Further Pure 1

1. Further Pure 1 Lesson 10 – Roots of Equations

2. Properties of the roots of cubic equations • Cubic equations have roots α, β, γ (gamma) • az3 + bz2 + cz + d = 0 a(z – α)(z – β)(z – γ) = 0 a = 0 • This gives the identity az3 + bz2 + cz + d = a(z - α)(z - β)(z – γ) • Multiplying out az3 + bz2 + cz + d = a(z – α)(z – β)(z – γ) = a(z2 – αz – βz + αβ)(z – γ) = az3 – a(α +β +γ)z2 + a(αβ + αγ + βγ)z - aαβγ

3. Properties of the roots of cubic equations Equating coefficients • -a(α +β +γ)= b α +β +γ = -b/a • a(αβ + αγ + βγ) = c αβ + αγ + βγ = c/a • -aαβγ = d αβγ = -d/a • Can you notice a pattern?

4. Properties of the roots of quartic equations • Quartic equations have roots α, β, γ, δ (delta) • az4 + bz3 + cz2 + dz + e = 0 a(z – α)(z – β)(z – γ)(z – δ) = 0 a = 0 • This gives the identity az4 + bz3 + cz2 + dz + e = a(z - α)(z - β)(z – γ)(z – δ) • Multiplying out (try this yourself) az4 + bz3 + cz2 + dz + e = a(z – α)(z – β)(z – γ)(z – δ) = a(z2 – αz – βz + αβ)(z2 – γz – δz + γδ)

5. Properties of the roots of quartic equations • = z4 – αz3 – βz3 – γz3 – δz3 + αβz2 + αγz2 + βγz2 + αδz2 + βδz2 + γδz2 – αβγz –αβδz – αγδz – βγδz + αβγδ • = z4 – (α+ β+ γ+ δ)z3 + (αβ+ αγ+ βγ+ αδ+ βδ+ γδ)z2 – (αβγ +αβδ + αγδ + βγδ)z + αβγδ

6. Properties of the roots of quartic equations • Remember the a • = a[z4 – (α+ β+ γ+ δ)z3 + (αβ+ αγ+ βγ+ αδ+ βδ+ γδ)z2 – (αβγ +αβδ + αγδ + βγδ)z + αβγδ] • = az4 – a(α+ β+ γ+ δ)z3 + a(αβ+ αγ+ βγ+ αδ+ βδ+ γδ)z2 – a(αβγ +αβδ + αγδ + βγδ)z + aαβγδ • Equating coefficients • -a(α+ β+ γ+ δ) = b α+ β+ γ+ δ = -b/a = Σα • a(αβ+ αγ+ βγ+ αδ+ βδ+ γδ) = c αβ+ αγ+ βγ+ αδ+ βδ+ γδ = c/a = Σαβ • -a(αβγ +αβδ + αγδ + βγδ) = d αβγ +αβδ + αγδ + βγδ = -d/a = Σαβγ • aαβγδ = e αβγδ = e/a

7. Example 1 • The roots of the equation 2z3 – 9z2 – 27z + 54 = 0 form a geometric progression. • Find the values of the roots. • Remember that an geometric series goes a, ar, ar2, ……….., ar(n-1) • So from this we get α= a, β = ar, γ = ar2 α +β +γ = -b/aa + ar + ar2 = 9/2 (1) αβ + αγ + βγ = c/a a2r + a2r2 + a2r3 =-27/2(2) αβγ = -d/a a3r3 = -27 (3) • We can now solve these simultaneous equations.

8. Example 1 • Starting with the product of the roots equation (3). a3r3 = -27 (ar)3 = -27 ar= -3 • Now plug this into equation (1) a + ar + ar2 = 9/2 (-3/r) + -3 + (-3/r)r2 = 9/2 (-3/r) + -15/2 + -3r= 0 (-9/2) -6 -15r – 6r2 = 0 (×2r) 2r2 + 5r + 2 = 0 (÷-3) (2r + 1)(r + 2) = 0 r = -0.5 & -2 • This gives us the arithmetic series 6, -3, 1.5 or 1.5, -3, 6

9. 2z3 – 9z2 – 27z + 54 = 0 This time because we know that we are going to use the product of the roots we could have the first 3 terms of the series as a/r, a, ar So from this we get α = a/r, β = a, γ = ar α + β + γ = -b/a a/r + a + ar = 9/2 (1) We have ignored equation 2 because it did not help last time. αβγ = -d/a a3 = -27 (3) We can now solve these simultaneous equations. Example 1 – Alternative Algebra

10. Example 1 – Alternative Algebra • Starting with the product of the roots equation (3). a3 = -27 a= -3 • Now plug this into equation (1) a/r + a + ar = 9/2 -3/r + -3 + -3r = 9/2 (-3/r) + -15/2 + -3r= 0 (-9/2) -6 -15r – 6r2 = 0 (×2r) 2r2 + 5r + 2 = 0 (÷-3) (2r + 1)(r + 2) = 0 r = -0.5 & -2 • This gives us the arithmetic series 6, -3, 1.5 or 1.5, -3, 6

11. Example 2 • The roots of the quartic equation 4z4 + pz3 + qz2 - z + 3 = 0 are α, -α, α + λ, α – λ where α & λ are real numbers. • i) Express p & q in terms of α & λ. • α+ β+ γ+ δ = -b/a • α + (-α) + (α + λ) + (α – λ) = -p/4 2α = -p/4 p = -8α • αβ+ αγ+ αδ+ βγ+ βδ+ γδ = c/a (α)(-α) + α(α + λ) + α(α - λ) + (-α)(α + λ) + (-α)(α - λ) + (α + λ)(α – λ) = q/4 -α2 + α2 + αλ + α2 – αλ – α2 – αλ – α2 + αλ + α2 – λ2 = q/4 – λ2 = q/4 q = -4λ2

12. Properties of the roots of quintic equations • This is only extension but what would be the properties of the roots of a quintic equation? • az5 + bz4 + cz3 + dz2 + ez + f = 0 • The sum of the roots = -b/a • The sum of the product of roots in pairs = c/a • The sum of the product of roots in threes = -d/a • The sum of the product of roots in fours = e/a • The product of the roots = -f/a • Now do Ex 4c pg 110, Ex 4d pg 113