UNIT 3 CUBIC AND BIQUADRATIC

Structure 3.1 Introduction Objectives 3.2 Let Us Recall Linear Equations Quadratic Equations 3.3 Cubic Equations Cardano's Solution Roots And Their Relation With Coefficients 3.4 Biquadratic Equations Ferrari's Solution Descartes' Solution Roots And Their Relation With Coeftlcients 3.5 Summary

- 3.1 INTRODUCTION

In this unit we will look at an aspect of that has exercised the minds of several ~nntllematicians through the ages. We are talking about the solution of equations over It. The ancient Hindu, Arabic and Babylonian mathematicians had discovered methods ni solving linear and quadratic equations. The ancient Babylonians and Greeks had also discovered methods of solviilg some cubic equations. But, as we have said in Unit 2, they had llol thought of complex iiumbers. So, for them, a lot of quadratic and cubic equations had no solutioit$. hl the 16th century various Italian mathematicians were looking into the geometrical prob- lein of trisectiiig an angle by straight edge and compass. In the process they discovered a inethod for solviilg the general cubic . This method was divulged by Girolanlo Cardano, and hence, is named after him. This is the same Cardano who was the first to iiilroduce coinplex numbers into algebra. Cardano also publicised a method developed by his contemporary, Ferrari, for solving quartic equations. Lake, in the 17th century, the French mathematician Descartes developed another method for solving 4th degree equrtioiis.

In this unit we will acquaint you with the solutions due to Cardano, Ferrari and Descartes. Rut first we will quickly cover methods for solving linear and quadratic equations. In the process we will also touch upon some general theory of equations. Tliere are several reasons, apart froin a mathematician's natural curiosity, for looking at cubic and biquadratic equations. The material covered in this unit is also useful for mathematicians, physicists, chemists and social scientists.

After going through the unit, please check to see if you have achieved the following c~bjectives.

Objectives After studying this unit, you should'be able. to solve a ; solve a quadratic equation; apply Cardano's method for solving a ; apply Femri's or Descartes' method for solving a ; use the relation between roots and the coefficients of a polynomial equation for obtaining solutions.

3.2 LET US RECALL

You may be hmiliar with expressions of the form 2x + 5, -5x2 + flx3 + x2 + 1, etc. I2 ' AU these expressions are in one variable with coefficients in R In general, we have the following definitions.

Dellnltlons :An expression of the form a. x 0 + al x 1 + a2 x2 + ...... + a, xn, where n E N and 4 E C Vi = 1 ,. . ,n ,is called a polynomial over C inthablex. a,,, al, ....., a, are the poemclents of the polynomial.

If a, d 0, we say that the degree of the polpomlel is n and the leadlng tenn is a, xn. While discuaing polynomials we will observe the foUowlng conventions.

Conventloas :We will 1) write x0 as 1, so that we will write so for ao xO, li) write.xl as x, lli) write xm instead of 1.P (i.e., when am = I), iv) omit tern of the type 0.5. Thus, the polynqmial2 + 3x2 - x3 is 2x0 + 0. x1 + 3x2 + (-1) x3. We usually denote polynomials in x by f (x) ,g (x), etc. If the variable x is understood, then we often write f instead off (x). We denote the degree of a polynomial f (x) by deg f (x) or deg f. Note that the degree off (x) is the highest power of x occurring in f (x). For example, 5 3x + 6x2 + ig3 is a polynomial of degree 3, i) -2 ii) xS is a polynomial of degree 5, and iii) 2 + i is a polynomial of degree 0, since 2 + i = ( 2 + i ) xO. Remark 1 :If f(x) and g(x) are two polynomials, then deg (f(x) + g(x) 4)s lnax (deg deg g(x)) deg (f(x) . g(x3) deg f(x) + deg g(x). We say that f (x) is a polynomial over R if its coefficients are real numbers, and f (x) is over Q if its coefficients are rational numbers. For example, 2x + 3 and x2 + 3 are polynomials over Q as well as R (of degrees 1 and 2, respectively). On the other hand, fi is a polynomial (of degree 0) over R but not over Q. In this course we shall almost always be deallug wlth polynomiels over R

Note that any non-zero element of R is a polynomial of degree 0 over R We defme the degree of 0 to be - =. Now, if we put a polynomial of degree n equal to zero, we get a polynomial equation of degree n, or an nth degree equation. For example, (i) 2x. + 3 - 0 is a polynomial equation of degree 1, and (ii) 3x2 + fix-1 = 0 is a polynomial equation of degree 2

If f(x) = a. + al x + ...... + a, xDis a polynomial and a E C, we can substitute x by r to 2 get f(a), the value of the polynomial at x = a . Thus, f (a) = a, + a, a + a2 r + ...... + r, a'. For example, iff (x) = 2x + 3, then f (1) = 21 + 3 = 5, f (i) = 2i + 3, and f(+) = 2 (3+3= 0. since f = 0, we say tbat 3 is a root off (x). ( ) 2 Definition :Let f (x) be a non-zero polynomial. a E C is called a root (or a zero) off (x) if f (a) = 0. In this case we also say that a is a solution (or a root) of the equation f (x) = 0. A polynomial equation can have several solutions. For example, the equation x2 - 1 - 0 has the two solutions x = 1 and x = - 1.

The set of solutions of an equation is called its solution set Thus, the solution set of x2 + 1 = 0 is {i, - i) . Another definition that you will need quite often is the following.

Definition :Two polynomials a. + al x + ..... + a, xu and bo + bl x .... + b, xmare alled . equal if 11 = m and ai = bi V i = 0, 1, ...., n.

Thus, two polynomials are equal if they have the same degree and their corresponding coef- ficients are equal. Thus ,2x3 + 3 = ax3 + bx2 + cx + d iff a = 2, b = 0, c = 0, d - 3. Let us now take a brief look at polynomials over R whose degrees are 1or 2, and their solutions sets. We start with degree 1 equatiod. 3.21 Linear Equations Consider any polynomial ax + b with a, b E R and a z 0. We call such a polynomial a linear polynomial. If we put it equal to zero, we get a linear equation. Thus, ax+b = 0, a,b ER, az0, is the most general form of a linear equation. / '-b You know that this equation has a solutioi~in R, namely, x = -; and that this is the only a solution. Sometimes you may come across equations that don't appear to be linear, but, after simplification they become linear.

Let us look at some examples.

3 -1 Example 1 :Solve - p. (Here we must assume p z 1.) 3 *p-1 -

Solution :At fmt glance, this equation in p does not appear to be linear. But, by cross- multiplying, we get the following equivalent equation : Two cqu&ions are 4v~tif their solutioa sets are equal.

On simplifying this we get 53 Solutions of Polynomial Equations 3p2-4p+1 -6p = 3p2-3p, that is, 7p - 1 = 0. The solution set of this equation is .Thus, this is the solutio~lset of the equation we started with.

Example 2: Suppose I buy two plots of land for Rs. 1,20,000, and then sdl them. Also suppose that I have made a profit of 15% on the first plot and a loss of 10% on the second plot. If my total profit is Rs. 5500, how much did I pay for each piece of land ?

