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Multiple Roots
Multiple Roots
If P(x), has a root, x = a, of multiplicity m,
then P’(x) has a root, x = a, of multiplicity m - 1
Multiple Roots
If P(x), has a root, x = a, of multiplicity m,
then P’(x) has a root, x = a, of multiplicity m - 1
Proof:

P x    x  a  Q x 
m

(m > 1, x = a is not a root of Q(x))
Multiple Roots
If P(x), has a root, x = a, of multiplicity m,
then P’(x) has a root, x = a, of multiplicity m - 1
Proof:

P x    x  a  Q x 
m

(m > 1, x = a is not a root of Q(x))

P x    x  a  Q x   Q x m x  a 

1
m 1
  x  a   x  a Q x   mQ x 
m

m 1
Multiple Roots
If P(x), has a root, x = a, of multiplicity m,
then P’(x) has a root, x = a, of multiplicity m - 1
Proof:

P x    x  a  Q x 
m

(m > 1, x = a is not a root of Q(x))

P x    x  a  Q x   Q x m x  a 

1
m 1
  x  a   x  a Q x   mQ x 
m1
  x  a  R x 
(where x = a is not a root of R(x))
m

m 1
Multiple Roots
If P(x), has a root, x = a, of multiplicity m,
then P’(x) has a root, x = a, of multiplicity m - 1
Proof:

P x    x  a  Q x 
m

(m > 1, x = a is not a root of Q(x))

P x    x  a  Q x   Q x m x  a 

1
m 1
  x  a   x  a Q x   mQ x 
m1
  x  a  R x 
(where x = a is not a root of R(x))
m

m 1

 P’(x) has a root, x = a, of multiplicity m - 1
e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a
double root
e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a
double root

P x   x 3  4 x 2  3 x  18
P x   3 x 2  8 x  3
e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a
double root
P x   x 3  4 x 2  3 x  18
P x   3 x 2  8 x  3
 3 x  1 x  3
1
x   or x  3
double root is
3
e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a
double root
P x   x 3  4 x 2  3 x  18
P x   3 x 2  8 x  3
 3 x  1 x  3
1
x   or x  3
double root is
3
NOT POSSIBLE
As (3x + 1) is not a factor
e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a
double root
P x   x 3  4 x 2  3 x  18
P x   3 x 2  8 x  3
 3 x  1 x  3
1
x   or x  3
double root is
3
NOT POSSIBLE
As (3x + 1) is not a factor

x 3  4 x 2  3 x  18  0
 x  32  x  2  0
e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a
double root
P x   x 3  4 x 2  3 x  18
P x   3 x 2  8 x  3
 3 x  1 x  3
1
x   or x  3
double root is
3
NOT POSSIBLE
As (3x + 1) is not a factor

x 3  4 x 2  3 x  18  0
 x  32  x  2  0
x  2 or x  3
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1

P0   1  0,   0
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1

P0   1  0,   0
P1  1  A  B  A  1
 2A  B  2
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1

P0   1  0,   0
P1  1  A  B  A  1
 2A  B  2

P 1  1  A  B  A  1
 2 A  B  2
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1

P0   1  0,   0
P1  1  A  B  A  1
 2A  B  2
BUT

2  B 2  4 A2

P 1  1  A  B  A  1
 2 A  B  2
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1

P0   1  0,   0
P1  1  A  B  A  1
 2A  B  2
BUT

2  B 2  4 A2

2  B  2 A
 2A  B  2  0

P 1  1  A  B  A  1
 2 A  B  2
(ii) (1991)
Let x   be a root of the quartic polynomial;
P x   x 4  Ax 3  Bx 2  Ax  1
where 2  B 2  4 A2
a) show that  cannot be 0, 1 or -1

