Further Pure 1
Lesson 9 –
Identities and roots of equations
Identities
 x3 – y3
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
(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
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Equating Co-efficient
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 Sometimes you will be given an identity with
unknown constants on one side, such as
3x2 + 11x + 18 (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
Equating Coefficient

+ 11x + 18  Ax
+ 11x + 18  Ax
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3x2 + 11x + 18 (Ax – B)(x+5) + C
 If you multiply out the right hand side you get
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2 – 5Ax – Bx – 5B + C
3x2
2 – (5A – B)x – 5B + C
3x2
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 + 18 (3x – 4)(x+5) + 38

Substituting in Values
 3x2 + 11x + 18
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 (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
32 = 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.
Substituting in Values
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 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
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zero.
So C = 75 - 55 + 18 = 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 – 24 + 38
18 = 6A
A=3
Note Identities are not always written using ``. Eg sin2θ + cos2θ = 1
Now do Ex 4A pg 100
Properties of the roots of
polynomial equations
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 In this chapter the variable x is replaced with a z to
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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)
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
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Task
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 Use the quadratic formula to prove the results from
the previous slide.
-b/a = α + β
 b  b 2  4ac
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2a
c/a = αβ
 b  b 2  4ac
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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
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  
 2
2
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2
a
2
a
4
a
4
a
a
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Properties of the roots of
polynomial equations
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 Find a quadratic equation with roots 2 & -5
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-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
Taking a = 1 gives us the easiest solution.
If b and c are fractions you might like to pick an
appropriate value for a.
Properties of the roots of
polynomial equations
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 The roots of the equation 3z2 – 10z – 8 = 0 are α & β
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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
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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
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 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
 Now do Ex 4B pg 104