Oxbridge: Properties of polynomial roots

1. Oxbridge: Properties of polynomial roots • Solve each of the following quadratic equations a) x2 + 7x + 12 = 0 b) x2 – 5x + 6 = 0 c) x2 + x – 20 = 0 d) 2x2 – 5x – 3 = 0 • Write down the sum of the roots and the product of the roots. • For a quadratic equation we use alpha (α) & beta (β) to denote these roots. • Can you see any relationships with the sums and products ?

2. Properties of the roots of polynomial equations • Given ax2+ bx+ c = 0 and since a is non-zero, then x2+ (b/a) x + (c/a) = 0 (1) If the roots are α and βthen (x - α)(x - β) = 0 • Multiplying out (x - α)(x - β) = x2– (α + β)x + αβ = 0 (2) Equating coefficients using (1) and (2) we see that α+ β = -b/a αβ = c/a

3. Task • Use the quadratic formula to prove the results from the previous slide. -b/a = α + β c/a = αβ

4. Properties of the roots of polynomial equations • Find a quadratic equation with roots 2 & -5 -b/a = α + β c/a = αβ -b/a = 2 + -5 c/a = -5 × 2 -b/a = -3 c/a = -10 • Taking a = 1 gives b = 3 & c = -10 • So z2 + 3z – 10 = 0 • Note: There are infinitely many solutions to this problem. • Taking a = 2 would lead to the equation 2z2 + 6z – 20 = 0 • But taking a = 1 gives us the easiest solution.

5. Properties of the roots of polynomial equations • The roots of the equation 3z2 – 10z – 8 = 0 are α & β 1 – Find the values of α + β and αβ. α + β = -b/a = 10/3 αβ = c/a = -8/3 2 – Find the quadratic equation with roots 3α and 3β. • The sum of the new roots is 3α + 3β = 3(α + β) = 3 × 10/3 = 10 • The product of the new roots is 9αβ = -24 • From this we get that 10 = -b/a & -24 = c/a • Taking a = 1 gives b = -10 & c = -24 • So the equation is z2 – 10z – 24 = 0

6. Properties of the roots of polynomial equations 3 – Find the quadratic equation with roots α + 2 and β + 2 • The sum of the new roots is α + β + 4 = 10/3 + 4 = 22/3 • The product of the new roots is (α + 2)(β + 2) = αβ + 2α + 2β + 4 = αβ + 2(α + β) + 4 = -8/3 + 2(10/3) + 4 = 8 • So 22/3 = -b/a & 8 = c/a • To get rid of the fraction let a = 3, so b = -22 & c = 24 • The equation is 3z2 – 22z + 24 = 0

7. Properties of the roots of polynomial equations: Transformation Method Alternative method to find the quadratic equation with roots α+ 2 and β + 2 • The equation was 3z2 – 10z – 8 = 0 satisfied by α and β • So 3 α 2– 10 α– 8 = 0 (1) • But if we let x = α+ 2 then α = x – 2 • So if we sub this in (1) we get 3(x-2)2– 10(x-2) – 8 = 0 • Or 3 (x2 -4x+4) – 10(x-2) – 8 = 0 • That is 3x2– 22x + 24 = 0 just as we had before

8. Properties of the roots of polynomial equations • The roots of the equation z2 – 7z + 15 = 0 are α and β. • Find the quadratic equation with roots α2 and β2 α + β = 7 & αβ = 15 (α + β)2 = 49 & α2β2 = 225 α2 + 2αβ + β2 = 49 α2 + 30 + β2 = 49 α2 + β2 = 19 • So the equation is z2 – 19z + 225 = 0

9. Properties of the roots of cubic equations • Cubic equations have roots α, β, γ (gamma) (x – α)(x – β)(x – γ) = 0 • x3+ (b/a) x2+ (c/a) x + (d/a) = 0 where a is non-zero • This gives the identity x3+ (b/a) x2+ (c/a) x + (d/a) = (x - α)(x - β)(x – γ) = 0 • We then proceed to multiply out in the same way as before :

10. Properties of the roots of cubic equations Equating coefficients as before: α+β +γ = -b/a αβ+ αγ + βγ = c/a αβγ= -d/a

11. Properties of the roots of quartic equations • Quartic equations have roots α, β, γ, δ (delta) • So z4+ (b/a) z3+ (c/a) z2+ (d/a) z + (e/a) = 0 since a is non-zero And (x – α)(x – β)(x – γ)(x – δ) = 0 • Equating coefficients α+ β+ γ+ δ = -b/a = Σα αβ+ αγ+ βγ+ αδ+ βδ+ γδ = c/a = Σαβ αβγ +αβδ + αγδ + βγδ = -d/a = Σαβγ αβγδ = e/a

12. 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.

13. 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

14. Example 1 – Alternative Algebra • 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.

15. 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

16. 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

17. Properties of the roots of quinticequations: Extension exercise • 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