5. 2x-9x+ 10 9x +2 · 2020. 12. 19. · Example 2. Find the quotient and remainder when 2 +3- 15+ 2x- 4 is divided by x + 5. The calculation is arranged as follows: 2 3 0 - 15 2 -

THEORY OF EQUATIONS 327

2. 3x - 3x - 1 = 0.

3.4x 20x+33x- 20x + 4 =0. (B.Sc.1990) 4. 60x"- 736x +1433x - 736x + 60 = 0. (B.Sc.1991)

5. 2x-9x+ 10x - 3x+ 10 9x +2 0.

6. 2x-x - 2x - x + 2 = 0.

7.x++++X+ 1 = 0. (B.Sc.1989) 8. 2x+x +x+2 = 12x (x +1).

9.6+11x 33x3 33+11x +6 =0. (B.Sc.1988) 10.x+4x +x*+*+4x +1 0.

11. 5x* +9x-9+5X 1 0.

12. -5x + 5x - 1 = 0.

13. x10 3x+5x 5x+3- 1 = 0. (B.Sc.1988)

14. 6x-25x+31x-31x+25x-6 = 0. (B.Sc.1990) 15. x+2x + 2x -2x 2X-1 = 0. 16.x-x - &+x + 1 = 0.

S 17. To increase or decrease the roots of a given equation

by a given quantity. Let the roots of the given equation

fx) = ax"+a,x-1+a-2 + +an 0

be a.a2, a3,.. , n and suppose we require the equation

whose roots are a1-h,a2-h, d3 -h,.. , an- h.

We have f (X) = ao (X - a1) (* - a2)... (X- an).

in this if we change x into y +h, we have f(y+h) = ao (y+h - a1) (0+h - a2)... (V+h -an).

The right-hand side vanishes when y ar-h, (r = 1,2, .,n).

Hence if an equation is to be transformed into another whose roots are those of the first diminished by h, substitute y +h for x in the given equation. Then we obtain the transformed equation as

ag (+h+a, (y+h'+a2 (y+h)"+...+a,=0 .(1)

I719. 24, b,

328 A TEXT BOOK OF ALGEBRA

Expanding and collecting the coefficients of the powers od let the equation be of y

Aov+A"+Ay-+ .. + An-1y + An 0 (2) Here Ao, A. A2... , An-1.An are functions of

Since y = X - h, this equation (2) is equivalent to

Ap (X h" + A (X- h +Az (x - h)n-2 4

+An-1( - h) + An 0. This equation (3) must be identical with the given equation (1). This form suggests an easy rule for calculating Ao A1.A2.. , An

..

(3)

We can easily see that Ao = ao If the poiynomial on the left-hand side (3) is divided byx - h, the remainder is An and the quotient is

Ao (X h)"-+ A (x -h-+Az (x - h-s+.. +An-2 X - h) +An-t

If the quotient again is divided by x - h, the remainder is An-1 and the quotient is

Ao (X h) +A x - h) + ... An-3 (x - h) +An-2 By continuing this process we can find all the coefficients of the

transformed equation (2).

Instead of diminishing the roots if we desire to increase them. we takeh negative.

18. Form of the quotient and remainder when a polyno mial is divided by a binomial.

Let the quotient when

f (x) = a0 X" +a1 X +az X +...+an

is divided by x - h be

boX'+ b1x+b2x + ... + bn-2X + Dn- This we shall represent by Q and the remainder by A.

We have then the following equation-t (x) = (x - h) Q + R.


(X-h) Q + R

(box+ b,x+. + bn-2X + bn-1) (X- h) +R

= bo +(b, - hbo) x+ (b2 -hb,) x

+(-1-hbn-2)x+ R- hbn-1 Equating the coefficients of corresponding powers ofx on both

