MEI FP1 Study Resources Algebra 1-2

MEI Further Pure Mathematics 1 Algebra Section 1: Identities Study Plan Background In this short section, you will learn the difference between an equation (true for one or more particular values of the variable(s)) and an identity (true for all values of the variable(s)). You will see how you can find unknown constants in an identity.

Detailed work plan 1. Read pages 97 – 99 on identities. There are some additional notes and two further examples in the Notes and Examples. 2. Exercise 4A Attempt Questions 1 to 6, and if you have time try the starred questions 7 and 8 as well.

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MEI Further Pure Mathematics 1 Algebra Section 1: Identities Notes and Examples These notes contain subsections on  The difference between an identity and an equation  Finding unknown constants in an identity

The difference between an identity and an equation The essential difference between an equation and an identity is that an identity is true for all values of the variable(s) concerned (provided the expression involved is defined for those values of the variable(s)). For example x( x  1)  x 2  x

(2 x  y )( x  y )  2 x 2  xy  y 2

2x2  2x , x

x0

is an identity because no matter what value of x you substitute in, the left hand side will equal the right hand side. is an identity because no matter what values of x and y you substitute in, the left hand side will equal the right hand side. is an identity because if you substitute in any value of x other than 0, the left hand side will equal the right hand side. When x = 0, the left hand side is not defined.

However,

x( x  1)  x

is NOT an identity.

To show that something is not an identity you only need to find one instance where it is not true. For example, if x = 1 in the above, the left hand side is 0 but the right hand side is 1. The only values of x for which the above is true are 0 and 2. It is an equation with solutions x = 0 and x = 2. With an equation, the objective is usually to solve it to find an unknown quantity, whereas an identity is usually used to prove a result or to write something in a different form. If you try to “solve” an identity, you will just end up with an expression such as 0 = 0.

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MEI FP1 Algebra Section 1 Notes and Examples Finding unknown constants in an identity In this section you need to be able to find unknown constants in an identity. An example is shown in the text book, another two are shown below. In Example 1 the method of equating coefficients is used. This is similar to the technique of equating real and imaginary parts used when working with complex numbers. The principle behind it is that since the identity is true for all possible values of x, then the x² term must be the same on both sides, the term in x must be the same on both sides, and the constant term must be the same on both sides. This method is also used in Solution 1 of Example 4.1 in the FP1 textbook.

Example 1 Find the values of the constants A, B and C in the identity 2 x 2  3x  2  A( x  B) 2  C Solution Multiplying out the right hand side gives 2 x 2  3x  2  A( x 2  2Bx  B 2 )  C

 Ax 2  2 ABx  AB 2  C Equating coefficients of x²: 2=A Equating coefficients of x: -3 = 2AB -3 = 4B B   34 Equating constant terms:

2  AB 2  C  2    34   C 2

 98  C C A  2, B   34 , C 

7 8

7 8

The example above is an alternative method of “completing the square”, which you have probably already met in C1. In the next example, the method of equating coefficients could be used, but it would result in quite a complicated set of simultaneous equations to solve. The method used instead is based on the idea that since an identity is true for all values of the variable, you can substitute any value you like and you will get a statement that is true. This method is also used in Solution 2 of Example 4.1 in the FP1 textbook. By choosing suitable values for the variable, you may be able to make the calculations very simple.

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MEI FP1 Algebra Section 1 Notes and Examples Example 2 Find the values of the constants P, Q and R in the identity x 2  1  P( x  1)( x  2)  Q( x  2)( x  3)  R( x  1) Solution Substitute x = -1

2  Q(3)(2) 2  6Q

Substitute x = 2

Q   13 5  3R

Substitute x = -3

R  53 10  P (2)(5)  2 R

10  10 P  2 R  10 P  2  53  10 P  103 1  P  13 P P  43 , Q   13 , R 

4 3

5 3

Notice how choosing x = -1 and x = 2 allows you to calculate the values of Q and R very easily. If you had chosen different values for x, you would have obtained three equations involving P, Q and R, which would have involved just as much work to solve as using the method of equating coefficients.

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MEI Further Pure Mathematics 1 Algebra Section 1: Identities Crucial points 1. Make sure that you understand the difference between an equation and an identity See textbook, Notes and Examples, and Glossary. 2. Make sure you understand both methods You should be able to use both the method of equating coefficients and the substitution method effectively. In particular, make sure that you choose sensible values for the variable when you use the substitution method.