Solution :Suppose the first piece of land cost Rs. x. Then the second piece cost 15 10 Rs. ( 1,20,000 x). Thus, my profit is Rs. - x and my loss is Rs. - ( 1,20,000 - x ) . - 100 100 15 10 - x -- (1,20,000-x) = 5500 "100 100

25~- 1,750,000 0 0 e x = 70,000. Thus, the fmt piece cost Rs. 70,000 and the second plot cost Rs. 50,000. You may like to try these exercises now.

El) Solve each of the following equations for the variable indicated. Assume that all denominators are non-zero.

a) J ( 2 + a ) = x for x, where J, k and a are amatants. 111 b, R = - + - for R, keeping rl and r2 constant. rl r2 5 c) C = ( F 32 ) for F, keeping C constant. -9 -

E2) An isosceles triangle has a perimeter of 30 cm. Its equal sides are twice as long as the third side. Find the lengths of the three sides.

E3) A student cycles from her home to the study centre in 20 minutes. The return journey is uphill and takes her half an hour. If her rate is 8 km per hour slower on the return trip, how far does she live from the study centre?

E4) Simple interest is directly proportional to the principal amount as well as the time for which the amount is invested. If Rs. 1000, left at interest for 2 years, earns Rs. 110, find the amount of interest earned by Rs. 5000 for 3 years. (Hint :S = kPt, where k is the constant of proportionality, S is the simple interest, P is the principal and t is the time. )

Now that we have looked at first degree equations, let us consider second degree equations, that is, equations of degree 2. 3.2.2 Quadratic Equations

Consider the general polynoinial in x over R of degree 2 :

ax2 + bx + c, where a, b, c E R, a * 0.

We call this polynomial a quadratic polynomial. On equating a quadratic polynomiar to The word 'quadratic' comes from zero, we get a quadratic equation ill standard form. the Latia-word 'quadraturn' meaning

Can you think of an example of a quadratic equation? One is x2 = 5, which is the same as x2 - 5 = 0. Another is the equation Cardano tried to solve, namely, x2 - lox + 40 = 0 (see Sec. 2.1). We are sure you can think of several others. Various methods for solving such equations have bee11 known since Babylonian times (WOO Cubif and ~~ratichw~ons B.C.). Brahmagupta, in 628 A.D. approximately ,also gave a rule for solving quadratic equations. The method that can be used for any quadratic equation is "completing the square". Using it we get the . Let us state this formula.

Quadratic Formula :The solutions of the quadratic equation ax2 + bx + c = 0, where a, b, c E Rand a # 0, are

The expression b2 - 4ac is called the of ax2 + bx + c = 0, Note that this fonnula tells us that a quadratic equation has only two roots. These roots may be equal or thcy lliity be distinct; they may be real or complex. Convention :Wc call a root that lies in C \ R a complex root, that is, a root of the form a + ib, a, b E R, b - 0, is a colnplex root. Let us co~lsidersolne examnples.

Example 3 :Solve i) x2-4x+1 = 0 ii) 4x2 + 25 = 20x iii) x2-lox + 40 = 0

Solution : i) This equation is in standard form. So we can apply the quadratic formula i~runediately.Here a = 1 , b = - 4 , c = 1. Substituting these values in the quadratic fonnula, we get the two roots of the equation to be

Thus, the solutio~lsare 2 + 6 and 2 - u'S , two distinct elements of R Note that in this case the discriminant was positive. ii) In this case let us first rewrite the equation in standard form as 4x' - 20x + 25 = 0. Now, puttinga = 4, b = -20, c = 25 in the quadratic formula, we find that

Here we find that both the roots coincide and are real. Note that in this case the discriminant is 0. iii) Using the quadratic for~nula,we find that the solutio~~sare

By a complex root we mean a roo1 inC\ R

Thus, in this case we get two distinct colnplex roots, 5 + i d% and 5 - i a. Note that in this case the discriminant is negative. Sdutbd-0mi.l Equtlonr In the example above do you see a relationship between the types of roots of a quadratic equation and the value of its discriminant? There is such a relationship, which we now state.

C The equation ax2 + bx + c = 0, a rr 0, a, b, c E R has two roots. They are i) real and distinct if b2- 4ac r 0 ; ii) real and equal if b2 - 4ac = 0 ; iii) complex and distinct if b2 - 4ac < 0. -

Now, is there a difference in the character of the roots of ax2 + bx + c - 0 and dax2 + dbx + dc = 0, where d is a non-zero real number? For example, if b2 - 4ac > 0, what is the sign of ( db )2 - 4 ( da ) ( dc ) ? It will also be positive. In fact, the character of the roots of equivalent quadratic equations is the same.

Now let us consider some important remarks which will be useful to you while solving quadratic equations.

Remark 2 :a and $ are roots of a quadratic equation ax2 + bx + c = 0 if any only if

Remark 3 :From the quadratic formula you can see that if b2 - 4ac < 0, then the quadratic equation ax2 + bx + c = 0 has 2 complex roots which are each other's conjugates. Remark 4 :Sometimes a quadratic equation can be solved without resorting to the quadratic formula. For example, the equation x2 = 9 clearly has 3 and - 3 as its roots. Similarly, the equation (x - 112 = 0 clearly has two coincident roots, both equal to 1(see Remark 2). Using what we have said so far, try and solve the following exercises.

E5) A quadratic equation over R can have complex roots while a linear equation over R can only have a real root True or false ? Why? E6) Solve the following equations: a) x2+5 -0 b) (x+~)(x-1)= 0

c) x2-fix = 1

E7) For what values of k will the equation kx2 + ( 2k + 6 ) x + 16 - 0 have coincident roots ? E8) Show that the quadratic equation ax2 + bx + c = 0 has equal roots if (2ax+b) I (ax2+bx+c).

E9) Find the values of b and c for which the polynomial x2 + bx + c has 1 + i and 1 - i as its roots.

Ell) Leta,fi€Csuchthata+$ - p ERanda(3 - qER Showthata and flarethe roots ofx2 - px + q = 0.

Ell is the converse of E10. We will use it in the next sectioll. Let us now consider some equations which are not quadratic, but whose solutiolls can be Cubicmd Biquadrstic Equations obtained fro111 related quadratic equations. Look at the following example.

Example 4 :Solve i) 2x4 + x2 + 1 = 0, and ii) x=-.

2 Solution :i) 2x4 + x2 + 1 = 0 can be written as 2y2 + y + 1 = 0, where y = x .Then, solving -1i ifi -1 * ifi this for y, we get y = ,that is, x2 = ,two polynomials over C. 4 4 Thus, the four solutions of the original equation are

ii) x = is not a polyilomial equation. We square both sides to obtain the 2 polynoinial equation x = 15 - 2x. Now, aoy root of x - is also a root of the equation x2 = 15 - 2x. But the converse need not be true, since x2 = 15 - 2x can also mean x = - m.