sides, we get the following series of equations to determine

Do b1,D2 ,Dn-1.R

b1- hbo a1, i.e., b, = a1 + hbo = a +hag

ba- hb = a2, ie., b2 = a2 + hb

b3- hb2-a3. 1.e., ba = a3 + hb2

Dn-2- nD-3an-2. 1e., Dn-2 an-2 t nDn-3

n-1-hb,-2 = an-1. 1., Dn-1an-1+ hb-2

R hbn-1 @n R = an +nDn-1

These equations supply a ready method of calculating in suc-

cession the coefficients bo. b1 ,b2. , Dn-1 of the quotient and

the remainder R. For this purp0se we can write the series of opera-

tions as follows:- an an-1 a1 a2 a3

n-2.h n-1.h agh b,h bh

b2 ba bn-1 R

In the first line the successive coefficients of the given equation

are written. The first term in the second line is obtained by multiplyingg

ao byh. The product agh is placed under a and then added to it in

order to obtain the termb, in the third line. This term, when obtained,

is multiplied in turr byh and placed under a2. The product is added

to a2 to obtain the second term b2 in the third line. The repetition of

this process furniches in succession all the coefficients of the

quotient, the last term thus obtained being the remainder.

b1


Example 1. Find the quotient and remainder when

3x+8x + 8x+ 12 is divided by x - 4.

The calculation is arranged as follows 8 3 8 12

4 12 80 352

3 20 88 364

The quotient is 3x + 20x +88 and the remainder is 364 Example 2. Find the quotient and remainder when

2 +3- 15+ 2x- 4 is divided by x + 5. The calculation is arranged as follows:

- 15 2 3 0 2 4 -- 5

10 35 - 175 875 4300 21490 2 - 7 35 175 860 - 4298 21486

The quotient is 2x- 7x+35x - 175x+860x- 4298 and the remainder is 21486.

Example 3.Diminish the roots of x-5x+7*-4x+5 = 0 by 2. The coefficients in the transformed equations are the remainders when the polynomial is divided by x - 2 in succession. The division of the polynomial by x -2 can be exhibited as foliows

1 -5 7 5 2 2 6 2 4

- 3 1 -2 1 The quotient is x - 3x+X - 2 and the remainder is 1. . 1 is the absolute term in the transformed equation. The coefficient of x is the remainder when x- 3x +X - 2

divided by x - 2.

The calculation is arranged as follows -3 2 2

2 2 2 -

1 - 1 1 4


The quotient is x -x - 1 and the remainder is -4.

When this is divided by x - 2, we get

1 -1 | 1 2 2 2

1 1

The quotient is x +1 and the remainder is 1.

When x +1 is divided byx -2, we get

2 1 1

2

1 3

The remainder is 3.

The coefficients of the transformed equation are

1, 3, 1, - 4 and1

The transformed equation is

+3 +*- 4x+ 1 0.

All these operations can be combined and exhibited as follows

7 4 5 1

- 5

2

3

2

Example 4. Increase by 7 the roots of the equation

3x+7x -15x+X -2 0.

Increasing by 7 the roots of the equation is the same as

diminishing the roots by- 7.

(B.Sc.1989)


2 3 7 15

4060 581 -580

-21 98 4058

-14 83

-21 245 -2296

-35 328 -2876

392 -21 -56 720

-21 -77


3x 77x + 720x - 2876X + 4058 0.

Example 5. Show that the equation

-3x3+ 4 2x+1 = 0 can be transformed into a reciprocal equation by diminishing the

roots by unity. Hence solve the equation.

The operation of diminishing the roots by 1 can be exhibited as

follows- 3 - 2

1

0

The transformed equation isx +x ++x + 1 = 0 which is a reciprocal equation.

The equation can be written as

() +1 0.

Writingx+=z, we get z-2+z+ 1 = 0


i.e., z+Z - 1 = 0

i.e., Z=1:V5

2

ie.x+=tv5 or x+1--1-V 2 X

i.e., 2x+ (1 - V5) x+ 2 = 0 or 2x + (1 + V5) x +2 = 0

V5 -1 V(1 -V5) - 16

i.e., x= 4

- (1 + V5) + V(1 + V5- 16

or 4

V5 1 t V- 10-2 V5 i.e., X= 4

-(1 V5) + V2 V510 or- 4

The roots of the original equation are these roots increased

by 1.

V5+3 V- 10 2 V5 3- V5 v2 V5 -10

They are 4 4 Exercises 48

1. Diminish by 3 the roots of the equation

4x+ 3x 4x+ 6 = 0. (B.Sc.1987)

2. Transform x- 5x+ 7x - 4x +5 = 0 into another

equation whose roots are less by 3.

3. Find the equation each of whose roots exceeds by 2 a root

of the equationx-

4x+3x1 = 0.