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Further Pure Mathematics 1 Algebra Section 1: Identities Exercise 1. Find values for a and b if (i) x 2 + 6 x + 3 ≡ ( x + a) 2 + b (ii) −4 x 2 − 9 x − 2 ≡ a( x + 1) 2 + b( x − 2) 2. Find values for a and b if (i) ( x + 3)( x − 2)( x 2 + ax + b) ≡ x 4 + x3 − 7 x 2 − x + 6 (ii) ( x + 1)( x − 4)(ax + b) ≡ 2 x 3 − 5 x 2 − 11x − 4 3. Find values for A and B in the identity 4 A B . ≡ + x( x + 3) x ( x + 3)

4. Find values for A,B and C in the identity 2x A Bx + C . ≡ + 2 2 ( x + 3)( x − 1) x − 1 x + 3

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Further Pure Mathematics 1 Algebra Section 1: Identities Solutions to Exercise 1. (i) x 2 + 6 x + 3 ≡ ( x + a )2 + b ≡ x 2 + 2ax + a 2 + b Equating coefficients of x: 6 = 2a a =3 Equating constant terms: 3 = a 2 + b 3 = 9 +b

b = −6 (ii) −4 x 2 − 9 x − 2 ≡ a( x + 1)2 + b( x − 2) −16 − 18 − 2 = 9a Putting x = 2: −36 = 9a

Putting x = -1:

a = −4

−4 + 9 − 2 = −3b 3 = −3b

b = −1 2. (i) ( x + 3)( x − 2)( x 2 + ax + b ) ≡ x 4 + x 3 − 7 x 2 − x + 6 Putting x = 0: −6b = 6 Putting x = 1:

b = −1 4 × −1(1 + a + b ) = 1 + 1 − 7 − 1 + 6 −4(1 + a + b ) = 0

a + b = −1 a =0 (ii) ( x + 1)( x − 4)(ax + b ) ≡ 2 x 3 − 5 x 2 − 11 x − 4 Putting x = 0: 1 × −4b = −4 Putting x = 1:

b =1 2 × −3(a + b ) = 2 − 5 − 11 − 4 −6(a + b ) = −18

a +b = 3 a =2

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Further Pure Mathematics 1 3.

4 A B ≡ + x( x + 3) x ( x + 3) 4 ≡ A( x + 3) + Bx Putting x = 0: 4 = 3 A Putting x = -3:

A=

4 3

4 = −3B

B = − 43 4.

2x A Bx + C ≡ + 2 ( x + 3)( x − 1) x − 1 x + 3 2 x = A( x 2 + 3) + (Bx + C )( x − 1) Putting x = 1: 2 = 4 A 2

Putting x = 0:

A=

1 2

0 = 3A − C

C = 23 Equating coefficients of x²: 0 = A + B B = − 21

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Further Pure Mathematics 1 Algebra Section 1: Identities Multiple Choice Test 1) One of the following “identities” is not actually an identity. Which one? (b) 2( x + 4 y ) ≡ 2 x + 8 y

(a) x 3 − 1 ≡ (1 − x − x 2 )( x − 1) (c) y 2 − 1 ≡ ( y − 1)( y + 1) (e) I don’t know

(d) ( x − 1) 2 ≡ (1 − x) 2

2) Which of the following are identities? (i) (2 x + 3) 2 = 2 x 2 + 6 x + 9 (ii) ( x − 1)3 = x 3 − 3x 2 + 3 x − 1 3x + 2 (iii) = 2x +1 x −1 (iv) (2 x − 1)( x 2 − 1) = 2 x 3 − x 2 − 2 x + 1 (v)

x2 − 1 =

2 x3 − x 2 − 2 x + 1 , 2x −1

(a) (i), (iii) and (v) (c) (i) and (ii) (e) I don’t know

x≠

1 2

(b) (ii) and (iv) (d) (ii), (iv) and (v)

Questions 3 and 4 are about the identity Ax 2 + 3x + B ≡ ( x + 1)( x + 2) + x 2 3) The value of the constant A is (a) 2 (c) 0 (e) I don’t know

(b) 1 (d) -1

4) The value of the constant B is (a) 2 (c) 3 (e) I don’t know

(b) 0 (d) 1

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Further Pure Mathematics 1 Questions 5, 6 and 7 are about the identity x3 + 2 x 2 − 3x + 1 ≡ ( x − 1)( x 2 + Ax + B) + C