So we will obtain the roots of x2 = 15 - 2x and see which of these satisfy x 5 -. Now, the roots of the quadratic equation x2 = 15 - 2x are x = - 5 and x = 3. We must put these values in the original equation to see if they satisfy it. Now, for x = -5, x--= (-5)-rn = (-5)-5 5 -10 * 0. So x = -5 is not a solution of the given equation. But it is a solution of x2 = 15 - 2x . We call it an extraneous solution.

What happens when we put x = 3 in the given equation? We get 3 = i.e., 3 = 3, which is true. Thus, x = 3 is the solution of the given equation. Now you may like to solve the following exercises, Remember that you must check if the solutions you have obtained satisfy the given equations. This will help you i) to get rid of extraneous solutions, if any, and ii) to ensure that your calculations are alright

E12) Reduce the following to quadratic equations and hence, solve them.

E13) Ameena walks 1 km per hour faster than Alka. Both walked fiom their village to the nearest library, a distance of 24 lun. Alka took 2 hours more thaa Ameena. What was Alka's average speed ?

In this section our aim was to help you recall the methods of solving linear and quadratic equations. Let us now see how to solve equations of degree 3.

3.3 CUBIC EQUATIONS

In this section we are going to discuss some inathematics to which the great 11th cen- tury Persian poet Omar Khayya~ngave a great deal of thought. He, and Greek mathe- lnaticians before him, obtail~edsolutions for third degree equations by considering geometric lnethods that involved the intersection of conics. But we will only discuss Solulions of Polynomial Eqlutioas algebraic methods of obtaining solutions of such equations, that is, solutio~lsobtained by using the basic algebraic operations and by radicals. Let us first see what an equa- tion of degree 3, or a cubic equation, is.

Definition :An equation of the form

is the most general fonn of a cubic equation (or a third degree equation) over R.

For example 2x3 = 0, \/-5 x3 + 5x2 = 0, -2x = 5x3 - 1 and x3 + 5x2 + 2x = - 7 are all cubic equations, since each of them can be written in the form ax3 + bx2 + cx + d r 0, with a # 0. On the other hand, x4 + 1 = 0, x3 + 2x2 = x3 - x and x3 + 6 = 0 are not cubic equations.

There are several situations in which one needs to solve cubic equations. For ex- ample, many problems in the social, physical and biological sciences reduce to obtai~~ingthe eigenvalues of a 3 x 3 matrix (which you can study about in the course). And for this you need to know how to obtain the solutions of a cubic equation.

For obtaining solutions of a cubic equation, or any polynomial equation, we need some results about the roots of polynomial equations. We will briefly discuss them one by one. We give the first one without proof.

Theorem 1 :The polyno~nialequation of degree n,

a. + al x + ..... + a, xn - 0, where ao, al, .... ,a, E Rand a, # 0, has 11 roots, which are real or non-real coinplex numbers. If x,, ...., x, are the n roots of the.equation in Theorem 1, then ao+alx+ .... +anxn a an(x-xl)(x-x2) .... (x-x,).

(Note that the roots need not be distinct. For example, 1 + 2x + x2 = ( x + 1 )? ) We will not prove this fact here; but we will now state a very important result which is used in the proof.

Theorem 2 (Division algorithm) : Given polynomials f (x) and g (x) ( # 0 ) over R, 3 polynolnials q(x) and r(x) over R such that f (x) = g (x) q (x) + r (x) and deg r (x) < deg g (x).

We will also use this theorem to prove the following result which tells us solnething about colnplex roots, that is, roots that are non-real colnplex wun~bers.

Theoren~3 :If a polynomial equation over R has complex roots, they occur in pairs. In fact. if a + ib E C is a root, then a - ib is also a root.

Proof: Let f(x) = a. + al x + ..... + a, xn be a polyl~omialover R of degree n. Suppose a+ibECisarootoff(x) = O,thatis(x-(a+ib)) I f(x). Wewanttoshowthat (x-(a-ib)) ( f(x)too. Now,(x-(a+ib))(x-(a-ib)) = (x-a)'+b2.

Also, by the division algorithm, 3 polynomials g (x) and r (x) over R such that f(x)lg(x) edeg f(x)sdegg(x) (see Remark 1). Since x - ( a + ib ) divides f (x) and ( x -a )2 + b2, it divides f (x) - {( x - a )2 + b2 ) g (x), that is, r (x).

But r(x) is linear over Ror a constant in R. So ( x - ( a + ib )) can only divide r (x) if r (x) = 0. So we find that Cubic and Biquadratic Equatious f(x) = {(~-a)~+b~)~(x). Siiice x - ( a - ib ) divides the right hand side of this equation, it must divide f(x). Thus, a - ib is a root of f(x) = 0 also. Note that Theorem 3 does not say that f(x) = 0 must have a complex root. It only says that if it has a complex root, then the coiijugate of the root is also a root.

Why don't you try the following exercises now ? In them we are just recalling some facts that you are already aware of.

- E14) How many complex roots can a linear equation over R have?

E15) Under what circumstances does the quadratic equation over R, x2 + px + q = 0, have complex roots? If it has complex roots, how many are they and how are they related?

Now let us look at Theorems 1 and 3 in the context of cubic equations. Consider the general cubic equation over R,

Any solution of this is also a solution of

and vice versa. Thus, we can always assume that the most general equation of degree 3 over R is

~~+gix'+~x+r= Owithy,q,rER Theorem 1 says that this equation has 3 roots. Theorem 3 says that either all 3 roots are real or one is real and two are coinplex. Let us find these roots algebraically. 3.3.1 Cardano's Solution

The algebraic method of solving cubic equations is supposed to be due to the Italian, del Ferro (1465-1526). But it is called Cardano's method because it became luiown to people after the Italian, Girolaino Cardano, published it in 1545 in his 'AnMagna'.

Let us see what the method is. We will first look at a particular case.

Example 5 :Solve 2x3 + 3x2 + 4x + 1 = 0.

Solution :We first remove the second degree tenn by completiiig the cube in the following way. Figure 1 : Cardano

1 Put x + = y. Then the equation becomes -2

Assume that % solution is y in + n, where in, 11 E C. Then Solutions dPolpo~nidEquations 5 1 (m+n)j+-(m+n)-- = 0 4 4 5 1 e m3+3mn(m+n)+n3+-(m+n)-- = 0 4 4 1 (m+n)-- = 0. 4 Let us add a further condition on m and n, namely, 5 5 3mn+- = 0, thatis,mn= -- 4 12 1 Then (1) gives us m3 + n3 = -4 '

125 and (2) gives us m3 n3 = - -1728'

Thus, using Ell, we see that m3 and n3 are roots of 1 125 t2--t-- = 0. 4 1728 Hence, by the quadratic formula we find that

m3 = i(l+fi) = assay,

andn3 = ;(I-e) - P,say.

From Unit 2 (536) you know that a and p have real roots, say u and v, respectively. Thus, m can take the values u, wu, u2U, and n can take the values v, ov, u%.

