Further Concepts for Advanced Mathematics - FP1

Unit 4 Identities and Roots of Equations

4c Cubic, Quartic and Quintic Equations

Cubic Equations

The three roots of the cubic equation ax 3  bx 2  cx  d  0 are called  ,  and  (alpha, beta and gamma). The solutions to the equation are x   , x   and x   .

To find the relationships between the coefficients in the original equation and the roots, we have to use a different technique.

Since the solutions to the equation are x   , x   and x   , the equation must have the factors (x ) , (x   ) and (x   ) . Multiplying these together would give the first term x 3 rather than ax 3 . It follows that the actual factorisation of ax 3  bx 2  cx  d is a(x )(x   )(x   )

This gives us the identity

ax 3  bx 2  cx  d  a(x )(x   )(x   ) multiplying out the right-hand side gives

ax 3  bx 2  cx  d  a(x )(x 2  x  x   ) ax 3  bx 2  cx  d  a(x 3  x 2  x 2  x x 2  x  x  ) ax 3  bx 2  cx  d  ax3  a(     )x 2  a(     )x  a

Equating coefficients of x 2 gives b  a(     )

b Hence        the sum of the roots a

Equating coefficients of x gives c  a(     )

c Hence       the sum of the pairs of roots a

Equating the constant terms gives d  a

d Hence    the product of the roots a

You should notice that there are some similarities between these formulae for cubic equations and those for quadratic equations. There is a formula for the sum and product just like the quadratic equations but there is also a formula for the sum of all the 1 possible different pairings of the roots. You will find that a new formula appears every time we increase the order of the equation we look at. For a quartic equation, there will be four formulae; the sum of the roots, the sum of the pairs of the roots, the sum of the triples of the roots and the product of the roots.

Notation

To cope with the confusion that could arise, the following notation is used

 is used to stand for the sum of the roots

For a cubic equation,        For a quartic equation,         

 is used to stand for the sum of the pairs of roots

For a cubic equation,        For a quartic equation,             

 is used to stand for the sum of the triples of roots

For a quartic equation,         

Examples with the roots of cubic equations

1. The roots of the cubic equation 2x3  3x 2  4x  5  0 are  ,  and  .

Find the cubic equations with roots i)   2 ,   2 and   2 1 1 1 ii) , and   

For the original equation

 3 b   the sum of the roots is   2 a

4 c    2 the sum of the pairs of the roots is  2 a

5 d    the product of the roots is  2 a

i) The sum of the new roots is   2    2    2        6

2 3 21 This is    6    6    2 2

The sum of the pairs of new roots is (  2)(  2)  (  2)(  2)  (  2)(  2)

which simplifies to

  2  2  4    2  2  4    2  2  4        4(     ) 12    4 12 3  2  4  12 2  2  6 12  20

The product of the new roots is (  2)(  2)(  2)

which simplifies to

(  2)(  2  2  4)    2  2  4  2  4  4  8    2  4  8 5 3    2 2  4   8 2 2 5    4  6  8 2 41   2

So the new equation is

 21  41 x3   x 2  20x     0  2   2  21 41 x3  x 2  20x   0 2 2 2x3  21x 2  20x  41  0

1 1 1       2 4         ii) The sum of the new roots is 5       2 5

1 1 1 1 1 1 1 1 1 The sum of the pairs of new roots is                 

Using a common denominator, this simplifies to

3 1 1 1                  3  3  2  5  2  5

1 1 1 1 2 The product of the new roots is          5

The new equation is

 4  3  2  x3   x 2  x     0  5  5  5  4 3 2 x3  x 2  x   0 5 5 5 5x 3  4x 2  3x  2  0

Compare this to the original equation 2x3  3x 2  4x  5  0

Do you notice anything?

Did this happen for quadratic equations (look back at the examples on 4b)

Can you make a generalisation about making a new equation using the reciprocals of the original roots?

Quartic and Quintic Equations

Rules for the roots of quartic and quintic equations can be found by following the same ideas as for quadratic and cubic equations.

For the quartic equation ax 4  bx3  cx 2  dx  e  0 , the roots are called  ,  ,  and  (alpha, beta, gamma and delta). The solutions to the equation are x   , x   , x   and x   .

For any quartic equation: b     a c   the sum of the pairs of roots i.e.             a d    this is the sum of the combinations of 3 roots i.e.         a e   a

4 For the quintic equation ax 5  bx 4  cx3  dx 2  ex  f  0 , the roots are called  ,  ,  ,  and (alpha, beta, gamma, delta and epsilon). The solutions to the equation are x   , x   , x   , x   and x   .

For any quartic equation: b     a c   the sum of the pairs of roots i.e.                     a d    this sum of the triples (work these out for yourself!)  a e   the sum of the combinations of 4 roots  a f    a

This can make working with equations of higher order difficult but there is a quick method for finding some of the new equations that are based on the roots of an original equation.

The Substitution Method

This method can be used to find new equations where the same thing is being done to each of the original roots.

E.G. The roots of 2z 3  3z 2  5z 1  0 are  ,  and  .

Find the cubic equation with roots 3  2 ,3  2 and3  2

If the original equation is written in terms of z , we can write a new equation in terms of another variable w , where w  3  2z .

3  w If w  3  2z then z  . 2

We can substitute this into the original equation giving

3 2  3  w   3  w   3  w  2   3   5  1  0  2   2   2  27  27w  9w2  w3 9  6w  w2 3  w 2  3  5 1  0 8 4 2 27  27w  9w2  w3 27 18w  3w2 5w 15   1  0 4 4 2 27  27w  9w2  w3  27 18w  3w2 10w  30  4  0  w3 12w2  35w  28  0 w3 12w2  35w  28  0