5) The value of the constant A is (a) 3 (c) -1 (e) I don’t know

(b) 1 (d) 2

6) The value of the constant B is (a) -3 (c) -1 (e) I don’t know

(b) 0 (d) 1

7) The value of the constant C is (a) 2 (c) 0 (e) I don’t know

(b) -1 (d) 1

Questions 8, 9 and 10 are about the identity x 2 − x + 3 ≡ L( x − 1) 2 + M ( x − 1)( x + 2) + N ( x + 2) 2 8) The value of the constant L is (a) 1

(b) 13 (d) -1

(c) − 13 (e) I don’t know

9) The value of the constant N is (b) − 13 (d) -1

(a) 13 (c) 1 (e) I don’t know

10) The value of the constant M is (a) − 13 (c) -1 (e) I don’t know

(b) 1 (d)

1 3

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MEI Further Pure Mathematics 1 Algebra Section 2: Roots and coefficients of polynomial equations Study Plan Background In this section you will look at the relationships between the coefficients of a polynomial and its roots. These relationships do not help you to solve polynomials, but they do give some other useful techniques, such as forming an equation whose roots you already know, or forming a new equation whose roots are related to the roots of a given equation.

Detailed work plan 1. Read the subsection 100 – 104 on quadratic equations, and do the Activities. There are some additional notes and a full solution to Activity 4.3 in the Notes and Examples. 2. Exercise 4B Attempt questions 1-5. If you have time try the “starred” question 6 and the enrichment questions 7 and 8. 3. For further practice, try the Roots of quadratic equations activity. 4. Read pages 105 – 109 on cubic equations. There is an additional example and a full solution to Activity 4.6 in the Notes and Examples. 5. Exercise 4C Attempt questions 1, 2(i),(iii) and(vi), 3(i),(iii), 4, 5, 6, 9. If you have time you may like to try the “starred” questions 10 and 11 and the enrichment questions 12 and 13. 6. Read pages 111 – 112 on quartic equations and also look at the enrichment Discussion point on page 113. There is a full proof of Activity 4.7 and two further examples in the Notes and Examples. 7. Exercise 4D Attempt questions 1, 2(i) and (iii), 3, 4 and 5. 8. For some extra practice, try the interactive questions Sums and products of roots and Polynomials with related roots.

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MEI Further Pure Mathematics 1 Algebra Section 2: Roots and coefficients of polynomial equations Notes and Examples These notes contain subsections on  Quadratic equations  Cubic equations  Quartic equations

Quadratic equations If you want to find a quadratic equation whose roots are, say, 2 and -5, a simple way to do this is to write the equation in the form (x – 2)(x + 5) = 0 and then multiply out to give the equation x² + 3x – 10 = 0. However, if the roots are complex, or if you want to find a cubic or quartic equation, this method is time consuming and it is easy to make mistakes in the algebra. Finding relationships between the roots and coefficients of polynomial equations gives an alternative approach which is often simpler. In Activity 4.2 you should notice that for a quadratic equation of the form ax2  bx  c  0 b the sum of the roots is given by  a c and the product of the roots is given by . a In the next section of the textbook, you will see the proof that this is the case for all quadratic equations. This is done by writing the equation in the form a( z   )( z   ) , multiplying out and equating coefficients. You may have done something similar in the Complex Numbers chapter, for the slightly simpler case where a is always 1 (Exercise 2B, Question 9). Activity 4.3 gives an alternative proof of these relationships for quadratic equations, using the quadratic formula. The full solution to this Activity is given below.

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MEI FP1 Algebra Section 2 Notes and Examples Solution to Activity 4.3

b  b2  4ac  , 2a

b  b 2  4ac  2a

b  b 2  4ac  b  b 2  4ac 2a 2b  2a b  a

  

b     



b 2  4ac b  b 2  4ac

b 2   b 2  4ac 



4a 2 using

(x + y)(x – y) = x² –

y²

4a 2

4ac 4a 2 c  a 

The other application of these relationships which you meet in this section is to find an equation whose roots are related to the roots of the original equation. The syllabus states that you are expected to be able to solve problems like this where the new roots are related to the original roots by a linear relationships, e.g. the new roots might be 2  1, 2  1 . Some more complicated examples are covered in enrichment material – these may involve rather more complicated algebra. Example 4.3 shows the sort of question you will be expected to be able to do. Example 4.4 in the enrichment section shows a slightly harder question.