Now, w and 02are non-real complex numbers such that u ( 0' ) = 1. 5 Also, from (2) we k~lowthat mn = ,a real number. - -12 Thus,ifm=u,nmustbev;if m=wu,n mustbew2 w;ifm=w2 u,nmustbewv. Hence, the possible values of y are u+v,wu+~%,~%+~v. To get tbe three~obtsof the original equation, we simply put these values of y in the relation

This example has given you some idea about Cardano's method for solving a general cubic equation. Let us outline this method for solving the general equation

Then (3) becomes

P P Now put y E x + - Then x y - 5.and the equation becomes 3' - Cubk and Blqudntic Equations

y3+~y+B = 0, ...... (4)

p2 2p3 Pq whereA = q-7 and B ---+r. - 27 3 Step 2 :Now let us solve (4). Let y = a + p be a solution. Putting this value of y in (4) we get (a+~)~+A(a+p)+~= 0 e a3+3af3(a+f3)+p3+A(a+p)+B= 0 e a3+p3+(3ap+A)(a+f3)+B = 0 ...... (5) Now, we choose a and fl so that, 3ap +A = 0. Then we have the two equations

and from (5) a3+p3 = -B...... (7) Thus, using Ell, we find that a3 and p3 are roots of the quadratic equation A3 t2+Bt-- = 0. (8) 27 ...... Hence, using the quadratic formula, we find that

-+ - = U, say, 4 27 ...... (9)

2 Now, from Unit 2 (E36) we know that any compleh number has three cube roots. We also know that if y is a cube mot, then the three rm&arey, o y and 02y. Therefore, if a and b denote a cube root each of u and v, respedively, then a can be a, a o or am2; and p can be b, bo or b2.Does this mean that y = a + fl can take on 9 values ? A Note that a and fl also satisfy the relation a fl - - E R - 3 Thus, since o E C, 02E C, o3 = 1E R, the only possibilities for y are a+b,ao+bo2,ao2+bo.

Step 3 :The 3 solutions of (3) are given by substituting each of these values of y in the equation x - y - 5.. So, what we hvejust shown is that

the.rootsof~~+~x~+~x+r= 0 are a+p--P ao+po2--,ao2+po--P P 3' 3 3 ' \ -1 + id3 where o = ,aisacuberootof ,pisacuberootof 2

The formula we have obtained is rather a complicated business. A calculator would certainly ease matters, as you may find while trying the following exercise. E16) Solve the followii~gcubic equations : a) 2x3+3x2+3x+1= 0

b) x3 + 21x + 342 = 0 c) x3 + 6x2 + 6x + 8 = 0 d) x3+29x-97 = 0

e) x3 = 30x - 133

B2 A3 In each of the equations in E16, you must have found that -+ - r 0. 4 27 B2 A3 But what happens if - + - c 0. 4 27 This case is known as the irreducible case. In this case (9) tells us that a3and fI3 are complex numbers of the form a + ib and a - ib, where b * 0. From Unit 2 you know that if the polar form of a + ib is r ( cos 9 + i sin 9 ), then its cube roots are

Similarly, the cube roots of a - ib are 9+2kn rl"(msT- i sin -

Hence, the 3 values of y in (4) are

V3 e+an 2r cos- , where k = 0, 1,2. 3 In the irreducible case a and All these are real numbers. Thus in this case all the roots of (3) are real, and are given by are not real, but x is real. 2r ~3 cos---, P 2r1"cos ---9+2n P 2r 1/3 cos -8+4n - -P 3 3 3 3' 3 3' This trigonometric form of the solution is due to Francois Vi'ete (1550-1603). Now try an exercise.

E17) Solve the equation x3 - 3x + 1 = 0

- - - - So far, we have seen that a cubic equation has three roots. We also know that either all the roots are real, or one is real and two are complex conjugates. Can we tell the roots or the character of the roots by just inspecting the coefficients? We shall answer this question now. 3.3.2 Roots And Their Relation With Coefiicients In this sub-section we shall first look at the cubic analogue of El0 and Ell. Over there we saw how closely the roots of a quadratic equation are linked with its coefficients. The same thing is true for a cubic equation. Why don't you try and prove the relationship that we give in the following exercise ?

E18) Show that a, b and y are the roots of the cubic equation ax3 + bx2 + cx + d = 0, a 0, if and only if - - (Hint :Note that the given cubic equation is equivalent to Cubic Biquadratic a(x-a)(x-$)(x-y) = 0.)

The relationship in El8 allows us to solve problems like the following.

Example 6 :If a, $, y are the roots of the equation 3 2 X -7x +x-5 =. 0, find the equation whose roots are a + $, $ + y, a + y. Solution :By El8 we know that

(~+B)($+Y)+($+Y)(~+Y)+(~+Y)(~+$) = (49-7(y+a)+ya) +{4Q-7(a+$)+a$)+{49-7(P+y)+$y) = 147 - 98 + 1, using (10) and (11). = 50, and ...... (12)

To evaluate the expression on the right hand side, we can use (10) or we call use the fact that

Therefore, ( a + P ) ( $ + y. ) ( a + y ) = 2...... (13) Now, E18, (ll), (12) and (13) give us the required equation, which is x3 - 14x2 + sox - 2 = 0. Why don't you try the following exercise now ?

E19) Find the sum of the cubes of the roots of the equation x3 - 62+ llx -6 = 0. Hence fmd the sum of the fourth powers of the roots.

Let us now study the character of the roots of a cubic equation. For this purpose we need to iiltroduce the notion of the discriminant. In the case of a quadratic equation x2 + bx + c = 0, you know that the discriiniilant is b2 - 4c. Also, if a and $ are the two roots of the equation, then a + $ = - b, a $ = c. Therefore,

Thus, the discriminant = ( a - p )', where a and p are the roots of the quadratic equation.

Now consider the general quadratic equation, ax2 + bx + c I 0. Let its roots be a and $ .Then its discriminant is b2 - 4ac - a2 ( a - p )'. We use this relationship to define the discriminant of any polyynoinial equation. Def'inltlon : The dlscrln~lnantof the 11th degree equation a0+a1x+a2x2 + ...... +a,xn= 0 is

where al, ...... a, are the roots of the polynomial equation. . .

SolutionsoCPolpon~~Equacons I11 ~articular.if wc considcr lie casc n = 3 and a,, = 1. we find that

Now consider Cardano's solution of the cubic equation (3), namely, x3 + px2 + qx + r = 0. B2 A3 -D The expression under the square root sign is _I +, = ,,, ,where D is the discriminant. Now, (9) tells us that the sign of the discri~ninantis closely related to the characters of the roots of the equation. Let us look at the different possibilities for the roots a, p and y of (3).

1) The roots of (3) are all real and distinct. Then ( u - P )2 ( fi - y )' ( a - y )2, that is D, must be positive.

2) Only one root of (3) is real. Let this root be u. Then P and yare colnplex co~~jugates. :. p - y is purely imaginary :. ( P - y )" 0. Also, a - P and u - yare col~jugates. Therefore, their product is positive. Hence, in this case D < 0.

3) Suppose a = P and y * a. Since a - fi = 0, D = 0. Also, B 0. Why? Because if B = 0,then A = 0 (since D = 0).