P0   1  0,   0
P1  1  A  B  A  1
 2A  B  2
BUT

2  B 2  4 A2

2  B  2 A
 2A  B  2  0
 P1  0, P 1  0
hence   1

P 1  1  A  B  A  1
 2 A  B  2
b) Show that

1



is a root
b) Show that

1



is a root

 1   1  A  B  A 1
P 
4
3
2
     
b) Show that

1



is a root

 1   1  A  B  A 1
P 
4
3
2
     


1  A  B 2  A 3   4

4
b) Show that

1



is a root

 1   1  A  B  A 1
P 
4
3
2
     



1  A  B 2  A 3   4

P 

4

4
b) Show that

1



is a root

 1   1  A  B  A 1
P 
4
3
2
     




1  A  B 2  A 3   4

P 

4

4
0

4

0

( P   0 as  is a root)
b) Show that

1



is a root

 1   1  A  B  A 1
P 
4
3
2
     




1  A  B 2  A 3   4
P 

4

4
0

4

0
1

 is a root of P x 



( P   0 as  is a root)
c) Deduce that if  is a multiple root, then its multiplicity is 2 and
4 B  8  A2
c) Deduce that if  is a multiple root, then its multiplicity is 2 and
4 B  8  A2
1
If  is a double root of P x , then so is , which accounts for 4 roots


c) Deduce that if  is a multiple root, then its multiplicity is 2 and
4 B  8  A2
1
If  is a double root of P x , then so is , which accounts for 4 roots



However P(x) is a quartic which has a maximum of 4 roots
c) Deduce that if  is a multiple root, then its multiplicity is 2 and
4 B  8  A2
1
If  is a double root of P x , then so is , which accounts for 4 roots



However P(x) is a quartic which has a maximum of 4 roots
Thus no roots can have a multiplicity > 2
c) Deduce that if  is a multiple root, then its multiplicity is 2 and
4 B  8  A2
1
If  is a double root of P x , then so is , which accounts for 4 roots



However P(x) is a quartic which has a maximum of 4 roots
Thus no roots can have a multiplicity > 2
P x   4 x  3 Ax  2 Bx  A
3

2

let the roots be  ,

1



and 
c) Deduce that if  is a multiple root, then its multiplicity is 2 and
4 B  8  A2
1
If  is a double root of P x , then so is , which accounts for 4 roots



However P(x) is a quartic which has a maximum of 4 roots
Thus no roots can have a multiplicity > 2
P x   4 x  3 Ax  2 Bx  A
1
3
   A
4

 1
1     B
 2
1
  A
4
3

2

let the roots be  ,

1



sum of roots  (1)
   (2)
   (3)

and 
Substitute (3) into (1)
Substitute (3) into (1)



1

1
3
 A A
 4
4
1
1
  A

2
Substitute (3) into (1)



1

1
3
 A A
 4
4
1
1
  A

2

Substitute (3) into (2)
Substitute (3) into (1)



1

1
3
 A A
 4
4
1
1
  A

2

Substitute (3) into (2)
1
1 1 1
1  A  A  B
4
4  2
1
1
1
1 - A    B


4 
 2
1
1
1  A2  B
8
2
8  A2  4 B
Substitute (3) into (1)



1

1
3
 A A
 4
4
1
1
  A

2

Substitute (3) into (2)
1
1 1 1
1  A  A  B
4
4  2
1
1
1
1 - A    B


4 
 2
1
1
1  A2  B
8
2
8  A2  4 B

Cambridge: Exercise 5B; 1 to 16
Patel: Exercise 5B; 6b, 7b, 8 a,c,e,g,h
Substitute (3) into (1)



1

1
3
 A A
 4
4
1
1
  A

2

Substitute (3) into (2)
1
1 1 1
1  A  A  B
4
4  2
1
1
1
1 - A    B


4 
 2
1
1
1  A2  B
8
2
8  A2  4 B

Cambridge: Exercise 5B; 1 to 16
Patel: Exercise 5B; 6b, 7b, 8 a,c,e,g,h
Note: tangent to a cubic has two solutions
only. A double root

and a single root

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11 x1 t01 01 algebra & indices (2014)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
11ext1 t08 08 products to sums etc (2013)
11ext1 t08 08 products to sums etc (2013)11ext1 t08 08 products to sums etc (2013)
11ext1 t08 08 products to sums etc (2013)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
11 x1 t15 05 polynomial results (2013)
11 x1 t15 05 polynomial results (2013)11 x1 t15 05 polynomial results (2013)
11 x1 t15 05 polynomial results (2013)
 