You can also try the Roots of quadratic equations activity.

Cubic equations The relationships between the roots and coefficients of a cubic equation b       a c       a d    a

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MEI FP1 Algebra Section 2 Notes and Examples can be proved using the same approach as for quadratic equations, by writing the equation in the form a( z   )( z   )( z   ) , multiplying out and equating coefficients. This is shown in the FP1 textbook, with some of the intermediate algebra left out – you can find the full working in the Answers section of the textbook.

Example 1 Find a cubic equation with roots 2, -1 and -3. Solution       2  1  3  2 b    2  b  2a a       (2  1)  (1 3)  (3  2)  2  3  6  5

c  5  c  5a a   2  1 3  6 d    6  d  6 a a Let a  1  b  2, c  5, d  6 The equation is z 3  2 z 2  5z  6  0 

Example 4.5 in the textbook shows how if you have additional information about the roots of a cubic equation, the relationship between the roots and coefficients can give a quick method of solving the equation. In the Quadratics section, you looked at finding a new equation whose roots are related to the roots of a given equation. Example 4.6 in the textbook shows a similar problem for a cubic equation. Two methods of solution are given. The first is a similar approach to the one used in the Quadratics section, but the second (the substitution method) is generally much simpler to use. As for the quadratics, the syllabus states that you are expected to be able to solve problems like this where the new roots are related to the original roots by a linear relationships, e.g. the new roots might be 2  1, 2  1, 2  1, so you should always be able to use the substitution method for these. The enrichment section shows some harder examples. In some cases the substitution method cannot be used (e.g. in Example 4.8). You can see that the algebra can be quite complicated. However, it is sometimes easier than it looks! Activity 4.6 asks you to prove some useful identities. The full solution to this Activity is given below.

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MEI FP1 Algebra Section 2 Notes and Examples Solution to Activity 4.6 (i)

           2

2

 (     )(     )   2         2         2   2   2   2  2(     )    2  2  (ii)

           2

2

             ( ) 2   2   2    2  ( ) 2   2   2    2  ( ) 2  ( ) 2  (  ) 2  ( ) 2  2( 2    2   2 )  ( ) 2  (  ) 2  ( ) 2  2 (     )   ( ) 2  2  

(iii)  

1 1 1            1

              

  

Quartic equations Proving the four relationships between the roots and coefficients of quartic equations is done in the same way as for quadratic and cubic equations, but of course the algebra gets progressively more complicated. The multiplying out is left for you to do in Activity 4.7 – the full solution to this is given below.

Solution to Activity 4.7 ( z   )( z   )( z   )( z   )  ( z 2   z   z   )( z 2   z   z   )  ( z 2  (   ) z   )( z 2  (   ) z   )

 z 4  (   ) z 3   z 2  (   ) z 3  (   )(   ) z 2   (   ) z   z 2  (   ) z    z 4  (       ) z 3  (  (   )(   )   ) z 2  ( (   )  (   ) ) z    z 4  (       ) z 3  (           ) z 2  (       )z  

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MEI FP1 Algebra Section 2 Notes and Examples az 4  bz 3  cz 2  dz  e  az 4  a(       ) z 3  a(           ) z 2  a(       ) z  a

      

b a

                      

c a

d a

e a

Be careful in particular when working out the sum of the product of the roots in pairs, to make sure that you include all six terms! The type of questions you will be asked on quartic equations are similar to those for cubic or quadratic, although of course they may involve a little more work! Here are two more examples.

Example 2 Find the quartic equation with roots 1, -2, 3 (repeated). Solution         1 2  3  3  5             (1 2)  (1 3)  (1 3)  (2  3)  (2  3)  (3  3)  2  3  3  6  6  9 1         (1 2  3)  (2  3  3)  (3  3 1)  (3 1 2)  6 18  9  6  21   1 2  3  3  18 b c d e   5,  1,   21,  18 a a a a Let a  1  b  5, c  1, d  21, e  18 The equation is z 4  5z 3  z 2  21z 18  0

Example 3 The roots of the quartic equation 2 z 4  z 3  3z 2  2 z  4  0 are α, β, γ, δ. Find the quartic equation with roots 2α+1, 2β+1, 2γ+1, 2δ+1.