4. Find the equation whose roots are the roots of

x-5x+ 7-17x + 11 0

each diminished by 2.

5. Find the equation whose roots are the roots of

4x-2x+7x-3=0 each increased by 2.

334 A TEXT B0OK OF ALGEBRA

6. Find the equation whose roots are the roots of x-- 10x4x + 24 = 0

increased by 2 and hence solve the equation. 7. Find the equation whose roots exceed by 2 the roots of the equation 4x" +32x+83 + 76x +21 = 0 and hence solve the equation. et the equattor wiiOSe roots are equat to those of the equation x-6x-x+30X-20 0 each diminished by 3. Given that two of the roots of the original equation differ by 3, solve it completely.

(B.Sc.1990)

9. Show that the equation x"+ 5x + 9+5x1 = 0 can be transformed into a reciprocal equation by diminishing the roots by 2. Hence solve the equation.

(B.Sc.1993) 10. Solve the equationx+ 4x- 2x 12x 2 =0 by trans forming this equation into another whose roots are increased by unity. 11. Find the equation whose roots are those of the equation 2x 25x +111x - 208x + 140 0 diminished by 3. Hence or otherwise solve the given equation. 12. Find the equation whose roots are the roots of the equation x+8x+ 12x 16X 28 =0 each increased by 2. Hence solve the equation.

S 19. Removal of terms. One of the chief uses of this transformation is to remove a certain specified term from an equation. Such a step always helps to find the solutions of an equation. Let the given equation be

+a,x'+ ax +... + an-1X +an= 0. 2

Then if y = X - h, we obtain the new equation ao +h) +a1 +h)+a2 (y+h +. + an = 0 which, when arranged in descending powers of y, becomes

ag+(nagh + a) y-1 + n (n-1 agh + (n 1) ah + aay-2+ = 2


If the term to be removed is the second, we put

a1 nagh +a1=0 so that h =

nao If the term to be removed is the third, we get

n(n ah+ (n - 1) ash +a2 0 2!

and so obtain a quadratic to find h ; and similarly we may remove

any other assigned term.

Example 1. Find the relation between the coefficients in the

equation x + px + qx + m +S = 0 in order that the coefficients

of 3 and x may be removable by the same transformation.

Let us reduce the roots of the equation by h. Instead of x

substitute x + h,the transformed equation is

(x+h) +p (x + h) + q (x + h) +r (* +h) +S =0.

i.e., x+ (4h +p)x+ (6h +3ph + q) X

+ (4h+3ph + 2qh + )x +h*+ ph + qh + rh +s =0.

The coefficients of x and x in the transformed equation are

zeros.

4h +p = 0. 4h+3ph+2gh +r= 0.

Eliminate h between these equations.

We get h -

. 4(-23 + 3p (2+29 (- +r=0

i.e., p-4pq +8 0.

Example 2.Solve the equation

x+20x-143 +430x+462=0

(B.Sc.1991) by removing its second term.

Let us assume that by diminishing the roots by h, the second

term is removed.


Then the transformed equation becomes

(K+h)"+ 20 (x+h) + 143 (x+h)+430 (x+h) + 462 = 0

i.e., x+ (4h + 20) x+.. = 0

4h +20 = 0, h= - 5.

Hence to remove the second term, increase the roots of the

equation by 5. 430 462 -5

1 20 143 - 450

- 340

90 75

15 68 12

50 90 10 18 0

5 -25 -7

The transformed equation is y" - 7 + 12 = 0

i.e. -3) (-4) = 0. The roots of the transformed equation are V3, +2.

These roots are greater than the roots of the original equation

by 5. . The roots of the original equation are

V3 -5, - v3 -5,2-5, -2-5

i.e.,-5 t V3, -3, - 7.

Exercises 49

1. Remove the second term from the following equations

(1) ax+ 10X 3 =0.

(2) +5x + 3x + *+x+ 1 = 0.

(B.Sc.1987)

2. Transform the equation x - 4x- 18- 3x + 2 0 into

one which shall want the third term.


3. Solve the following equations by removing the second term

in each:

(1) x- 12x +48x - 72x + 35 0. (B.Sc. 1987)

(2) x+4x + 5x + 2x -6 = 0

(3)x+16x+83x+ 152x +84 = 0. (4)x- 12x + 48 72 = 0.

