Roots: discriminant, sum and product
Without solving, Δ = b² − 4ac tells the nature of the roots, and α + β = −b/a, αβ = c/a build new equations.
The discriminant and the nature of roots
| Value of Δ = b² − 4ac | Nature of roots | Example |
|---|---|---|
| Δ > 0 | Real and unequal | x² − 5x + 6 = 0: Δ = 25 − 24 = 1; roots 2 and 3 |
| Δ = 0 | Real and equal | x² − 6x + 9 = 0: Δ = 36 − 36 = 0; roots 3 and 3 |
| Δ < 0 | No real roots | x² + x + 1 = 0: Δ = 1 − 4 = −3 |
Useful identities for α and β
| Expression | In terms of α + β and αβ |
|---|---|
| α² + β² | (α + β)² − 2αβ |
| (α − β)² | (α + β)² − 4αβ |
| α³ + β³ | (α + β)³ − 3αβ(α + β) |
| 1/α + 1/β | (α + β)/αβ |
| α/β + β/α | (α² + β²)/αβ |
Worked example 1: find k for equal roots
Find the values of k for which x² − 2kx + (7k − 12) = 0 has real and equal roots.
- Here a = 1, b = −2k and c = 7k − 12.
- Real and equal roots need Δ = b² − 4ac = 0.
- (−2k)² − 4(1)(7k − 12) = 0, so 4k² − 28k + 48 = 0.
- Divide by 4: k² − 7k + 12 = 0, so (k − 3)(k − 4) = 0.
- k = 3 or k = 4.
- Check k = 3: the equation becomes x² − 6x + 9 = (x − 3)², which has equal roots.
k = 3 or k = 4
Worked example 2: a new equation from α and β
α and β are the roots of x² − 5x + 3 = 0. Find α² + β². Then form the equation whose roots are α² and β².
- Here a = 1, b = −5, c = 3. So α + β = −b/a = 5 and αβ = c/a = 3.
- α² + β² = (α + β)² − 2αβ = 25 − 6 = 19.
- Sum of the new roots = α² + β² = 19.
- Product of the new roots = α²β² = (αβ)² = 9.
- Formula: x² − (sum of roots)x + (product of roots) = 0.
α² + β² = 19, and the required equation is x² − 19x + 9 = 0
Worked example 3: is the garden possible?
A school plans a rectangular garden with a 60 m fence around it and an area of 250 m². Is this possible?
- Perimeter 60 m, so length + breadth = 30 m. Let the length be x m; the breadth is (30 − x) m.
- Area: x(30 − x) = 250, so x² − 30x + 250 = 0.
- Δ = b² − 4ac = (−30)² − 4(1)(250) = 900 − 1000 = −100.
- Δ < 0, so the equation has no real roots.
- Check the idea: even a 15 m by 15 m square gives only 225 m².
Not possible: no real length gives an area of 250 m² with a 60 m fence
Before building, a planner can check whether a design is possible. The discriminant shows in one line whether a fence length and an area can ever fit together.
2 marks: “Determine the nature of the roots of 4x² − 12x + 9 = 0.” Δ = (−12)² − 4(4)(9) = 144 − 144 = 0. So the roots are real and equal.
2 marks: “Form the quadratic equation whose roots are 3 + √2 and 3 − √2.” Sum = 6, product = 9 − 2 = 7. Equation: x² − 6x + 7 = 0.
5 marks: “If α and β are the roots of x² + 3x − 2 = 0, find α/β + β/α.” α + β = −3 and αβ = −2. α² + β² = (α + β)² − 2αβ = 9 + 4 = 13. So α/β + β/α = (α² + β²)/αβ = −13/2.
Sum: minus b over a. Product: c over a. Then Δ decides: positive, real and unequal; zero, real and equal; negative, no real roots.