X2 t07 06 roots of functions (2013)
X2 t07 06 roots of functions (2013)X2 t07 06 roots of functions (2013)
X2 t07 06 roots of functions (2013)
 
11 x1 t12 03 second derivative (2013)
11 x1 t12 03 second derivative (2013)11 x1 t12 03 second derivative (2013)
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12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t01 03 factorising (2014)
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12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
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Goodbye slideshare UPDATE
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Goodbye slideshare UPDATE
 

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X2 t02 02 multiple roots (2013)

  • 2. Multiple Roots If P(x), has a root, x = a, of multiplicity m, then P’(x) has a root, x = a, of multiplicity m - 1
  • 3. Multiple Roots If P(x), has a root, x = a, of multiplicity m, then P’(x) has a root, x = a, of multiplicity m - 1 Proof: P x    x  a  Q x  m (m > 1, x = a is not a root of Q(x))
  • 4. Multiple Roots If P(x), has a root, x = a, of multiplicity m, then P’(x) has a root, x = a, of multiplicity m - 1 Proof: P x    x  a  Q x  m (m > 1, x = a is not a root of Q(x)) P x    x  a  Q x   Q x m x  a  1 m 1   x  a   x  a Q x   mQ x  m m 1
  • 5. Multiple Roots If P(x), has a root, x = a, of multiplicity m, then P’(x) has a root, x = a, of multiplicity m - 1 Proof: P x    x  a  Q x  m (m > 1, x = a is not a root of Q(x)) P x    x  a  Q x   Q x m x  a  1 m 1   x  a   x  a Q x   mQ x  m1   x  a  R x  (where x = a is not a root of R(x)) m m 1
  • 6. Multiple Roots If P(x), has a root, x = a, of multiplicity m, then P’(x) has a root, x = a, of multiplicity m - 1 Proof: P x    x  a  Q x  m (m > 1, x = a is not a root of Q(x)) P x    x  a  Q x   Q x m x  a  1 m 1   x  a   x  a Q x   mQ x  m1   x  a  R x  (where x = a is not a root of R(x)) m m 1  P’(x) has a root, x = a, of multiplicity m - 1
  • 7. e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a double root
  • 8. e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a double root P x   x 3  4 x 2  3 x  18 P x   3 x 2  8 x  3
  • 9. e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a double root P x   x 3  4 x 2  3 x  18 P x   3 x 2  8 x  3  3 x  1 x  3 1 x   or x  3 double root is 3
  • 10. e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a double root P x   x 3  4 x 2  3 x  18 P x   3 x 2  8 x  3  3 x  1 x  3 1 x   or x  3 double root is 3 NOT POSSIBLE As (3x + 1) is not a factor
  • 11. e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a double root P x   x 3  4 x 2  3 x  18 P x   3 x 2  8 x  3  3 x  1 x  3 1 x   or x  3 double root is 3 NOT POSSIBLE As (3x + 1) is not a factor x 3  4 x 2  3 x  18  0  x  32  x  2  0
  • 12. e.g. (i) Solve the equation x 3  4 x 2  3 x  18  0 , given that it has a double root P x   x 3  4 x 2  3 x  18 P x   3 x 2  8 x  3  3 x  1 x  3 1 x   or x  3 double root is 3 NOT POSSIBLE As (3x + 1) is not a factor x 3  4 x 2  3 x  18  0  x  32  x  2  0 x  2 or x  3
  • 13. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1
  • 14. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1 P0   1  0,   0
  • 15. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1 P0   1  0,   0 P1  1  A  B  A  1  2A  B  2
  • 16. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1 P0   1  0,   0 P1  1  A  B  A  1  2A  B  2 P 1  1  A  B  A  1  2 A  B  2