(5)+6x+12x 19 = 0. 6)x-8x+19- 12x +2 =0.

(B.Sc.1993)

(B.Sc.1992)

(7) x- 21x+144x- 320 = 0

(8)x- 8x - x+ 68x +60 = 0. 4. Show that it is possible to remove the second and the third

terms simulteneously of an equation of the n degree, provided n

times the sum of the squares of the roots is equal to square of the

sum of the roots. Discuss the particular case when n = 3.

S 20. To form an equation whose roots are any power of

the roots of a given equation.

The method of forming such equations is illustrated in the

following examples.

Example 1. Find the equation whose roots are the squares of

the roots of the equation

+px-"+ px +... + Pn-1X+Pa0. Let a1, a2, ... , an be the roots of equation. Then we have

+px+px'+.. +Pn-1X+pn=(*-a) -a2)...(-n).

If we transform the equation into another whose roots are the

negatives of the roots of this equation, we get

X'-px''+px -... = (+X+a1) (+X*a2)...(+X+an). -2

-a -a.. -ah)

+p-2+p*+... - (p,x+p +...

It is evident that the left-hand side when expanded contains only

even powers of x.

A-22


22. If the roots of the equation x - ax + bx" - abx + 1 = 0

are a, ,Y, ð, show that (a+p+ v) (a +B +d) (a + y + ð) (ß +y +d) = 1

S23. S23. Without actually solving an equation, it is possible to determine the nature of the roots of that equation. The following articles wHgive the various methods of determining the nature of the rogis of the equation.

24. Descartes' Rule of signs. An equation í (x) =0 cannot have more positive roots than

there are changes ef sign in f (x).

Let f () be a polynomial whose signs of the terms are

++ +- +++- + -.

In this there are seven changes of sign including changes from +to - and from - to +. We shall show that if this polynomial be

multiplied by a binomial (corresponding to a positive root) whose

signs of the terms are +-, the resulting polynomial will have atleast one more change of sign than the original. Writing down only the

signs of the terms in the multiplication, we have

++ +- + + + - ++

+++- + +t- +++ -+t+ - + - +

Here in the last ine tha ambiguous sign is placed wherever there are two different signs to be added. Here we see in the product

(1) an ambiguity replaces each continuation of sign in the

original polynomial (2) the sign before and after an ambiguity or a set of ambiguities

are unlike ; and

(3) a change of sign is introduced in the end.

Let us take the most unfavourable case and suppose that all

the ambiguities are replaced by continuations, then the sign of the

terms become

++- - + - +++ -

The number of changes of sign is 8. Thus even in the most

unfavourable case there is one more change of sign than the number of changes of sign in the original polynomial. Therefore we may


conclude in general that the effect of multiplication of a binomial

tactorx- a is to introduce at least one change of sign. Suppose the product of all the factors corresponding to nega

tive and imginary roots of f (X) = 0 be a poiynomial .Ihe effect of multipluing F () by each of the factors x -a,X - B, x - y,.

Corresponding to the positive roots, a , B,y is to introduce at least

one change of sign for each, so that when the complete product is formed containing all the roots, we have the resulting polynomial

which has at least as many changes of signs as it has positive roots. This is Descartes' rule of signs.

S 24.1. Descartes' rule of signs for negative roots.

Let f () = (- a1) (* - ag).. ( - Gn).

By substituting x instead of x in the equations, we get

f(-x)= (-x - a1) (-x- ag).. (-X - an). The roots of f (-x) = 0 are - a, a2 .,n

. The negative roots of f (X) = 0 become the positive roots of

f(-x) = 0.

Hence to find the maximum number of negative roots of f(x) = 0, it is enough to find the maximum number of positive roots

of f (-x) = 0.

So we can enunciate Descartes rule for negative roots as

follows. No equation can have a greater number of negative

roots then there are changes of sign in the terms of the polynomial f (-X).

24.2. Using Descartes' rule of signs we can ascertain whether an equation f (*) = 0 has imginary roots or not.

We can find the maximum number for positive roots and also for negative roots. The degree of the equation will give the total numbers of roots of the equation. So if the sum of the maximum numbers of positive roots and negative roots is less than the degree of the equation, we are sure of the existence of imginary roots. Take

for example the equation x +8x - x +9 =0. The series of signs of the terms are as follows:

++ - +

The number of changes of signs is 2 and the equation canno have more than two positive roots.


