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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 ? 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 β then (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 Task ¥ Use the quadratic formula to prove the results from the previous slide. -b/a = α + β c/a = αβ 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. 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 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 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 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 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 : Properties of the roots of cubic equations Equating coefficients as before: α + β + γ = -b/a αβ + αγ + βγ = c/a αβγ = -d/a 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 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/a a + 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. 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 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. 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 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 Properties of the roots of quintic equations: 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