  • 17. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1 P0   1  0,   0 P1  1  A  B  A  1  2A  B  2 BUT 2  B 2  4 A2 P 1  1  A  B  A  1  2 A  B  2
  • 18. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1 P0   1  0,   0 P1  1  A  B  A  1  2A  B  2 BUT 2  B 2  4 A2 2  B  2 A  2A  B  2  0 P 1  1  A  B  A  1  2 A  B  2
  • 19. (ii) (1991) Let x   be a root of the quartic polynomial; P x   x 4  Ax 3  Bx 2  Ax  1 where 2  B 2  4 A2 a) show that  cannot be 0, 1 or -1 P0   1  0,   0 P1  1  A  B  A  1  2A  B  2 BUT 2  B 2  4 A2 2  B  2 A  2A  B  2  0  P1  0, P 1  0 hence   1 P 1  1  A  B  A  1  2 A  B  2
  • 21. b) Show that 1  is a root  1   1  A  B  A 1 P  4 3 2      
  • 22. b) Show that 1  is a root  1   1  A  B  A 1 P  4 3 2        1  A  B 2  A 3   4 4
  • 23. b) Show that 1  is a root  1   1  A  B  A 1 P  4 3 2         1  A  B 2  A 3   4 P  4 4
  • 24. b) Show that 1  is a root  1   1  A  B  A 1 P  4 3 2          1  A  B 2  A 3   4 P  4 4 0 4 0 ( P   0 as  is a root)
  • 25. b) Show that 1  is a root  1   1  A  B  A 1 P  4 3 2          1  A  B 2  A 3   4 P  4 4 0 4 0 1  is a root of P x   ( P   0 as  is a root)
  • 26. c) Deduce that if  is a multiple root, then its multiplicity is 2 and 4 B  8  A2
  • 27. c) Deduce that if  is a multiple root, then its multiplicity is 2 and 4 B  8  A2 1 If  is a double root of P x , then so is , which accounts for 4 roots 
  • 28. c) Deduce that if  is a multiple root, then its multiplicity is 2 and 4 B  8  A2 1 If  is a double root of P x , then so is , which accounts for 4 roots  However P(x) is a quartic which has a maximum of 4 roots
  • 29. c) Deduce that if  is a multiple root, then its multiplicity is 2 and 4 B  8  A2 1 If  is a double root of P x , then so is , which accounts for 4 roots  However P(x) is a quartic which has a maximum of 4 roots Thus no roots can have a multiplicity > 2
  • 30. c) Deduce that if  is a multiple root, then its multiplicity is 2 and 4 B  8  A2 1 If  is a double root of P x , then so is , which accounts for 4 roots  However P(x) is a quartic which has a maximum of 4 roots Thus no roots can have a multiplicity > 2 P x   4 x  3 Ax  2 Bx  A 3 2 let the roots be  , 1  and 
  • 31. c) Deduce that if  is a multiple root, then its multiplicity is 2 and 4 B  8  A2 1 If  is a double root of P x , then so is , which accounts for 4 roots  However P(x) is a quartic which has a maximum of 4 roots Thus no roots can have a multiplicity > 2 P x   4 x  3 Ax  2 Bx  A 1 3    A 4   1 1     B  2 1   A 4 3 2 let the roots be  , 1  sum of roots  (1)    (2)    (3) and 
  • 33. Substitute (3) into (1)  1 1 3  A A  4 4 1 1   A  2
  • 34. Substitute (3) into (1)  1 1 3  A A  4 4 1 1   A  2 Substitute (3) into (2)
  • 35. Substitute (3) into (1)  1 1 3  A A  4 4 1 1   A  2 Substitute (3) into (2) 1 1 1 1 1  A  A  B 4 4  2 1 1 1 1 - A    B   4   2 1 1 1  A2  B 8 2 8  A2  4 B
  • 36. Substitute (3) into (1)  1 1 3  A A  4 4 1 1   A  2 Substitute (3) into (2) 1 1 1 1 1  A  A  B 4 4  2 1 1 1 1 - A    B   4   2 1 1 1  A2  B 8 2 8  A2  4 B Cambridge: Exercise 5B; 1 to 16 Patel: Exercise 5B; 6b, 7b, 8 a,c,e,g,h
  • 37. Substitute (3) into (1)  1 1 3  A A  4 4 1 1   A  2 Substitute (3) into (2) 1 1 1 1 1  A  A  B 4 4  2 1 1 1 1 - A    B   4   2 1 1 1  A2  B 8 2 8  A2  4 B Cambridge: Exercise 5B; 1 to 16 Patel: Exercise 5B; 6b, 7b, 8 a,c,e,g,h Note: tangent to a cubic has two solutions only. A double root and a single root