Now change x into - x, we get

- x+8x+x + 9 = 0

i.e., x+ & - x - 9 = 0.

The series of signs of the terms are ++

and the number of changes of sign is only one and so the equation cannot have more than one negative root.

Hence in the equation there cannot exist more than three real roots. Since it is a seventh degree equation, it has seven roots real

or imaginary. Therefore the given equation has at least four imginary roots.

S24.3. An equation f (x) = 0 is called complete when all

powers ofx from n to the constant term are present. In an complete equation, we can easily see that the sum of the number of changes

of sign in f (x) and f(-x) is exactly equal to the degree of the

equation. Hence this rule can be used to detect the imginary roots

only in incomplete equations.

Example. Determine completely the nature of the roots of the

(B.Sc.1994) equationx - 6x - 4x + 5 = 0.

The series of signs of the terms are+ -

Here there are two changes of sign.

Hence there cannot be more than two positive roots.

Changingx into x, the equation becomes

- - 6 + 4x +5 =0

i.e., + 6x 4x - 5 = 0.

The series of the signs of the terms aree

+ +

Here there is only one change of sign.

There cannot be more than one negative root.

So the equation has got at the most three real roots. The total

number of roots of the equation is 5. Hence there are at least two

imginary roots for the equation. We can also determine the limits

between which the real roots lie.

A-23


6. Determine the value of a such that the equation

x- 12x +a = 0 has only one real root.

7. Find the range of values of k for which the followinn

equations have real roots

(1) x+ 4x+ 5x + 2 +k = 0.

(2) 2x- 9+ 12x-k = 0.

(3) 3x 4x - 12x+k = 0.

(4) x- 14x+24x- k = 0.

(5) x+4x - a+ k = 0.

(6) 3+& - 6x- 24x + k =0.

(7) 2x3 15 + 36x + k = 0.

8. Discuss the nature of the roots of the equation

3x+8x - 30x - 72x+k = 0

for different values of k. 3 9. Show that the equation 3x+8x - 6x - 24x+ r = 0 has

four real roots if 13 <r< - 8, two real roots if 8<r< 19or

k< 13 and no real root ifr> 19.

26. Multiple roots. If f(x) is a polynomial in x and the equation f (x) = 0 has m

roots equal to a, then f() must be of the form (- a)#) where (a) * 0.

m f()= - a)¢' ()+m x - a) ) = (X - a)"{x - a) o' (X) + mo () }.

Hence (x - a is a common factor of f (x) and f (x) and t

is easily seén that (X- a) m-will not be a common factor unless f (x) is divisible by ( m Hence the multiple roots off (x), t an

are to be deducted by finding the greatest common factors of and f X) by the usual algebraic process. We may then state a

for finding the multiple roots of an equation f (x)0 as folloWS (1) Find f' (*).

(2) Find the H.C.F. of f (x) and f (). (3) Find the roots of the H.C.F.


Each different root of the H.C.F. will occur once more in 1 X)

than it does in the H.C.F.

Example 1. Find the multiple roots of the equation - 9x + 4x + 12 =0. (B.Sc.1988)

f (x) = x - 9x + 4x + 12

f (x) = 4x - 18x + 4.

The H.C.F. of f (X) and f' (x) is easily found to be x- 2.

Hence (x - 2) is a factor of f (X)

The remaining factors are easily ascertained.

Thus we find f (X) = (X- 2) (x + 1) (x +3).

The roots of f (X) = 0 are 2, 2, - 1 and - 3.

Example 2. Find the values of a for which

ax- 9x + 12x - 5 0

has equal roots and solve the equation in one case.

f (x) = ax - 9x + 12x - 5.

f (X)= 3ax - 18x+ 12.

Find the H.C.F. of

ax 9x+12x-5

and 3ax - 18x + 12.

The process of finding the H.C.F, is exhibited below

3ax-18x+12 -a xax 9x+12x -5

3ax- 8ax + 5a ax-6+ 4x

-3x+8x 5 (8a-18)x+12-5a

-3(8a-18)x +8(8a-18)x-5(Ba-18)

3 (8a 18) x + 3 (12 - 5a) x

(49a-108) x - 5 (8a - 18)

Iff (X) and (X) have a common linear factor

49a- 108 5 (8a- 18) 12- 5a 8a- 18B


2 (4) 8x - 20x+6x+ 9 = 0. (5) x -x - 4x +7x-3 = 0 (6) 27x 72x +64x16 0. (7) 12x+ 40x +39x+9= 0. 2. Find the value of k for whichx +4x +5X +2 +k = 0 hasS equal roots. Also find those roots.