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MEI FP1 Algebra Section 2 Notes and Examples Solution Use the substitution method. w 1 w  2z 1  z  2 2 z 4  z 3  3z 2  2 z  4  0

 w 1   w 1   w 1   w 1  2     3   2 40  2   2   2   2  2( w4  4w3  6w2  4w  1) ( w3  3w2  3w  1) 3( w2  2 w  1)    w 1 4  0 16 8 4 w4  4w3  6w2  4w  1  w3  3w2  3w  1  6( w2  2 w  1)  8( w  5)  0 4

3

2

w4  4w3  6w2  4w  1  w3  3w2  3w  1  6w2  12w  6  8w  40  0 w4  3w3  3w2  19w  46  0

You should now be able to see how the pattern of relationships between the roots and coefficients of polynomial equations develops. The enrichment section invites you to predict the relationships for quintic (degree 5) equations, and you could of course continue this to equations of higher degree. However, you will not be asked to deal with equations of any higher degree than quartic.

For practice on the work in this section, try the interactive questions Sums and products of roots and Polynomials with related roots.

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Further Pure Mathematics 1 Algebra Section 2: Roots and coefficients of polynomial equations Crucial points 1. Check your algebra The most common problem in this topic is mistakes with algebra. The algebra can be quite complicated, but if you can do this then it will help you in all areas of mathematics. 2. Make sure that you have learnt the relationships between roots and coefficients In particular, make sure that you remember the pattern of alternating signs in the relationships between roots and coefficients. 3. Be careful with the Σ notation In particular, make sure that you know how many terms are involved in each case: for example, for a cubic then ∑ αβ has three terms, but for a quartic

∑ αβ

has six terms.

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Further Pure Mathematics 1 Algebra Section 2: Roots and coefficients of polynomial equations Exercise 1. One root of 2 x 2 − kx + k = 0 is twice the other. Find k. You may assume that k ≠ 0. 2. The two roots of x 2 + (7 − p) x − p = 0 differ by 5. Find the possible values for p. 3. If α and β are the roots of ax2 + bx + c = 0 prove that (i) if β = 4α then 4b2 = 25ac (ii) if β = α + 1 then a2 = b2 – 4ac 4. If α and β are the roots of 2x2 – x – 4 = 0 find 2 (i) α 2 + β 2 (ii) (α − β ) (iii) α 3 + β 3 (iv)

1

α

+

1

β

(v)

1

α

2

+

1

β2

5. If p + q = 5 and p2 + q2 = 19 find the value of pq and hence write down a quadratic equation with roots p and q. 6. Given that –1 and 4 are two roots of x3 + 5x2 + ax + b = 0 find the third root and values for a and b. 7. If the roots of x3 + 5x2 + hx + k = 0 are α, 2α, and α + 3 find α, h and k. 8. Solve the equation 24x3 + 28x2 – 14x – 3 = 0 given that the roots are in a geometric progression. 9. The equation 6x3 + 11x2 + kx – 9 = 0 has roots α,

1

α

and β.

Find the value of k and solve the equation. 10. The roots of x3 − 2 x 2 − x + 2 = 0 are α, β, and γ. Find equations which have roots 1 1 1 (i) , ,

α β γ (ii) α − 3, β − 3, γ − 3

11. The roots of x4 + ax3 + bx2 + cx + d = 0 are α, β, γ, and δ. Given that α + β = γ + δ , show that a3 + 8c = 4ab.

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Further Pure Mathematics 1 Algebra Section 2: Roots and coefficients of polynomial equations Solutions to Exercise 1. Let the roots of the equation 2 x 2 − kx + k = 0 be α and 2α. Sum of roots:

α + 2α =

Product of roots: α × 2α = 4α = k 2

k

2 k

⇒α =

k

6

2

2

k 4 ⎛⎜ ⎞⎟ = k ⎝6⎠ 2 k = 9k k(k − 9) = 0 Since k ≠ 0 , the value of k must be 9. 2. Let the roots of the equation x 2 + ( 7 − p )x − p = 0 be α and α + 5. Sum of roots: α + α + 5 = −( 7 − p ) 2α + 5 = −7 + p 2α = p − 12 Product of roots: α(α + 5 ) = − p ⎛ p − 12 ⎞ ⎛ p − 12 + 5 ⎞ = − p ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ ( p − 12)( p − 12 + 10) = −4 p ( p − 12)( p − 2) = −4 p

p 2 − 14 p + 24 = −4 p p 2 − 10 p + 24 = 0 ( p − 4)( p − 6) = 0 p = 4 or p = 6. 3. (i) β = 4α Sum of roots:

α + 4α = −

b a

⇒α = −

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Further Pure Mathematics 1 α × 4α =

Product of roots:

c a

2

b ⎞ c 4 ⎜⎛ − ⎟ = a ⎝ 5a ⎠ 2 4b c = 2 25 a a 2 4b = 25 ac (ii) β = α + 1

α +α + 1 = −

Sum of roots:

b a

⇒ 2α = −

b −1 a

c a b 1 b 1 c ⎛− − ⎞⎛ − + 1 ⎞⎟ = ⎜ ⎟⎜ − ⎝ 2a 2 ⎠⎝ 2a 2 ⎠ a ⎛ b + 1 ⎞⎛ b − 1 ⎞ = c ⎜ ⎟⎜ ⎟ ⎝ 2a 2 ⎠ ⎝ 2a 2 ⎠ a b2 1 c − = 2 4a 4 a 2 2 b − a = 4ac α(α + 1) =

Product of roots:

a 2 = b 2 − 4ac 4. α + β =

1 2

αβ = −2 (i) α 2 + β 2 = (α + β ) − 2αβ 2

= ( 21 ) − 2 × −2 2

= 41 + 4 =

17 4

(ii) (α − β ) = (α + β ) − 4αβ 2

2

= ( 21 ) − 4 × −2 2

= 41 + 8 =

33 4

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Further Pure Mathematics 1 (iii) α 3 + β 3 = (α + β ) − 3α 2 β − 3αβ 2 3

= (α + β ) − 3αβ (α + β ) 3

= ( 21 ) − 3 × −2 × 21 3

= 81 + 3 = (iv)

1

α

+

1

β

=

25 8

α +β αβ 1 2

=

−2 = − 41 (v)

1

α2

+

1

β2

= = =

5. p + q = 5 ,

α2 + β2 α 2β 2

from (i)

17 4

( −2)2 17 16

p 2 + q 2 = 19

( p + q )2 = p 2 + q 2 + 2 pq 5 2 = 19 + 2 pq 6 = 2 pq pq = 3

b b ⇒5 =− ⇒ b = −5 a a a c c Product of roots = ⇒3= ⇒ c = 3a a a Putting a = 1 gives b = -5 and c = 3 A quadratic equation with roots p and q is x 2 − 5 x + 3 = 0 Sum of roots = −

6. Sum of roots = -5 −1 + 4 + α = −5

α = −8 The third root is -8.

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Further Pure Mathematics 1 ∑ αβ = a ( −1 × 4 ) + ( 4 × − 8 ) + ( − 8 × − 1 ) = a

−4 − 32 + 8 = a a = −28

αβγ = −b −1 × 4 × −8 = −b

b = −32 7.

∑ α = −5

α + 2α + α + 3 = −5 4α = −8 α = −2 The roots of the equation are -2, -4 and 1. ∑ αβ = h

( −2 × −4) + ( −4 × 1) + (1 × −2) = h 8 −4−2 =h

h=2

αβγ = −k −2 × −4 × 1 = −k

k = −8

8. Let the roots be

αβγ = 243 α × α × αr = r α 3 = 81 α=

1 2

α ,α ,α r r

1 8

28

∑ α = − 24

7 α + α + αr = − 6 r 1⎛1 7 ⎞ ⎜ + 1 +r ⎟ = − 2 ⎝r 6 ⎠ 2 3 + 3r + 3r = −7 r

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Further Pure Mathematics 1 3r 2 + 10r + 3 = 0 (3r + 1)(r + 3) = 0

r = − 31 or − 3 Roots are − 61 ,

9. (i) αβγ = −

d a b

∑α = − a

1 2

and − 23

⇒α ×

1

α

×β =

⇒β =

3 2

11 6 α 1 11 3 10 ⇒α + = − − =− 6 2 3 α 2 ⇒ 3α + 10α + 3 = 0 ⇒α +

1

9 6

+β =−

⇒ (3α + 1)(α + 3) = 0 ⇒ α = − 31 or − 3 The roots of the equation are − 31 , -3 and

3 2

.

