With FAQ etc

12.12.2012

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Algebra and Trigonometry 31

Theory of Equations - Part 5

ObjectivesFrom this unit a learner is expected to achieve the following

1. Familiarize with some methods of finding the roots of a cubic equation.2. Study the nature of roots of a cubic.3. Familiarize with Ferrari’s method for finding the roots of a quartic equation.

Sections

1. Introduction

2. Depressing the cubic equation

3. Solving the Depressed Cubic

4. Cardan’s Solution of the Standard cubic

5. Nature of the roots of a cubic

6. Quartic (or Biquadratic) Equation

7. Solution of Quartic Equations

1

Introduction

Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving

a cubic equation, on the other hand, was done by Renaissance mathematicians in Italy.

In this session we discuss some methods to find one root of the cubic equation

…(1)

so that other two roots (real or complex) can then be found by polynomial division

and the quadratic formula. The solution proceeds in two steps. First, the cubic

equation is depressed; then one solves the depressed cubic. In this session we also

discuss the solution of quartic equations.

Depressing the cubic equation

This trick, which transforms the general cubic equation into a new cubic equation

with missing x2-term is due to Italian mathematician Nicolo Fontana Tartaglia (1500-

1557). We apply the substitution

to the cubic equation (1), and obtain

Multiplying out and simplifying, we obtain

a cubic equation in which x2-term is absent.

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Solving the Depressed Cubic

How to solve a depressed cubic equation of the form

…(2)

had been discovered earlier by Italian mathematician Scipione dal Ferro (1465-1526).

The procedure is as follows:

First find s and t so that

…(3)

and

…(4)

Then will be a solution of the depressed cubic. This can be verified as

follows: Substituting for A, B and y, equation (2) gives

This is true since we can simplify the left side using the binomial formula to obtain

Now to find s and t satisfying (3) and (4), we proceed as follows: From Eq.(3), we

have and substituting this into Eq.(4), we obtain

Simplifying, this turns into the tri-quadratic equation

which using the substitution becomes the quadratic equation

From this, we can find a value for u by the quadratic formula, then obtain t,

afterwards s. Hence the root can be obtained.

Example 1 Using the discussion above, find a root of the cubic equation

Solution

Comparing with Eq. (1), we have and Hence the

substitution will be

;

expanding and simplifying, we obtain the depressed cubic equation

Now to find the solution of depressed equation we proceed as follows:

We need s and t to satisfy

…(5)

and

…(6)

Solving for s in (5) and substituting the result into (6) yields:

which multiplied by t3 becomes

Using the substitution the above becomes the quadratic equation

Using the quadratic formula, we obtain that

We take the cube root of the positive value of u and obtain

By Equation (6),

and hence

Hence the solution y for the depressed cubic equation is

Hence the solution to the original cubic equation

is given by

Example 2 Find one real root of the cubic equation

Solution

Since the term is absent, the given equation is in the depressed form. Here

and hence and

Now substituting in we obtain

or

or

Take Then the above becomes the quadratic equation with

We take the cube of root of the value with largest absolute value, and obtain

.

Putting this value in

we obtain .

Hence one of the roots of the given equation is

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Cardan’s Solution of the Standard cubic

Italian Renaissance mathematician Girolamo Cardano (1501 –1576) published the

solution to a cubic equation in his Algebra book Ars Magna.

Usually we take the cubic equation as

.

But it has been found it is more advantageous to take the general cubic as

… (7)

This method of writing is referred to as the cubic with binomial coefficients.

Taking the form (7) and putting or and multiplying throughout by

, we obtain

.

i.e.,

i.e., , … (8)

where and . The equation (8), where the term in is

absent, is the “standard form” of the cubic.

Now to solve (8) using Cardan’s method, assume that the roots are of the form ;

where p and q are to be determined.

Putting , we get

.

Hence

… (9)

Comparing the coefficients in (8) and (9), we have

and

i.e.,

and

Now

i.e.,

Solving for p and q, we get

Then the solution is given by .

Remark We notice that has three values, viz., and where m is a

cube root of p and is one of the imaginary cube roots of unity. But we cannot

take the three values of independently, for we have the relation .

Thus if , are the three values of where n is a cube root of q and is

one of the imaginary cube roots of unity, we have to choose those pairs of cube roots

of p and q such that the product of each pair is rational. Hence the three admissible

roots of equation (8) are

Example 3 Solve the cubic

by Cardan’s method.

Solution

Let

be a solution. Then

Hence

Comparing this with the given cubic equation, we get

… (10)

Hence

Now

Hence

… (11)

From (10) and (11), we get

and .

Hence

and

;

where is one of the imaginary cube roots of unity.

Hence the roots of the given cubic are

and

i.e., .

Aliter: Since is a root of the given cubic, is a factor of the polynomial in the given

cubic equation. Removing the factor , the cubic equation yields the quadratic equation

Hence

Hence the roots of the given cubic are

Example 4 Solve

Solution

To reduce to standard form, put . i.e, and obtain

i.e.,

is the standard form of the cubic.

Putting taking the cube and a rearrangement yields

.

Comparing this with the standard form of the cubic, we obtain

… (12)

and

Hence

and

… (13)

From (12) and (13),

and .

