1. Lesson 9 – Identities and roots of equations
Further Pure 1
Lesson 9 –
Identities and roots of equations

2. Identities x3 – y3 (x-y)(x2 + xy + y2) is an example of an identity.
The 3 lined equals sign means identically equal to.
This means that both sides of the equation will always be equal whatever the values of x and y are.
Here are some more examples of identities
2(x+3) 2x + 6
a2 – b2 (a+b)(a-b)
In an identity all possible values of the variable(s) will satisfy the identity.
With an equation only certain values satisfy the equation.
Example : x2 – 5x + 6 = 0, only has two values that work, 2 & 3.
What would happen if you tried to solve the identity
(x+3)2 x2 + 6x + 9

3. Equating Co-efficient
Sometimes you will be given an identity with unknown constants on one side, such as 3x2 + 11x (Ax – B)(x+5) + C
There are two methods to finding out the values of these constants.
Method 1 – Equating Coefficients
Method 2 – Substituting in values

4. Equating Coefficient 3x2 + 11x + 18 (Ax – B)(x+5) + C
If you multiply out the right hand side you get
3x2 + 11x Ax2 – 5Ax – Bx – 5B + C
3x2 + 11x Ax2 – (5A – B)x – 5B + C
As both expressions are identically equal then we can equate the Coefficients.
The terms in front of the x2 are equal so: A = 3
The terms in front of the x will be equal so: 5A – B = 11
15 – B = 11
B = 4
Finally the constants at the ends of both equations must be equal:
-5B + C = 18
-20 + C = 18
C = 38
Now 3x2 + 11x (3x – 4)(x+5) + 38

5. Substituting in Values
3x2 + 11x (Ax – B)(x+5) + C
Since the expressions are equal we can plug in any values we like for x to form equations in A,B and C that we can solve.
Example if x = 1, then 3(1)2 + 11(1) + 18 = (A(1) – B)(1 + 5) + C = 6A - 6B + C
The problem is though that you now have one equation with 3 unknowns.
When you pick your value of x to plug in try to pick values that will cancel out some of the unknowns.

6. Substituting in Values
3x2 + 11x + 18 = (Ax – B)(x+5) + C
If you pick x = -5 then all of the expression (Ax – B)(x+5) is equal to zero.
So C = = 38
Now if x = 0 then the A term will go and we are left with B and C, however we already know what C is.
So 3(0) + 11(0) + 18 = (A(0) – B)(0 + 5) + C
18 = -5B + 38
-20 = -5B
B = 4
Finally we could use the example of x = 1 on the previous slide because we now know B & C.
If x = 1, 32 = 6A – 6B + C
32 = 6A –
18 = 6A
A = 3
Note Identities are not always written using ` `. Eg sin2θ + cos2θ = 1
Now do Ex 4A pg 100

7. Properties of the roots of polynomial equations
In this chapter the variable x is replaced with a z to emphasize that the results could be complex or real.
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.
Roots of polynomial equations are usually denoted by Greek letters.
For a quadratic equation we use alpha (α) & beta (β)

8. Properties of the roots of polynomial equations
az2 + bz + c = 0
a(z - α)(z - β) = 0 a = 0
This gives the identity
az2 + bz + c = a(z - α)(z - β)
Multiplying out
az2 + bz + c = a(z2 – αz – βz + αβ)
= az2 – aαz – aβz + aαβ
= az2 – az(α + β) + aαβ
Equating coefficients
b = – a(α + β) c = aαβ
-b/a = α + β c/a = αβ

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

10. Properties of the roots of polynomial equations
Find a quadratic equation with roots 2 & -5
-b/a = α + β c/a = αβ
-b/a = 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
Taking a = 1 gives us the easiest solution.
If b and c are fractions you might like to pick an appropriate value for a.

11. 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 α 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

12. 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 = αβ + 2(α + β) = -8/3 + 2(10/3) = 8
So 22/3 = -b/a & = c/a
To get rid of the fraction let a = 3, so b = -22 & c = 24
The equation is 3z2 – 22z + 24 = 0

13. 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 = & αβ = 15
(α + β)2 = 49 & α2β2 = 225
α2 + 2αβ + β2 = 49
α β2 = 49
α2 + β2 = 19
So the equation is z2 – 19z = 0
Now do Ex 4B pg 104