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Oxbridge_ Properties of polynomial roots.pptx

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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

				
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posted:2/7/2014
language:English
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Lingjuan Ma Lingjuan Ma
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