3. Find the condition that ax - 9+ 12x 2 = 0 may have a double root and find that root.

4. Prove that the equation x - 3qx +2r = 0 will have twoequal roots if g' =

5. Show that 3 +qx + r = 0 has two equal roots if

27+4q = 0. Verify that this equation 4x- 27x +27 = 0 satis-

(B.Sc.1990)

fies the condition and find all its roots.

6. Show that the equation x" - ngx + (n - 1)r = 0 will have a pair of equal roots if q" =

7. Show that the binomial equation x - a" = 0 cannot have equal roots.

8. If any root of the equationf' (x) = 0 be subtracted from each of the roots of the equation f (X) = 0, show that the sum of the reciprocals of the remainder thus obtained is zero, provided that the roots of f (x) = 0 are all different.

9.If the equation x"+p,x+px"+...+pn-X+Pn=0 has a double root a, show that a is a root of the equation

1+2p +3p + ... + npn = 0 10. If the equation*'+p,x+pX+... +Pn-1XtP= has three roots each equal to a, show that a is a root of

n +(n-1) P +(n-2)p>+...+Pn-1=0. 5 27. Strum's Theorem. Let f (x)= 0 be an equation having no equal roots. Letf1 () De the first derived function of f (x). Let the process of finding tn8 greatest common measure of f (x) and f (X) be performed. Let 41 be the quotient and f2 (*) the remainder with the sign changed.

HEORY OF EQUATIONS 363

f() = qf ()12 ).

Similar operation can be performed between f (X) and f2 (*)

) = 92/2 ()- fa ()

If we continue the operations, we get

Then

and we get

fr-1 )= q,f, (X) - f+1() and so on.

The successive remainders with their signs changed, i.e., the

functions f2 (X) , f3 (),.. will go on diminishing in degree till we

reach a numerical remainder, say, fn ).

These functions f().f, (*). fz (). Ím ) are called

Strum's functions.

The difference between the number of changes of sign in

the series of Strum's functions when a is substrtuted for x and

the number whenbis substituted for x express exactily the

number of real roots of the equation f (X) = 0 between a and b.

This is known as Sturm's Theorem and the proof of this theorem is

beyond the scope of this book.

Substitute 0 and o in the series of Strum's functions and the

difference between the number of changes of sign will give the

positive roots.

Substitute - o and 0 in the series of Sturm's functions and the

difference between the number of changes of signs will give the

negative roots.

The difference betvween the number of changes of sign when

- oo and+ 0o are substituted in the series of Sturm's function will

give the number of real roots of the equation.

Example 1. Find the number of real roots of the equation

x-14x+ 16x +9 0.

fx) = X*- 14x + 16x +9

f1 (x) = 4x - 28x + 16

2 ()= 7* - 12x 9

fa (X) = 17x 38

4 (X) = 35


7.96x 16x 6x + 1 0. 8. Solve the equationx - 4x+ 4x +x -2 = 0.

9. Transform the equation 12x+ 16x- 5X 3 = 0 to an

equation having the coefticient of highest power of X unity and the other coefficients integral. Solve the original equation.

10. Solve the equation 3x 23x+35x +31x 30 =0 given that it has more than two integral roots.

11. Solve the equation 2x 7x+6x11x+4x + 6 = 0 given that two of the roots are integral. 12. In an equation f() = 0 in which the coefficient of the

highest term is unity and the other coefficients integers, if f (0) and f(1) be both odd integers, show that the equation cannot have a rational root.

30. Horner's Method. This method can be used to determine both the commen-

surable and the incommensurable roots of a numerical equation. First we shall explain the method for obtaining the positive root. The

procedure is to determine the root figure by figure, first the integra part and then the first decimal place, then the second decimal place and so on until the root terminates or the root has been obtained to the required degree of apprOximation. The main principle invoved in this method is diminishing the roots by certain known quantties by successive transformations. In this method the successive trans formations can be exhibited in a compact form and the roots can be obtained to any number of places of decimals required.