[Over here we have used tlv relationsl~ipbctwccn the roots,

sincep = -(a+p+y) = -(2u+y) and q = afi+By+ay = u(a+2y)]. On simplifying we get u = y, a co~ilradiclio~~.

Thus, B t 0. So, if exactly two roots of (3) are equal, then D = 0 and B 0, and hence, A * 0.

4) If all the roots of (3) are equal, then D = 0,B = 0, and hence A = 0. Let us summarize the different vossibilities for the character of the roots now.

Consider the cubic equation x3 + px2 + qx + r = 0. p, q, r E R, 2p3 p2 and let B = -- -+ r and A = q - - . Then 27 3 3 B~ 1) all its roots are real and distinct iff - + - < 0. 4 27 B2 A' 2) exactly one root is real iff - + - > 0 4 27 3) exactly two In this case all the roots are real.

You inay now like to try the followi~~gproblen~ to see if you've understood what we have just discussed. Cubic adBiquadmtir Equntions E20) Ui~derwhat coi~ditionson the coefficients of ax3 + 3bx2 + 3cx + d - 0, a * 0, will the equation have complex roots? E21) Will all the root5 of x3 = 15x + 126 be real ? Why ?

-- -- - So far we have introduced you to a method of solving cubic equatioils and we have studied the solutions in some depth. We shall study them some more in Unit 6, as an application of I the Cauchy- Schwarz inequality. Now let us go on to a discussion of polynomial equatioi~s of degree 4. ;I S 3 f 3.4 BIQUADRATIC EQUATIONS

As in the case of cubic equations, biquadraticequations have been studied for a long time. The ailcieilt Arabs were known to have studied them from a geometrical point of view. In this section we will discuss two algebraic inethods of solving such equations. Let us first see what a biquadratic equation is. t Definition :An equation of the forin I ax4 + hx3 + CX* + dx + e = 0, where a, b, c, d, e E R and a t 0 , is the most general form of a biquadratic equation (or a quartlc equation, or a fourth degree equation) over R.

Call you think of examples of quartic equations over K? What about x4 + 5 = fix- x2 ? This certainly is a quartic equation, as it is eq"iva1ent to x4 + x2 -fix + 5 = 0. Wl,ilt about fi = x4 + I? This isn't even a polynomial equation. So it can't be a quartic. Let us now consider various ways in which we call solve an equation of degree 4. In some cases, as you have seen in Exainyle 4, such an equation call be solved by solvii~grelated quadratic equations. But most biquadratic equations can't be solved in this mamler. Two al- gebraic methods for obtaining the roots of such equations were developed in the 16th a-ld 17th centuries. Both these methods depend on the solviilg of a cubic equation. Let us see what they are,

3.4.1 Ferrari's Solution

The first inethod for solvii~ga biquadratic equation that we will discuss is due to the 16th century Italian mathematician Ferrari, who worked with Cardano. Let us see what the inethod is with the help of an example. Example 7 :Solve the equation

Solution :We will solve this in several steps.

Step 1 : Add the quadratic polyi~omial( ax + b )2 = a2x' + 2abx + b2 to both sides. We get

Step 2 :Choose r and b in R so that the left hand side of (14) becomes a perfect square, say ( xZ- x + k )', wherc k is an unknown. Thus, we need to choose a and b so tbat

Equating the coefticiei~tsof x 2, x and the corstant term on both sides, we get a2-5 = 2kt1 ...... (15) 2(ab+S) - -2k ...... (16)

b2-3 = k2 ...... (17) Q-lutioasof Polynomid Equations (15) * a2 = 2k+6

(k+5)2 Then (17) * k2 + 3 = 2k +6

This cubic equation is called the resolvent cubic of the given biquadratic equation. We have obtained it by eliminating a and b from the equations (15), (16) and (17). We choose any one root of the cubic. One real solution of (18) is k = - 1. (It is easy to see this by inspection. Otherwise you can apply Cardano's method.) Then, from (15), (16) and (17) we get

a = 2 and b = -2 (or a = -2 and b = 2)satisfy these equations. Weneed only one set of values of a and b. Either will do. Let us take a = 2 and b = - 2. Step 3 :Put these values of k, a and b in ( x2 - x + k )2 = ( ax + b )2. On taking square roots, we get lwo quadratic equations, namely,

x2-3~+1= 0 and x2+x-3 = 0. Applying t11e quadratic fonnula to these equations we get

Does Example 7 give you some idea of the general method developed by Ferrari? Let us see what it is. We want to solve the general 4th degree equation over K,namely,

x4 + p~3+ q~2 + rx + s = O, p, q, r, s E R...... (19) The idea is to express this equation as a difference of squares of tw~polynomials. Then this difference call be split into a product of two quadratic factors, and we can solve the two quadratic equations that we obtain this way. Let us write down the steps involved. Step 1 : Add ( ax + b )%o each side of (19), wl:!,re a and b will be chosen so as to make the left hand side a perfect square. So (19) becomr.~

~~+~x~+(~+a~)x~+(r+2ab)x+s+b~= (a~+b)~ ...... (20) Step 2 : We want to choose a and b so that the left hand side is a perfect square, say ( x 2 + - x + k )2, where k is an unknown. 2 P Notc that the coefficient of x is necessarily -, since the coefficient of x3 ill (20) is p. 2 So we see that

Colnparil~gcoefficients of x2,x and the constant tenn, we have n

Eliminating a and b from these equations, we get the resolvent cubic

From Ser. 3.3 you lalow tliat this cubic equatiol~has at least one real root, say u Then, wt c.an find a and b in terms of a. Cubic and Biquadratic Equations Step 3 :Our assumption was that

Now, putting k = a and substituting the values of a and b , we get the quadratic equations

Then, using the quadratic fonnula we can obtain the 4 roots of these equations, which will be &e roots of (20), and hence of (19). The following exercise gives you a chance to try out this method for yourself.

-- - E22) Solve the followi~~gequations:

------Let us now coilsider the other classical method for solving quartic equations. 3.4.2 Descartes' Solution

The second method for obt;iining an for a quartic was given by the malhematician and philosopher Rent? Descartes i 1637. In this method we write the biquadratic polynomial as a product of two quadratic polynomials. Then we solve the resultallt quadratic equations to get the 4 roots of the original quartic. Let us consider an example. In fact, let us solve the problem .in Example 7 by this method. Thus, we want to solve x4-2x3-5x2+ lox-3 = 0 ...... (21)

Step 1: Remove the cube term For this we rewrite x4 - 2x3as

Thus, the given equation beconles

1 Now, put x - - = y. We get 2

Step 2 :Write the left hand side of (22) as a product of quadratic polynomials. For this, let us assume that

(Note that th~coefficients of y in each of these factors are k and - k, respectively, since the product does not contain any term with y3. ) Equatiiig coefficients, we get 13 9 m+n- k2 = -- k(n-m) ;= 4,mn = - 2 ' 16 Elhninating m and n from these equations, we get 13 4 13 4 9 (k~-~i-)(P-~+ j-) = a, thatis,

If we put k2 = t, then this becomes the resolvent cubic t3 - 13t2+ 40t - 16 = 0. This has one real root; in fact, it has a positive real root, because of the followi~~gresult, that we give without proof.

