This set has eleven original questions, ordered from easier to harder, covering all five lessons in quadratic structure and discriminants. Questions 1 to 4 are warm-ups, 5 to 8 need a full method, and 9 to 11 combine skills.
Attempt each question on paper first. Give exact answers unless told otherwise, and use the substitution check in each answer. The quadratic structure explorer and the mistake log and retest queue are useful companions while you work.
Questions
1. Write x² − 10x + 7 in the form (x + p)² + q.
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Half of −10 is −5, and (−5)² = 25. So x² − 10x + 7 = (x − 5)² − 25 + 7 = (x − 5)² − 18. Check at x = 0: 25 − 18 = 7.
2. Write 4x² + 8x + 9 in the form a(x + p)² + q.
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4(x² + 2x) + 9 = 4[(x + 1)² − 1] + 9 = 4(x + 1)² − 4 + 9 = 4(x + 1)² + 5. Check at x = 1: 4 + 8 + 9 = 21 and 4 × 4 + 5 = 21.
3. Find the value of the discriminant of 2x² − 3x + 5 and say what it tells you about the roots.
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a = 2, b = −3, c = 5. The discriminant is (−3)² − 4(2)(5) = 9 − 40 = −31. It is negative, so the equation has no real roots.
4. Find a quadratic equation with roots 5 and −2.
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Sum = 3, product = −10. So x² − 3x − 10 = 0. Check at x = 5: 25 − 15 − 10 = 0. Check at x = −2: 4 + 6 − 10 = 0. The equation is x² − 3x − 10 = 0.
5. Find the turning point of y = x² + 6x + 4 and state whether it is a maximum or a minimum.
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x² + 6x + 4 = (x + 3)² − 9 + 4 = (x + 3)² − 5. The turning point is (−3, −5). Since the coefficient of x² is positive, it is a minimum. Check: 9 − 18 + 4 = −5.
6. Find the value of k for which kx² + 12x + 9 = 0 has equal roots.
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The discriminant is 144 − 4(k)(9) = 144 − 36k. Set it to 0: k = 4. Check: 4x² + 12x + 9 = (2x + 3)², a repeated root. So k = 4.
7. The line y = x + c and the curve y = x² − 3x + 1 are given. Find the values of c for which the line meets the curve at two distinct points, and the point of contact when it is a tangent.
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x² − 3x + 1 = x + c gives x² − 4x + (1 − c) = 0. The discriminant is 16 − 4(1 − c) = 12 + 4c. Two points when c > −3. Tangent when c = −3: x² − 4x + 4 = 0, so x = 2, y = 2 − 3 = −1. Contact point (2, −1). Check: the curve at x = 2 gives 4 − 6 + 1 = −1.
8. The roots of x² − 6x + 4 = 0 are α and β. Find the equation whose roots are α² and β².
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α + β = 6 and αβ = 4. Then α² + β² = (α + β)² − 2αβ = 36 − 8 = 28, and α²β² = 16. The equation is x² − 28x + 16 = 0. Check: the original roots are 3 ± √5, whose squares are 14 ± 6√5, with sum 28 and product 196 − 180 = 16.
9. For f(x) = −2x² + 12x − 5, find the greatest value of f and the range of f.
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−2(x² − 6x) − 5 = −2[(x − 3)² − 9] − 5 = −2(x − 3)² + 18 − 5 = −2(x − 3)² + 13. The greatest value is 13 at x = 3, so the range is f(x) ≤ 13. Check: f(3) = −18 + 36 − 5 = 13.
10. Find the range of values of k for which x² + kx + k + 3 = 0 has no real roots.
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The discriminant is k² − 4(k + 3) = k² − 4k − 12 = (k − 6)(k + 2). No real roots needs (k − 6)(k + 2) < 0, which gives −2 < k < 6. Check k = 0: x² + 3 = 0 has no real roots. Check k = 7: 49 − 28 − 12 = 9, positive, so real roots exist outside the range.
11. The curve y = 2x² − 8x + k has a minimum value of 3. Find k, and find the equation of the line parallel to the x-axis that touches the curve.
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2(x² − 4x) + k = 2[(x − 2)² − 4] + k = 2(x − 2)² − 8 + k. The minimum value is k − 8, so k − 8 = 3 and k = 11. The line touching the curve at its lowest point is y = 3. Check: at x = 2, 8 − 16 + 11 = 3. As a further check, 2x² − 8x + 11 = 3 gives 2x² − 8x + 8 = 0, discriminant 64 − 64 = 0.
If you got these wrong
- Questions 1, 2, 9 or 11 (completing the square, turning points): revisit complete the square without losing a coefficient and connect a turning point with a minimum value. The usual error is forgetting to multiply the added constant by a.
- Questions 3, 7 or 10 (discriminant and intersections): revisit use the discriminant to classify intersections. Check that the equation equals zero before reading a, b and c.
- Questions 4 or 8 (roots): revisit recover a quadratic from roots. Write the template x² − (sum)x + (product) before filling in.
- Questions 6 or 10 (conditions on k): revisit solve a parameter condition for repeated roots. Expand (k + something)² in full.
Write each error in the mistake log with its type, so your next attempt targets the habit. If the same type keeps returning, our teachers can look at your working in online one-to-one Additional Mathematics tuition.