First we have to find by trial two consecutive integers betwee which a real positive root of the equation lies. This will give the integral part of the root. Let it be a. First diminish all the roots of tnE equation by a. Then the transformed equation will have a roo between 0 and 1. In order to avoid decimals in the working, all ne roots of this transtormed equation are multiplied by 10. Then the n transformed equation has a root between 0 and 10. By trial find tn integers between which the root lies and thus find the integra of the root. Let it be b. Then diminish the roots be band again multpy the roots by 10 and continue the process till we get the root to number of decimals we required. The method and presentation be clear from the examples below.

an


Example 1. The equation x - 3x + 1 = 0 has a root between

1 and 2. Calculate it to three places of decimals. Since the root lies between 1 and 2, the integral part of the root

is 1. Diminish the root of the equation by 1.

(B.Sc.1988)

to an nd the

1 - 3 (1

B0= 0 -2 1

6 0 2

of the ) and

The transformed equation is x+3x 1 0.

This equation has therefore a root between 0 and 1.

Multiply the roots of this equation by 10.

ave a

Then the equation transforms into x+ 30x 1000 = 0.

We can easily see that a root of this equation lies between 5

and 6. Diminish the roots of the equation by 5.

men-

ation. . The egral blace 30 0 -1000 (5 1

875 ed to lved

175 175

5 35

- 125

tities ans- n be

200 375

5 40

5 45

een the The transformed equation is x + 45x +375x 125 = 0.

This equation has therefore a root between 0 and 1.

Multiply the roots of the equation by 10.

Then the equation transforms into

2

the root the

new the +450x + 37500x - 125000 = 0.

We can easily see that a root of this equation lies between 3 partiply the will

and 4.


Diminish the roots of this equation by 3. 1 450 37500 - 1250000 (3

3 1359 116577 - 8423 453 38859

3 1368 456 40227

3 459


+459x + 40227x - 8423= 0. Multiply the roots be 10. Then the equation transforms into

x+4590x+ 4022700x 8423000 = 0. We can easily see that a root of this equation lies between 2

and 3.

Diminish the root be 2. 4590 40227000 - 8423000 (2

8063768 - 359232

2 9184 4592 4031884

2 9188

4594 4041072

2 4596


x+4596x+ 4041072X 359232 = 0. Multiply the roots by 10.

Then the equation transforms into

45960x+ 404107200x 35923200 = 0. We can easily see that a root of this equation lies between 0

and 1. We can stop with this since we require the root correct to three decimal places. Thus the root correct to three decimal places is 1.532. In the actual presentation we need write only the coefficients of the various transformed equations omitting completely the powers of x. The series of arithmetical operations is represented as

follows:


0 (1.5320 3

2

1 2 - 1000

875 2 0 - 125000

116577 - 8423000

175

30 175 8063768

359232000 5 200

35 37500

5 1359 40 38859

1368 4022700

5

450

3 9184 453 4031884

9188 3

456 404107200

3

4590

2 4592

2 4594

2

45960

Example 2. Find the positive root of the equation

- 2- 3x 4 = 0

(B.Sc.1994) correct to three places of decimals.

By Descartes' rule of signs, there can be at the most onlv one

positive root and we can easily see tnat t lies between 3 and 4. The

process is exhibited as follows:


4 (3.2842 2 3

4000

12 2688 1200 - 1312000

1242752 - 69248000

3 144 70 1344

148 64746224 72 149200 4501776000 3243903688

1257872312 2 6144 74 155344 2 760

6208 16155200

31356 8 768

16186556 8

31392 776 1621794800

8 157044

1621951844 7840

7844 4

7848

4 78520

2 78522

The roots correct to three decimal places is 3.284. 30.1. The above two examples show that it is necessary in

each case to determine the required figure by actual substitutioi The numerical work involved is considerable. In the two examples we note that after two transformations the tigures in the roots are obtained by dividing the first coefficient from the end by tne second coefficient from the end. This hint of practical use is base

Newton's method of approximation.

Documents

5. 2x-9x+ 10 9x +2 · 2020. 12. 19. · Example 2. Find the quotient and remainder when 2 +3- 15+ 2x- 4 is divided by x + 5. The calculation is arranged as follows: 2 3 0 - 15 2 -