These eight original questions follow the order of the lessons in quadratic functions. Check each answer by substituting back, as the explained answers do. Use the timed original practice session builder if you want to work under a time limit.
Questions
Question 1
The roots of 2x² − 7x + 3 = 0 are α and β. Find α + β and αβ.
Answer
a = 2, b = −7, c = 3.
α + β = −b/a = 7/2, and αβ = c/a = 3/2.
Check: 2x² − 7x + 3 = (2x − 1)(x − 3), so the roots are 1/2 and 3. Sum = 7/2, product = 3/2.
Question 2
One root of x² − 8x + k = 0 is three times the other. Find k.
Answer
Let the roots be α and 3α. The sum is 4α = 8, so α = 2. The roots are 2 and 6.
k = 2 × 6 = 12.
Check: x² − 8x + 12 = (x − 2)(x − 6).
Question 3
The roots of x² − 5x + 2 = 0 are α and β. Form the quadratic equation whose roots are 1/α and 1/β.
Answer
α + β = 5 and αβ = 2.
New sum: 1/α + 1/β = (α + β) ÷ αβ = 5/2. New product: (1/α)(1/β) = 1/αβ = 1/2.
Equation: x² − (5/2)x + 1/2 = 0, so 2x² − 5x + 1 = 0.
Question 4
The equation x² + (k − 1)x + 4 = 0 has equal roots. Find the possible values of k.
Answer
Equal roots means b² − 4ac = 0, so (k − 1)² − 16 = 0.
Then (k − 1)² = 16, so k − 1 = 4 or k − 1 = −4.
k = 5 or k = −3.
Check k = 5: x² + 4x + 4 = (x + 2)². Check k = −3: x² − 4x + 4 = (x − 2)².
Question 5
Write f(x) = 2x² − 12x + 7 in the form a(x + p)² + q. State the minimum value and where it occurs.
Answer
f(x) = 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11.
The minimum value is −11, at x = 3.
Check: f(3) = 18 − 36 + 7 = −11.
Question 6
Find the maximum value of g(x) = −2x² + 8x − 3 and the value of x at which it occurs.
Answer
g(x) = −2(x² − 4x) − 3 = −2[(x − 2)² − 4] − 3 = −2(x − 2)² + 8 − 3 = −2(x − 2)² + 5.
Since a is negative, the graph opens downward. The maximum value is 5, at x = 2.
Check: g(2) = −8 + 16 − 3 = 5.
Question 7
Solve 2x² + 5x − 3 ≥ 0.
Answer
Factorise: (2x − 1)(x + 3) ≥ 0. The roots are x = 1/2 and x = −3.
The graph opens upward, and ≥ 0 means on or above the axis, which is outside the roots, including the roots.
x ≤ −3 or x ≥ 1/2.
Check x = 0: −3, which is not ≥ 0, so the middle is correctly excluded.
Question 8
The line y = 4x − k is a tangent to the curve y = x² + 2x + 3. Find k and the coordinates of the point where the line touches the curve.
Answer
Set equal: x² + 2x + 3 = 4x − k, so x² − 2x + (3 + k) = 0.
A tangent means b² − 4ac = 0: (−2)² − 4(1)(3 + k) = 0, so 4 − 12 − 4k = 0 and k = −2.
Then x² − 2x + 1 = 0 gives x = 1. The curve gives y = 1 + 2 + 3 = 6, so the point is (1, 6).
Check: the line is y = 4x + 2, and at x = 1 it gives y = 6.
If you got these wrong
| Question | Skill to revisit |
|---|---|
| 1, 2 and 3 | Relating roots to coefficients |
| 4 and 8 | Using the discriminant to classify roots |
| 5 and 6 | Completing the square to find a turning point |
| 7 | Solving quadratic inequalities using intervals |
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