Relationship between zeroes and coefficients
For ax² + bx + c with zeroes α and β: the sum α + β = −b/a and the product αβ = c/a.
The two rules
Let α (alpha) and β (beta) be the zeroes of p(x) = ax² + bx + c, where a ≠ 0. Then sum of zeroes α + β = −b/a and product of zeroes αβ = c/a. Watch the minus sign in the sum. There is no minus sign in the product. Why it works: ax² + bx + c = a(x − α)(x − β) = ax² − a(α + β)x + aαβ. Compare the terms: b = −a(α + β) and c = aαβ.
Example 1: find the zeroes and verify
Find the zeroes of x² + 2x − 15 and verify the relationship between the zeroes and the coefficients.
- Two numbers with product −15 and sum 2 are 5 and −3.
- x² + 2x − 15 = x² + 5x − 3x − 15
- = x(x + 5) − 3(x + 5) = (x + 5)(x − 3)
- Zeroes: x + 5 = 0 gives −5; x − 3 = 0 gives 3.
- Here a = 1, b = 2, c = −15.
- Sum: −5 + 3 = −2, and −b/a = −2/1 = −2. Verified.
- Product: (−5) × 3 = −15, and c/a = −15/1 = −15. Verified.
The zeroes are −5 and 3, and both relationships hold.
Example 2: when a is not 1
Find the zeroes of 5x² + 13x − 6 and verify the relationship between the zeroes and the coefficients.
- ac = 5 × (−6) = −30. Two numbers with product −30 and sum 13 are 15 and −2.
- 5x² + 13x − 6 = 5x² + 15x − 2x − 6
- = 5x(x + 3) − 2(x + 3) = (5x − 2)(x + 3)
- Zeroes: 2/5 and −3.
- Sum: 2/5 + (−3) = −13/5, and −b/a = −13/5. Verified.
- Product: (2/5) × (−3) = −6/5, and c/a = −6/5. Verified.
The zeroes are 2/5 and −3, and both relationships hold.
Example 3: use the rules without finding the zeroes
If α and β are the zeroes of x² − 5x + 4, find (i) α² + β² and (ii) 1/α + 1/β − 2αβ.
- α + β = −b/a = 5 and αβ = c/a = 4.
- (i) α² + β² = (α + β)² − 2αβ = 25 − 8 = 17
- (ii) 1/α + 1/β = (α + β)/(αβ) = 5/4
- So 1/α + 1/β − 2αβ = 5/4 − 8 = 5/4 − 32/4 = −27/4
- Check with the actual zeroes 1 and 4: 1 + 16 = 17, and 1 + 1/4 − 8 = −27/4.
(i) α² + β² = 17 (ii) 1/α + 1/β − 2αβ = −27/4
Useful identities
- α² + β² = (α + β)² − 2αβ
- 1/α + 1/β = (α + β)/(αβ)
- (α − β)² = (α + β)² − 4αβ
- α/β + β/α = (α² + β²)/(αβ)
Sometimes only the total and the product of two lengths are known. A plot in a Gujarat village has length + breadth = 25 m and area 150 m². Its sides, 10 m and 15 m, are the zeroes of x² − 25x + 150.
1 marks: “One zero of x² − 7x + k is 2. Find the other zero.” Sum of zeroes = 7, so the other zero is 7 − 2 = 5 (and k = 2 × 5 = 10).
2 marks: “Find the zeroes of 4x² − 9 and verify the relationship between the zeroes and the coefficients.” 4x² − 9 = (2x − 3)(2x + 3), zeroes 3/2 and −3/2. Sum 0 = −0/4; product −9/4 = c/a.
3 marks: “If α and β are the zeroes of 2x² − 6x + 3, find α² + β².” α + β = 3 and αβ = 3/2. α² + β² = (α + β)² − 2αβ = 9 − 3 = 6.
Sum: minus b over a. Product: c over a. Only the sum wears the minus sign.