1 1 k ⇒ ⎛⎜ α × ⎞⎟ + ⎛⎜ × β ⎞⎟ + αβ = 6 α ⎠ ⎝α ⎝ ⎠ 1 1 3 3 k ⇒ ⎛⎜ − × −3 ⎞⎟ + ⎛⎜ − × ⎞⎟ + ⎛⎜ −3 × ⎞⎟ = 2⎠ 6 ⎝ 3 ⎠ ⎝ 3 2⎠ ⎝ 1 9 ⇒ k = 6 ⎛⎜ 1 − − ⎞⎟ = −24 2 2⎠ ⎝ The value of k is -24.

c

∑ αβ = a

10. (i) Let y =

1

x

so x =

1

y

Substituting into x 3 − 2 x 2 − x + 2 = 0 : 3

2

⎛1⎞ ⎛1⎞ 1 ⎜⎜ ⎟⎟ − 2 ⎜⎜ ⎟⎟ − + 2 = 0 y ⎝y⎠ ⎝y⎠ 1 − 2y − y2 + 2y3 = 0 2y3 − y2 − 2y + 1 = 0 (ii) Let y = x − 3 so x = y + 3 Substituting into x 3 − 2 x 2 − x + 2 = 0 : ( y + 3)3 − 2( y + 3)2 − ( y + 3) + 2 = 0

y 3 + 9 y 2 + 27 y + 27 − 2 y 2 − 12 y − 18 − y − 3 + 2 = 0 y 3 + 7 y 2 + 14 y + 8 = 0

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Further Pure Mathematics 1 11.

∑ α = −a

α + β + γ + δ = −a 2(α + β ) = −a a α +β =− 2

∑ αβ = b

αβ + αγ + αδ + βγ + βδ + γδ = b αβ + (α + β )(γ + δ ) + γδ = b αβ + γδ = b − (α + β ) = b − 2

a2 4

∑ αβγ = −c

αβγ + βγδ + γδα + δαβ = −c αβ(γ + δ ) + γδ(α + β ) = −c αβ(α + β ) + γδ (α + β ) = −c (αβ + γδ )(α + β ) = −c a 2 ⎞⎛ a ⎞ ⎛ − b ⎜ − ⎟ = −c ⎜ 4 ⎟⎠ ⎝ 2 ⎠ ⎝ (4b − a 2 )a = 8c a 3 + 8c = 4ab

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Roots of quadratic equations Cut out all the cards on this page and the following page. Sort the cards into four sets, each containing one of the quadratic equations.

x − 3x − 7 = 0 x2 − 7 x + 3 = 0 2

3 x − 21x + 1 = 0 3 x − x − 21 = 0 2

2

α +β =7 αβ = 3 1 α+β =3 αβ = − 7

α +β = 2

2

127 9

α + β = 23 2

2

α +β =7 αβ = 13 α +β =3 αβ = − 7

α + β = 43 2

2

α +β = 2

© Susan Wall, MEI 2005

2

145 3

1/2

Roots of quadratic equations

α β + αβ = −21 α β + αβ = − 73 2

2

2

α β + αβ = 2

2

1

+

1

+

α

α

1

= − 211

1

=−

β

β 3

αβ 3

αβ

7 3

2

α β + αβ = 21 2

2

+

1

α

1

3 7

α

=1

3

αβ

= − 73

4α + 4 β =

+

3

αβ 4 3

4α + 4 β = 28

= 21

1

β

1

β

=

7 3

=9 = − 73

4α + 4 β = 12 4α + 4 β = 28

© Susan Wall, MEI 2005

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Further Pure Mathematics 1 Algebra Section 2: Roots and coefficients of polynomial equations Multiple Choice Test 1) The quadratic equation 2 z 2 + 3 z − 4 = 0 has roots α and β. Which of the following is true? (a) α + β = −3, αβ = −4

3 (b) α + β = , αβ = 2 2 (d) α + β = 3, αβ = 4

3 (c) α + β = − , αβ = −2 2 (e) I don’t know

2) The quadratic equation z 2 − 5 z + 1 = 0 has roots α and β. The quadratic equation with roots 2α + 1, 2β + 1 is (a) w2 − 8w − 5 = 0 (c) w2 − 12w + 15 = 0 (e) I don’t know

(b) w2 + 12 w + 15 = 0 (d) w2 + 8w − 5 = 0

3) A quadratic equation with roots 0.5 and -2 is (a) 2 z 2 − 3z − 2 = 0 (c) 2 z 2 + 3z + 2 = 0 (e) I don’t know

