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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