Every polyt~omialequation, whose leading coeffiiient is 1 and degree is an odd number, has at least one real root whose siga is opposite to that of its last term.

So, usi~lgthis result, we see that we car1 expect to get one positive value oft. By trial, we see that t = 4 is a root, that is, k2 = 4, that is, k = * 2. Any one of these values is sufficient for us. So let us take k = 2. Then, from the equations in (23) we get

Thus, (22) is equivalent to

Step 3 :Solve the quadratic equations 9 1 y2+2y-- = 0andy2-2y-- = 0. 4 4 By the quadratic for~nulawe get

1 Step 4 : Put these values inx = y + -to get the four roots of (21). 2 Thus, the roots of (21) are :

Let us write dawn the steps in this method of solution for the general quartic equation x4 + ax3 + bx2 + cx + 3 = 0, a, b, c, d E R ...... (24) Step 1: Reduce the equation to the fonn x~+~x~+~xts = 0...... (25) Step 2 :Assume that x~+~x~+Tx+s= (x2+kx+m) (x2-kx+n). Then, on equating coefficients, we get m+n-k2 = q, k(n-m) R r, mn - s. From these equations we get

2 r m+a = k +q, n-in - - k '

2 r r Therefore, 2g = k + q - c, 211 = k2 t q t - k ' Substituting in mn = s, we get Cubic and Biquadratic Equli~u~

k6+2qk4+(q2-4s)k2-r2 = 0, thatis,

t3+2qt+(q2-4s)t-r2 3 ~,~uttin~k~=t. This is a cubic with at least one positive real root. Then, with a known value oft, we can determine the values of k, m and n. So, (25) is equivalent to (x2+kx+m) (x2-kx+n) = 0 Step 3 :Solve the quadratic equations

x2+kx+m = 0andx2-kx+n = 0. This will give us the 4 roots of (25), and hence, the 4 roots of (24). Now, why don't you try the following exercises to see if you have grasped Descartes' lnethod ?

E23) Solve the followilig equations by Descartes' method:

a) x4-2x2+8x-3 = O b) x4+8x3+9x2-8x = 10 2 c) x4-3x -6x-2 = 0 d) x4+4x3-7x2-22x+24 = 0.

E24) Reduce the equation 2xs + 5x6 -5x2 = 2 to a biquadratic equation. Hence solve it.

- While solving quartic equations you may have realised that the methods that we have dis- cussed appear to be very easy to use; but, ui practice, they can become quite cumbersome. This is because Cardano's lnethod for solving a cubic often requires the use of a calculator. Well, so far we have discussed methods of obtaining algebraic solutions for polynomial equations of degrees 1, 2,3 and 4. You may think that we are going to do something similar for quintic equations, that is, equations of degree 5. But, in 1824 the Norwegian algebraist Abel(1802-1829) published a proof of the followiilg result:

1 There call be no general formula, expressed in explicit algebraic operations on the coefficients of a polynomid equation, for the roots of the equation, if the degree of the equation is greater than 4.

This result says that polynomial equations of degree > 4 do not have a general algebraic solu- tion. But, there are methods that can give us the value of ally real root to any required degree of accuracy. We will discuss these methods in our course on Numerical Analysis. There are, of course, special polynomial equations of degtee s 5 that can be solved (as in E24). Let us now look a little closely at the roats of a biquadratic equation. We shall see how they are related to the coefficients of the equation, just as we did in the case of the cubic. 3.43 Roots And Their Relation With Coefficients

In the two previous sub-sections we have shown you how to explicitly obtain the 4 roots of a bi- quadratic equation.Let us go back to Theorems 1 and 3 for a moment. Theorem 1tells us that a quartic has 4 roots, which may be real or comp~kx.By Theorem 3, the possibilities are

i) all the roots are real, or ii) two are real and two are c~mplexconjugates of each other, or iii) the roots are two pairs of complex conjugates, that is, a + ib, a - ib, c + id, c -id for some a, b, c, d E R

Now, if rl, r2, r3, r4 are the roots of the quartic ax4 + bx3 + cx2 + dx + c 0, then of Po'ynomial Equations ax4 + bx3 + cX2+ dx + t: = a ( x - r1 ) ( x - r2 ) ( x - r3 ) ( x - r4 ) 4 3 = x - ( rl + 12 + 13 + r4 ) x + ( rl r2 + rl r3 + rl r4 +,r2 r3 + r2 r4 + r3 r4 ) x 2

- ( r, r2 13 + rl r2 14 + rl 13 14 + r~ 13 14 ) x + rl r2 r3 14. Con~pariilgthe coefficients, we see that b rl+r2+r3+r4 = - - a

C rlr2 + rlr3 + rlr4 + r2r3 + r2r4 + 1314 =- a d rl 12 r3 + r1 r2 r4 + rl 13 r4 + r2 13 r4 = - - a e rl r2 13 r4 = - a This means that coeff. of x3 suln of the roots = - coeff. is short for coefficient. coeff. of x4 coeff. of x2 sum of the roots taken two at a time = coeff. af x4 coeff. of x sum of the roots taken three at a time = - coeff. of x4

coeff. of xo constant term product of the roots = .at is, coeff. of x4 coeff, of x4 ' These f~urequations are a particular case oft@ followi~igresult that relates the roots of a polynoomial equation with its coeficients. . Theorem 4: Let a,, ...... , a, be the 4 roots of the equation aoxn+alxn-'+ .... +an = 0, a,ERVi 0,17...., n,aO # 0.Then

n a1 ,iA. - A, + .... +A" Za, = -- I - I a0 n Z a, a, = -a2 1.j-1 a0 '

n Za,, a,2... a,, = (-I)t -at 1, < 12< ..

n a n a, = 1-1 a0 In El0 and El8 you have already seen that this result is true for n = 2 and 3.

Theoreill 4 is very useful in several ways. Let us consider an application in the case 11 = 4.

Example 8: If the sun^ of two rpots of the equation 4x4-24x3+31x2+6x-8 = 0 is zero, find all the roots of the equation. Solution : Let the roots be a, b, c, d, where a+b = 0. 24 Thena+b+c+d= - = 6 4 :. :. c+d = 6 31 Also ab + ac + ad + bc + bd + cd = (a + b) (c + d) + ab + cd = - 4 31 (27) ab cd = ...... :. + -4 3 Further, (a + b) cd + ab (c + d) = acd + bcd +abc + abd = - -2 1 (26) ab = - - :. * 4 Finally, abcd = - 2 / / :. (28) * cd = 8

Now using Ell, (26) and (25) tell us that c and d are roots of x2 - 6x + 8 = 0. Thus, by the quadratic formula, c = 2, d = 4. 1 1 Similarly, a and b are roots of x 2 - -1 - 0. :. a 4 = TP = --2 ' Thus, the roots of the given quartic are

Try the following problems now.