(b) 2 z 2 + 3z − 2 = 0 (d) 2 z 2 − 3z + 2 = 0

4) The cubic equation 3z 3 + 2 z 2 − z − 3 = 0 has roots α, β, γ. Which of the following is true? 2

1

∑ α = − 3 , ∑ αβ = − 3 ,αβγ = 1 (c) ∑ α = −2, ∑ αβ = −1, αβγ = 3 (a)

2

1

∑ α = 3 , ∑ αβ = − 3 ,αβγ = −1 (d) ∑ α = 2, ∑ αβ = −1, αβγ = −3

(b)

a. I don’t know

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Further Pure Mathematics 1 5) A cubic equation with roots -2, 1 + 2j, 1 – 2j is (a) z 3 − z + 10 = 0 (c) z 3 + z + 10 = 0 (e) I don’t know

(b) z 3 + z − 10 = 0 (d) z 3 − z − 10 = 0

6) The roots of the cubic equation z 3 + z 2 − z − 1 = 0 are α, β, γ. The cubic equation with roots 2α, 2β, 2γ is (a) w3 + 2 w2 − 2w − 4 = 0 (c) 2w3 + 2w2 − 2w − 2 = 0 (e) I don’t know

(b) 8w3 + 4 w2 − 2 w − 1 = 0 (d) w3 + 2w2 − 4w − 8 = 0

Questions 7 and 8 are about the cubic equation 3z 3 + pz 2 + qz + 15 = 0 , which has 1 roots α ,1 − 2α , .

α

7) The roots of the equation are (a) 3, −5, 13 (c) −3, 7, − 13 (e) I don’t know

(b) −2,5, − 12 (d) 2, −3, 12

8) The values of p and q are (a) p = -5, q = 47 (c) p = 5, q = 47 (e) I don’t know

(b) p = -5, q = -47 (d) p = 5, q = -47

9) One root of the equation 4 z 3 − 13 z + 6 = 0 is three times another. The roots of the equation are (a) 12 , 23 , −2 (c) 1, 3, −4 (e) I don’t know

(b) − 12 , − 23 , 2 (d) −1, −3, 4

10) The roots of the quartic equation z 4 + 3 z 3 − 2 z + 1 = 0 are α, β, γ and δ. The value of αβγ + βγδ + γδα + δαβ is (a) 0 (c) 2 (e) I don’t know

(b) -2 (d) 1

© MEI, 04/01/07

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Further Pure Mathematics 1 Algebra Chapter assessment 1. Find the values of the unknown constants in each of these identities. (i) 3 x 2 − 6 x − 1 ≡ A( x + B) 2 + C

[3]

(ii) 3 x − 2 ≡ P( x + 1) 2 + Q( x + 1)( x − 3) + R( x − 3) 2

[3]

(iii) x 3 + 2 x − 3 ≡ ( x − 2)( Ax 2 + Bx + C ) + D

[4]

2. The quadratic equation 2 z 2 − 4 z + 5 = 0 has roots α and β. (i) Write down the values of α + β and αβ. (ii) Find the quadratic equation with roots 3α – 1, 3β – 1. (iii)Find the cubic equation which has roots α, β and α + β.

[2] [3] [4]

3. The equation z 3 + kz 2 − 4 z − 12 = 0 has roots α, β and γ. (i) Write down the values of αβ + βγ + γα and αβγ, and express k in terms of α, β and γ. [3] [4] (ii) For the case where γ = –α, solve the equation and find the value of k. [4] (iii)For the case k = 5, find a cubic equation with roots 2 – α, 2 – β, 2 – γ.

4. The cubic equation 2 z 3 + pz 2 + qz + r = 0 has roots

α k

, α , kα .

(i) Express p, q and r in terms of k and α. (ii) Show that 2q 3 = p 3 r .

[3] [3]

(iii)Solve the equation for the case where p = q = -3.

[4]

5. The equation x 4 − 6 x3 − 73x 2 + kx + m = 0 has two positive roots, α, β and two negative roots γ, δ. It is given that αβ = γδ = 4. [5] (i) Find the values of the constants k and m. [4] (ii) Show that (α + β)(γ + δ) = -81. [2] (iii)Find the quadratic equation which has roots α + β and γ + δ. [3] (iv) Find α + β and γ + δ. (v) Show that α 2 − 3(1 + 10)α + 4 = 0 , and find similar quadratic equations satisfied by β, γ and δ. [6]

Total 60 marks

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