + 1) satisfies since 2x − 6x = −4x. Swapping the signs of the 3 and 1, or using 4x and x instead of 2x and 2x, produces a different middle term.

Simplify (x² - 9)/(x + 3) fully.

• Ax - 3✓
• Bx² - 3
• Cx - 9
• Dx + 3

Answer: The numerator x² − 9 factorises as (x − 3)(x + 3) using the difference of two squares, and the common factor (x + 3) cancels with the denominator, leaving x − 3. Forgetting to factorise fully, or cancelling the wrong term, produces the other results.

Simplify (2x² + 6x)/(4x) fully.

• A(x + 6)/2
• B(x + 3)/4
• C(2x + 3)/2
• D(x + 3)/2✓

Answer: Both terms in the numerator share a factor of 2x, so 2x² + 6x = 2x(x + 3); dividing by 4x cancels the common 2x, leaving (x + 3)/2. Cancelling only part of the coefficient, or missing a factor of x, produces the other results.

Write 3/(x+1) + 2/(x-1) as a single fraction.

• A(x - 1)/(x² - 1)
• B(5x - 1)/(x² - 1)✓
• C(5x + 1)/(x² - 1)
• D(5x - 1)/(x² + 1)

Answer: Combining over the common denominator (x + 1)(x − 1) gives [3(x − 1) + 2(x + 1)]/(x² − 1), and expanding the numerator gives 3x − 3 + 2x + 2 = 5x − 1. A sign slip when expanding, or an error on the denominator, produces the other fractions.

Solve the equation 3/x + 1/(x+2) = 2, where x ≠ 0 and x ≠ -2.

• Ax = ±√3✓
• Bx = ±√6
• Cx = ±3
• Dx = √3

Answer: Multiplying through by x(x + 2) gives 3(x + 2) + x = 2x(x + 2), which simplifies to 4x + 6 = 2x² + 4x and then 2x² = 6, so x² = 3. Taking both square roots gives x = ±√3; a slip in the constant, or forgetting the negative root, produces the other results.

Simplify (x² + 5x + 6)/(x² - 4) fully.

• A(x - 3)/(x - 2)
• B(x + 2)/(x - 2)
• C(x + 3)/(x - 2)✓
• D(x + 3)/(x + 2)

Answer: The numerator factorises as (x + 2)(x + 3) and the denominator as (x − 2)(x + 2), so the common factor (x + 2) cancels, leaving (x + 3)/(x − 2). Cancelling the wrong bracket, or a sign slip in one of the factors, produces the other results.

Solve the simultaneous equations y = x² and y = x + 6.

• Ax = -3, y = 9 or x = 2, y = 4
• Bx = 3, y = 9 or x = 2, y = 4
• Cx = 3, y = 9 or x = -2, y = 4✓
• Dx = 3, y = 3 or x = -2, y = -2

Answer: Substituting y = x² into y = x + 6 gives x² − x − 6 = 0, which factorises as (x − 3)(x + 2) = 0, so x = 3 or x = −2. Substituting back gives y = 9 and y = 4 respectively; sign errors on x, or using the wrong equation to find y, produce the other pairs.

Solve the simultaneous equations y = x² - 4 and y = 3x.

• Ax = 4, y = 12 or x = -1, y = -3✓
• Bx = -4, y = 12 or x = 1, y = -3
• Cx = 4, y = 8 or x = -1, y = -3
• Dx = 4, y = 12 or x = 1, y = 3

Answer: Substituting y = 3x into y = x² − 4 gives x² − 3x − 4 = 0, which factorises as (x − 4)(x + 1) = 0, so x = 4 or x = −1. Substituting into y = 3x gives y = 12 and y = −3; a sign error on x, or substituting into the wrong equation, produces the other pairs.

Solve the simultaneous equations x² + y² = 25 and y = x + 1.

• Ax = -4, y = -3 or x = 3, y = -4
• Bx = 4, y = -3 or x = -3, y = 4
• Cx = -4, y = 3 or x = 3, y = 4
• Dx = -4, y = -3 or x = 3, y = 4✓

Answer: Substituting y = x + 1 into x² + y² = 25 gives x² + (x + 1)² = 25, which simplifies to 2x² + 2x − 24 = 0 and then x² + x − 12 = 0, factorising as (x + 4)(x − 3) = 0. This gives x = −4 or x = 3, and y = x + 1 gives y = −3 or y = 4; sign slips when substituting back produce the other pairs.

Solve the simultaneous equations y = 2x² and y = 5x - 2, giving both pairs of solutions.

• Ax = 1/2, y = 1/2 or x = 2, y = 8✓
• Bx = 1/2, y = 1/2 or x = 2, y = 10
• Cx = -1/2, y = 1/2 or x = -2, y = 8
• Dx = 1/2, y = 2 or x = 2, y = 8

Answer: Substituting y = 2x² into y = 5x − 2 gives 2x² − 5x + 2 = 0, which factorises as (2x − 1)(x − 2) = 0, so x = 1/2 or x = 2. Substituting into y = 5x − 2 gives y = 1/2 and y = 8; a sign error on x, or an arithmetic slip finding y, produces the other pairs.

Which algebraic statement correctly proves that the sum of two consecutive odd numbers is always even?

• A(2n + 1) + (2n + 3) = 4n + 3, which is stated to be an even number
• B(2n + 1) + (2n + 3) = 4n + 4 = 2(2n + 2), an even number✓
• C(2n + 1) + (2n + 3) = 4n + 4 = 2(2n + 4), which is stated to be an even number
• D(2n) + (2n + 2) = 4n + 2 = 2(2n + 1), which is stated to be an even number

Answer: Two consecutive odd numbers can be written as 2n + 1 and 2n + 3, and their sum is 4n + 4, which factorises as 2(2n + 2) — a multiple of 2, so it is even. Using an incorrect sum, starting from even rather than odd expressions, or a slip when factorising, produces the other statements.