 Solve each of the following quadratic equations a)x 2 + 7x + 12 = 0 b)x 2 – 5x + 6 = 0 c)x 2 + x – 20 = 0 d)2x 2 – 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 ?

 Given ax 2 + bx + c = 0 and since a is non-zero, then x 2 + (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 ) = x 2 – ( α + β )x + αβ = 0 (2) Equating coefficients using (1) and (2) we see that α + β = -b/a αβ = c/a

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

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

 The roots of the equation 3z 2 – 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 3 α and 3 β.  The sum of the new roots is 3 α + 3 β = 3( α + β ) = 3 × 10/3 = 10  The product of the new roots is 9 αβ = -24  From this we get that 10 = -b/a & -24 = c/a  Taking a = 1 gives b = -10 & c = -24  So the equation is z 2 – 10z – 24 = 0

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 + 2 α + 2 β + 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 3z 2 – 22z + 24 = 0

Alternative method to find the quadratic equation with roots α + 2 and β + 2  The equation was 3z 2 – 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 (x 2 -4x+4) – 10(x-2) – 8 = 0  That is 3x 2 – 22x + 24 = 0 just as we had before

 The roots of the equation z 2 – 7z + 15 = 0 are α and β.  Find the quadratic equation with roots α 2 and β 2 α + β = 7 & αβ = 15 ( α + β ) 2 = 49& α 2 β 2 = 225 α αβ + β 2 = 49 α β 2 = 49 α 2 + β 2 = 19  So the equation is z 2 – 19z = 0

 Cubic equations have roots α, β, γ (gamma) (x – α )(x – β )(x – γ ) = 0  x 3 + (b/a) x 2 + (c/a) x + (d/a) = 0 where a is non-zero  This gives the identity x 3 + (b/a) x 2 + (c/a) x + (d/a) = (x - α )(x - β )(x – γ ) = 0  We then proceed to multiply out in the same way as before :

Equating coefficients as before: α + β + γ = -b/a αβ + αγ + βγ = c/a αβγ = -d/a

 Quartic equations have roots α, β, γ, δ (delta)  So z 4 + (b/a) z 3 + (c/a) z 2 + (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

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

 Starting with the product of the roots equation (3). a 3 r 3 = -27 (ar) 3 = -27 ar = -3  Now plug this into equation (1) a + ar + ar 2 = 9/2 (-3/r) (-3/r)r 2 = 9/2 (-3/r) + -15/2 + -3r = 0(-9/2) r – 6r 2 = 0(×2r) 2r 2 + 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

 2z 3 – 9z 2 – 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/aa 3 = -27(3)  We can now solve these simultaneous equations.

 Starting with the product of the roots equation (3). a 3 = -27 a = -3  Now plug this into equation (1) a/r + a + ar = 9/2 -3/r r = 9/2 (-3/r) + -15/2 + -3r = 0(-9/2) r – 6r 2 = 0(×2r) 2r 2 + 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

 The roots of the quartic equation 4z 4 + pz 3 + qz 2 - 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 2 α = -p/4 p = -8 α  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 - α 2 + α 2 + αλ + α 2 – αλ – α 2 – αλ – α 2 + αλ + α 2 – λ 2 = q/4 – λ 2 = q/4 q = -4 λ 2

 This is only extension but what would be the properties of the roots of a quintic equation?  az 5 + bz 4 + cz 3 + dz 2 + 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