Oxbridge: Properties of polynomial roots

Solve each of the following quadratic equations
a)
b)
c)
d)
x2 + 7x + 12 = 0
x2 – 5x + 6 = 0
x2 + x – 20 = 0
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

Multiplying out
(x - α)(x - β)
then
(x - α)(x - β) = 0
= 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
 b  b 2  4ac

2a
 b  b 2  4ac

2a
 b  b 2  4ac  b  b 2  4ac
2b
b
  



2a
2a
2a
a
  b  b 2  4ac   b  b 2  4ac  b 2  (b 2  4ac) 4ac c


  
 2
2



2
a
2
a
4
a
4
a
a



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
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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

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
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
2 = 9/2
α + β + γ = -b/a
a
+
ar
+
ar
(1)

αβ + αγ + βγ = c/a  a2r + a2r2 + a2r3 = - 27/2
(2)
3r3 = -27
αβγ = -d/a
a
(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
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
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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
