Review: complex numbers, the discriminant, the sum and product of the roots, the sign of a function and quadratic inequalities
Mathematics Sec 1 Term 1, Units 1–4, Review
GoalRevise Unit 1 with new mixed problems.
New words
conjugate · a − bi is the conjugate of a + bidiscriminant · b² − 4ac tells the type of rootssolution set · all values that make an inequality true
Explanation
Checklist for Unit 1. (1) i² = −1, and the powers of i repeat every 4: divide the exponent by 4 and use the remainder. (2) Multiply complex numbers like brackets; divide by multiplying the top and bottom by the conjugate of the bottom. (3) For ax² + bx + c = 0 find the discriminant D = b² − 4ac: D > 0 two real roots, D = 0 equal roots, D < 0 two complex conjugate roots. (4) Without solving: sum of roots = −b/a, product = c/a; build new equations from the sum and product of the new roots. (5) The sign of f(x) = ax² + bx + c: find the roots, draw the parabola (up if a > 0) and read where it is above or below the axis. (6) For an inequality, give the solution as an interval or a union of intervals.
Unit
Key tools
Complex numbers
i² = −1, conjugate, D = b² − 4ac, sum −b/a, product c/a
Solve x² − 6x + 13 = 0. D = 36 − 52 = −16 < 0, so x = (6 ± 4i)/2 = 3 ± 2i. For 2x² − kx + 8 = 0 to have equal roots, k² − 64 = 0, so k = 8 or k = −8.
L and M are the roots of 2x² − 5x + 1 = 0. L + M = 5/2, LM = 1/2, L² + M² = 25/4 − 1 = 21/4. The equation with roots 1/L and 1/M has sum (L + M)/LM = 5 and product 2: x² − 5x + 2 = 0.
Solve x² − 5x − 6 ≤ 0. Roots −1 and 6, the parabola is up, so f ≤ 0 between the roots: [−1, 6]. And 2x² − x − 3 > 0: roots −1 and 3/2, f > 0 outside: x < −1 or x > 3/2.
A rectangular plot has perimeter 40 m and area at least 91 m². The length x satisfies x(20 − x) ≥ 91, so x² − 20x + 91 ≤ 0, roots 7 and 13: 7 ≤ x ≤ 13 m.
Class activity
Error hunt. A friend says x² + x + 3 > 0 has no solution because it ‘has no roots’. Use the discriminant to show that your friend is wrong, and say what the solution set is.
Practice
1
Compute i^53 + (2 − i)(2 + i).
i^53 = i (53 = 4 × 13 + 1) and (2 − i)(2 + i) = 4 + 1 = 5: 5 + i.
2
Find k so that x² + kx + 9 = 0 has equal roots.
k² − 36 = 0, k = ±6.
3
The roots of x² − 7x + 3 = 0 are L and M. Find L² + M² and the equation with roots L² and M².
L + M = 7, LM = 3: L² + M² = 49 − 6 = 43; product 9: x² − 43x + 9 = 0.
4
Solve x² − 4x − 5 > 0.
Roots −1 and 5: x < −1 or x > 5.
5
Show that x² + x + 3 > 0 for every real x.
D = 1 − 12 < 0 and a > 0: the parabola is above the x-axis everywhere.