E25) Solve the equation

x4 + 15x3 + 70x2 + 120x + 64 = 0, given that the roots are in G.P., i.e., geometrical progression. (Hint: If four numbers a,b,c,d are in G.P., then ad = bc.)

E26) Show that if the sum of two roots of x4 - px3 + qx2 - rx + s = 0 (where p, q, I, s € R) equals the suin of the other two, then p3 - 4pq + 81 = 0.

We have touched upon relations between roots and coefficients for n = 2, 3,4. But you can apply Theorem 4 for any n E N. So, in future whenever you need to, you can refer to this tbeorein and use its result for equations of degree 2 5.

Let us now wind up this unit with a suinmary of what we have done in it.

3.5 SUMMARY

In this unit we have introduced you to the theory of lower degree equations. Specifically, we have covered the following points: -b The linear equation ax + b = 0 has one root, namely, x = - ' 1) a 2) The quadratic equation ax2 + bx + c = 0 has 2 roots given by the quadratic formula -b*G. X = 2a /3)* Every polynomial equation of degree n over R has n roots in C. 4) If a + ib € C is a root of a.real polynomial, then so is a - ib. 5) Cardano's method for solving a cubic equation. 6) A cubic equation can have: i) three distinct real roots, or ii) one real root and two coinplex roots, which are coiljugates, or iii) three real roots, of which exactly two are equal, or iv) three real roots, all of which are ~qual.

7) Methods due to Ferrari and Descartes for solviilg a quartic equation. Both these methods require the solviilg of one cubic and two quadratic equations. Solu1ionsofPol~nomidhutions 8) A quartic equation can have four real roots, or two real and two complex roots, or 4 complex roots.

9) If the 11 roots of the nth degree equation a. xn + al xn- ' + .... + a, - x + a, = 0, are PI, P27 ...-.,Pn, the11

That is, the sum of the product of the roots taken k at a ti~neis

As in our other units, we have given our solutions and/or answers to the exercises in the unit in the followi~~gsection. You can go through them if you like. After that please go back to Section 3.1 and see if you have achieved the objectives.

El) a) This has a solution provided J * k. Ja Jak +Ja=O exa--- J-k (t-1)

E2) Let the third side be x cln. Then the other two sides are each 2x cm long. Therefore, x + 2x + 2x = 30 * x = 6. Thus, the lengths of the sides are 6 cm, 12 un and 12 cm.

E3) Let her rate of travel to the study centre be x km per hour. Thus, the distance from her X home to the study centre is lun. While returning, her rate is (x - 8) knl12lr. -3

24 Thus, the distance is - km = 8 km. 3

E4) S = k t. 11 We know that 110 k x 1000 x 2 * k = - - 200' 11 S 3 - Pt. :. 200

So, the required interest is

11 - x 5000 x 3, that is, Rs. 8251-. 200 Cubic and Biquadratic E~~IL~~UIIS

E5) True. For example, x2 + 1 = 0 has co~nplexroots. Any linear equation ax + b = 0 -b over R has only one root, namely, -E R. a

E6) a)x2=-5 *x=ifiand-ifi. b) This is (x - (-9)) (x - 1) = 0. Thus, by Remark 2, - 9 and 1 are the roots. c) We rewrite the given equation in standard form as x2-Gx -1 = 0

fi+3 a-3 :. X = - and ,'

E7) The roots are -(2k+6),:d(2k+6)2-64k X = 2k The roots will coincide if the discriminant is zero, that is, ( 2k + 6 )2 - 64 k = 0. This will happen when k2 - 10 k + 9 = 0, that is, ka1 or k=9.

2 the root of 2ax + bx = 0 is a root of ax + bx + c = 0.

ax2 + bx + c = 0 has coincidental roots.

E9) By Remark 2, we must have

x2+bx+c = (x-(I+ i)) (x-(I-i)) 2 = x -2x+2. Thus, coinparing the coefficients of x1 and xO,we get b=-2, C-2.

E10) a and are roots of ax2 + bx + c = 0 - ax2+bx+c = a(x-a) (x-p) e ax2+bx+c = a {x2-(a+p)x+ap) e. b = -a(a+p) and C= aaP

? Ell) Substitutingx = a in x'-px + q, we get a2-pa+q = a2 --(a+p)a+up,sincea+$ = p and a@= q. = 0 :. :. aisarootofx2 -px+q = 0. Similarly, fl is a root of x2 - px + q = 0. solu~nsdP''b"'~h"'hs E12) a) 4p4 - 16p2 + 5 r 0. Put p2 = X. Then the equation becomes 4x2- 16x + 5 = 0. fl Itsrootsare2+-m and 2-- . 2 2

These 4 values of p are the required roots.

Every root of this is a root of 5x2- 6 = x4 * x4-5x2+6 = 0. Put x2 = y. Then y2-5y+6 = 0. 5*4'ZEa 5*1 Its roots are y = I- = 3, 2. 2 2

Putting tbese 4 values of x in the giver quation, we find tbat fl and fi are its solutions. c) Separating the radicals, we get

Squaring both sides, we get

Again squaring both sides, we get

x2-2x-3 = 0. Its roots are x = 3 and x = - 1. Substituting these values of x in the given equation, we get 1/2(3)+3 = l,and\/2(-1)+3 -\r-i-+i = 1. Thus, both x =-- 3 and x = - 1 are roots of the given equation. E13) Let Alka's rate be x km per hour. Then Ameena's is (x + 1) km per hour. 24 The time taken by Ameena to walk to the library = -hours. Thus, the &metaken x+l by Alka = (5+ 2) hours.

Shlce (-4) can't be the rate, it is an extraneous solution. Thus, the required speed must be 3 km per hour. There will be two such roots, and they will be conjugates.

Referring to Cardano's fonnula, we see that in this case

2 1 0-o2 1 o -0 1 :. therootsare------, ---, that is, 22222 1 - -, W, o2 (since 1+o+02= 0). 2 b) x3 + 21x + 342 = 0. Here we don't need to apply Step 1 of Cardano's method, since there is no term containing x2. Now, with reference to Cardano's formula, A=21-0 = 21, B- 0-O+342= 342.

1

-1 and P = (-171 -172)~ - -7. Thus, the roots of the equation are 1-7, o-7u2, w2-7o,tbatis, -6, 01-70', 02-7o. c) x3 +6x2+6x+8 = 0. Here p = 6, q = 6, r = 8.

1 and $ = (-6-2fi)' - -2.243. (We have used a calculator to evaluate a and $ to 3 decimal places.) Then the required roots are a+P-2, ao+$o2 -2, ao2 +Po-2. d) x3 + 29x - 97 = 0. Here p = 0, q = 29, r = - 97.

Then the roots are

u+F, ao+flo2, ao2+@w. ~o~utiousof PoIynomid Equalioos e) X' - 30~+ 133 = 0. Here p = 0, q = - 30, r = 133. :. A = -30, B = 133: -1 3 , -1 a = [-?+(-} = (-8). = -2,a.d

-1 fJ = (-66.5 -58.5 )3 = -5. :. therootsare-7, -2w-5w,2 -2w 2 -5w.