Hence

or

Hence the roots of the given equation are

.

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5. Nature of the roots of a cubic

Let be the roots of the cubic

… (14)

Then the equation whose roots are

and

Is

Hence

… (15)

Then nature of the roots of Eq. (14) can be obtained by a consideration of the product

in Eq. (15). Since imaginary roots occur in pairs, equation (14) will have either all real

roots, or one real and two imaginary roots. The following cases can occur.

Case 1: The roots are all real and different. In this case

is positive. Therefore by Eq. (15), is negative.

Case 2: One root, say , is real and the other two imaginary. Let and be . Then

,

which is negative, whatever may be. Therefore by Eq.(15), is positive

in this case.

Case 3: Two of the roots, say are equal. Then , and therefore

, is zero.

Case 4: are all equal. In this case all the three roots of equation (14) are zero. This

will be so if .

Conversely, it is easy to see that

(i) when , the roots of the cubic in Eq. (14) are all real;

(ii) when , the cubic in Eq. (14) has two imaginary roots;

(iii) when , the cubic in Eq. (14) has two equal roots; and

(iv) when , all the roots of the cubic in Eq. (14) are equal.

Remark On substituting the values of G and H, it can be seen that

.

The expression in brackets is called the discriminant of the general cubic in Eq. (7), and is

denoted by . It is evident that the discriminant of the standard cubic in Eq. (8) is

itself.

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6. Quartic (or Biquadratic) Equation

A quartic function is a polynomial of degree four and is of the form

where a is nonzero. Such a function is sometimes called a biquadratic function, but the

latter term can occasionally also refer to a quadratic function of a square, having the form

or a product of two quadratic factors, having the form

Setting results in a quartic equation of the form

where Quartic equation is some times called biquadratic equation.

Solution of Quartic Equations

Shortly after the discovery of a method to solve the cubic equation, Lodovico Ferrari

(1522-1565), a student of Cardano, found a similar method to solve the quartic

equation. In this method the solution of the quartic depends on the solution of a

cubic. We now describe the Ferrari’s method.

Let the given quartic equation be

… (16)

We write (16) to the form

where is arbitrary at present.

i.e.,

… (17)

Now, choose in such a way that the quadratic expression in x within the double

bracket may be a perfect square.

So … (18)

Equation (18) is a cubic in giving three values for , say and .

Replacing by in (17), it takes the form

.

Now the left hand side can be factorised into quadratic factors and thus the equation can be

completely solved.

Remarks

Equation (18), which is a cubic in , is known as the reducing cubic.

The reducing cubic gives three values of . These do not however lead to three

different sets of roots for the quartic equation. They only give three different methods

of factorizing the left hand side of the quartic. Hence it is enough to find any one

root of the reducing cubic.

Example 5 Solve

Solution

We write the given equation as

i.e., .

The expression within the double bracket will be a perfect square if

i.e., if .

By inspection, is a root of this cubic in .

So the quartic equation reduces to

i.e.,

.

Example 6 Solve

Solution

We write the given equation as

i.e., … (19)

The expression within the double brackets, is a perfect square if

i.e., if … (20)

Multiplying the roots of Eq. (20) by 2, we obtain the cubic

… (21)

is evidently a root of equation (21).

Hence is a solution of the reducing cubic (20)

Putting in (19), we get

\ .

i.e.,

Hence the roots of the given equation are

.

Summary

In this session we have discussed some methods to find roots of cubic equations. The

solution proceeds in two steps. First, the cubic equation is depressed; then one solves

the depressed cubic. In this session we have also discussed the solution of quartic

equations.

Assignments

1. Show that y = 2 is a solution of our depressed cubic

Then find the other two roots by polynomial division and then using the quadratic formula. Which of the roots equals the solution

obtained in Example 1.

2. Solve using Cardan’s method .3. Solve using Cardan’s method .

4. Solve using Ferrari’s method

5. Solve using Ferrari’s method

6. When a quartic has two equal roots, show that its reducing cubic has two equal roots and

conversely.

Quiz 1. The cubic equation with rational coefficients whose roots are and is

given by _____

(a)

(b)

(c)

(d)

Ans.(c) 2. 3 is a double root of the equation . Third root is

____________

(a)

(b)

(a) 8

(a)

Ans. (a)

3. If are roots of then

(a)

(b)

(c)

(d)

Ans.(c)

FAQ

1. What can you say about the nature of the roots of a cubic equation

.

Answer

Every cubic equation

with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant,

The following cases hold:

(i) If Δ > 0, then the equation has three distinct real roots.

(ii) If Δ = 0, then the equation has a multiple root and all its roots are real.

(iii) If Δ < 0, then the equation has one real root and two nonreal complex conju-gate roots.

Glossary

Polynomial function of degree n in the indeterminate x: A function defined by where is a positive integer or zero and are fixed complex numbers, is called a polynomial function of degree n in the indeterminate x. The numbers are called the coefficients of Zero of a Polynomial: A complex number is called a zero (or root) of a polynomial if Polynomial equation of nth degree: Let

Then is a polynomial equation of nth degree. Root of a polynomial equation: A complex number is called a root (or solution) of a polynomial equation if