~17)x3-3x+1 = 0. Herey = 0, q - -3, r = 1. :. A = -3, B = 1. B' . - 3 < O. .. 4 +v=-4 So we are in the irreducible case.

Now,_B+/m -1 id-3 211 -+- = -+- = cos 3+ i sin -. 2 427 2 2 3 3

Thus, the solutiol~sof the given equation are 2 cos ["f z], where k = 0, 1,2, that is, 211 8 x 14 x 2 cos -, 2cos -, 2cos - 9 9 9 '

E18) a, fJ, y are the roots iff

ax3+bx2+cx+d = a(x-a)(x-f3)(x-y)

= a {~~-(a+~+~)x~+(afJ+fJ~+a~)x+afJ~). On comparil~gcoefficients, we get b a+f3+y = -- a

C crfJ+fJy+ay= - a d afJy= --. a E19) Let the roots be a, fJ,y. The11 a+fJ+y= 6,afJ+f3y+ay= 11, af3y = 6. :. a2+fJ2+y2= (a+fJ+y)2-2(afJ+fJY+aY) = 36-22 = 14.

:.d+p3+y3 = (a+p+y)3-3a2(p+y)-3~2(a+y) -3y2(a+fJ)-6clfJy 3 2 = 6 -3a (6-a)-3fJ2(6-fJ)-3y2(6-y)-6x6 = I~o-Is(~~+~J~+~')+~(~~+P~+~~)

:. 4(a3+fJ3+y3)= 180-18x 14 = -72 :. a3+ p3 + y3 = - 18. Now, each of a, 0, y satisfy 2 x3-6x +llx-6 = 0. Thus, they will satisfy x4 - 6x3 + 11x2- 6x = O also. :. a4-6a3+ lla2-601 = 0 ~~-6p~+llf3~-68= 0 Y4-6y3+lly2-6y -- 0. Adding these equations, we get

(a4+~4+Y4)-~(a3+~3-~Y3)+11(a2+~2+Y2)-6(a+fl+y)= 0

3 a4+p4+y4= 6(-18)-11(14)+6(6) = -226.

3b 3c d E20) Here p = - a ' as1 = -.a

Therefore, the equation has coinplex roots if B' - + - > 0, that is, 4 27

( 2b3 - 3 abc + a2 d )' + ( ac -)2 l3 ,O, that is, 4a6 a

E21) Here B = - 126, A r -15.

Thus, the equation has 1 real and 2 coinplex roots.

Adding ( ax + b )2to both sides, we get x4+(a2-3)x2+(2ab-42)x+b2-40 - (a~+b)~ Assume that the left hand side is ( x2 + k )2. (Note that the coefficient of x3 in the given equation is 0.)

Then x4+(a2-3)x2+(tb. 4?)x+b2-40 = x4+k2+2kx2. Co~nparingcoefficients, we get

Eliminating a and b we get (21)~ (2k+3) (k2+40) = 2k3+3k2+80k+120

:. :. 2k3 t 3k2 + 8Ok - 321 = 0. 3 is a root of this equation. With this value of k we get

a2 1: 9, b2 - 49, ab - 21. These equations are satisfied by a - 3, b = 7. Thus, solving the given quartic reduces to solving the followi~lgquadratic equations : xZ+3= 3x+7andx2+3*= -(3x+7),thatis, x2-3x-4 = Oand x2+3x+10 = 0.

-3*iCK Thus, the required roots are 4, -1 and 2 a Solutions ot Polynomial Equations b) The given equation is equivalent to

The resolve~ltcubic is 8k3 - 33k2 + 42k - 17 - 0. One real root is 1. With this value of k, we find that a *O, b=0. Thus, the given equation becomes 5 (x2--x+ I)~0. 2 - Therefore, the given equation has the roots 11 2, 2, 2, - 7 that is, two pairs of equal roots. 2 c) x4 + 12x - 5 = 0. The resolvent cubic is k3 + Sk - 18 = 0. A real root is k = 2. Then, solving the given equation reduces to solving (x2+2)= f(2x-3),thatis

xZ-2x+5 = 0andx2+2x-1 = 0. Thus, the required roots are 2 -2tm , that is, 2 2

E23) a) x4-2x2+8x-3 = 0. Since there is no x3 term, we don't need to apply Step 1. Now assume x4 -2x2+8x-3 = (x2+kx+m) (x2-kx+n).

Thus, eliminating m and n, we get k6-4k4+ 16k2-64 = 0.

k2 a 4 is a root of this cubic in k2. Thus, k = 2 is a solution. For this value of k, we get n = 3, m = -1. Thus, the roots of the given equation are the roots of x2 + 2x - 1 = 0 and x2-2x+3 = ~,thatis,therootsare-l*fi and -1 kifi.

b) This equation can be rewritten as (~+2)~-15~~-40x-26= 0. Putting x + 2 = y, we get y4-15y2+20y-6 = 0. Then the cubic in k2 is k6 - 30k4 + 249k2- 400 = 0, that is, t3 - 30$ + 249t - 400 = 0, putting k2 = t. One real positive root is t = 16. So we can take k = 4. Then we need to solve the quadratic equations = O and y2+4y-2 = 0.

Thus,y = 3, 1, -2tG. Thus, the roots of the given equation are (y-2), that is, 1, - 1, - 4 fi. Cubii and

Thecubicink2is k6-6k4+ 17k2-36 = 0.

kZ= 4 is a root. So we can take k = 2 Then we need to solve the equations 2 x2+2x+2 = 0, x -2x-1 = 0. :. x =-l*i,l*fi. d) 1,2, -3, -4.

E24) Putting x2 = y in the equation, we get

2y4+5y3-5y-2=0. , Then, by either Ferrari's or Descartes' method, we can find the four values of y, -1 which are 1, -1, -2, - 2 .

putt& these values in x2 = y, and solving, we get the 8 rbots of the given equation. Thus, the required roots are

E25) Let the roots be a, b, c, d. Then ad = bc. Now, we know that i) a+b+c+d = -15 - (a+d)+(b+c) = -15 ...... (30) ii) ab+ac+ad+bc+bd+cd = 70 - (a+d) (b+c)+ad+bc = 70 ...... (31) iii) abc + abd + acd + bcd - 120 ...... (32) iv) abcd = 64 Now,(32) =s ad(b+c)+bc(-a+d)= - 120 =s ad(a+b+c+d) = -120 - - 15 ad = - 120 =s ad = 8. Thus, ad = 8 = bc. Then(31) - (a+d) (b+c) = 70-16 = 54. This, with (30) tells us that a + d and b + c are roots of x2 + 15x + 54 = 0. Thus, by the quadratic formula,

Then, ad = 8 and bc = 8 tell us that a and d are zeros of x2 + 6x + 8 = 0, and b and c are zeros of x2 + 9x + 8 = 0.

E26) Let the roots be a, b, c, d, where

a+b = c+d. We know that

abcd = s P 2r Then (36)* -(ab+cd) =r a ab+cd = -. 2 P