Page images
PDF
EPUB

9. Find the equation whose roots are greater by 4 than the corresponding roots of 3+ +16x3 + 72x2 +643 — 129=0.

10. Solve the equation 3x3 -- 22x2 +48%-32=0, the roots of which are in harmonical progression.

11. The roots of 23 – 11x2 +362 — 36=0 are in harmonical progression; find them.

Remove the second term from the equations: 12. 28 — 6x2 + 10x - 3=0. 13. 24 +4x3 + 2x2 - 4x —2=0. 14. Transform the equation 28. =0 into one whose roots

4 4 exceed by

3

[ocr errors]

15. Diminish by 3 the roots of the equation

25 — 4x+ + 3x2 - 4x+6=0. 16. Find the equation each of whose roots is greater by unity than a root of the equation 23 — 5x2 +6x—3=0.

17. Find the equation whose roots are the squares of the roots of

204 + x3 +222 ++1=0. 18. Form the equation whose roots are the cubes of the roots of

23 +3x2+2=0. If a, b, c are the roots of 23+qx+r=0, form the equation whose roots are 19. ka-1, kb-1, kc-1.

20. boca, ca?, a2b2. 21. b+c, c+a, a+b.

22 a? 62 Solve the equations : 23. 2x4 +- 23 — 6x2+x+2=0. 24. 24 — 10x3 +26x2 — 10x+1=0. 25. 205 — 5x4 +9x3 — 9x2 + 5x – 1=0. 26. 426 — 24xc5 +574 — 73.03 +5722 – 24x+4=0.

DESCARTES' RULE OF SIGNS.

562. When each term of a series has one of the signs + and – before it, a continuation or permanence occurs when the signs of two successive terms are the same : and a change or

variation occurs when the signs of two successive terms are opposite.

[ocr errors]

DESCARTES' RULE. 563. In any equation, the number of positive roots cannot exceed the number of variations of sign, and in any complete equation the number of negative roots cannot exceed the number of permanences of sign.

Suppose that the signs of the terms in a multinomial are ++ .t -+-+-; we shall show that if this multinomial is multiplied by a binomial whose signs are +-, there will be at least one more change of sign in the product than in the original multinomial.

Writing down only the signs of the terms in the multiplication, we have ++

+-+-
+
++

-+-
+ +++ + +

++-Ft - FF+-+-+ a double sign, spoken of as an ambiguity, being placed wherever there is a doubt as to whether the sign of a term is positive or negative.

Examining the product we see that

(i) an ambiguity replaces each continuation of sign in the original multinomial;

(ii) the signs before and after an ambiguity or set of ambiguities are unlike;

(iii) a change of sign is introduced at the end.

Let us take the most unfavorable case and suppose that all the ambiguities are replaced by continuations; from (ii) we see that the number of changes of sign will be the same whether we take the upper or the lower signs; let us take the upper; thus the number of changes of sign cannot be less than in

++--+---+-+-+, and this series of signs is the same as in the original multinomial with an additional change of sign at the end.

If then we suppose the factors corresponding to the negative and imaginary roots to be already multiplied together, each factor x- a corresponding to a positive root introduces at least one

change of sign; therefore no equation can have more positive roots than it has changes of sign.

To prove the second part of Descartes' Rule, let us suppose the equation complete and substitute - y for x; then the permanences of sign in the original equation become variations of sign in the transformed equation. Now the transformed equation cannot have more positive roots than it has variations of sign, hence the original equation cannot have more negative roots than it has permanences of sign.

Whether the equation f(x)=0 be complete or incomplete its roots are equal to those of f(-x) but opposite to them in sign; therefore the negative roots of f(x)=0 are the positive roots of f(-x)=0; but the number of these positive roots cannot exceed the number of variations of sign in f(-x); that is, the number of negative roots of f(x)=0 cannot exceed the number of variations of sign in f(-x).

We may therefore enunciate the Rule of Signs as follows:

An equation f(x)=0 cannot have more positive roots than there are variations of sign in f(x), and cannot have more negative roots than there are variations of sign in f(-x).

Example. Consider the equation 29 +5x8 — 23+73+2=0.

Here there are two changes of sign, therefore there are at most two positive roots.

Again f(-x) = –29+5x8 +33 — 7x+2, and here there are three changes of sign, therefore the given equation has at most three negative roots, and therefore it must have at least four imaginary roots.

564. It is very evident that the following results are included in the preceding article.

(i) If the coefficients are all positive, the equation has no positive root; thus the equation 205+23 +2x+1=0 cannot have a positive root.

(ii) If the coefficients of the even powers of x are all of one sign, and the coefficients of the odd powers are all of the contrary sign, the equation has no negative root; thus the equation

27 +25 — 2x4 + x3 – 3x2+72-5=0 cannot have a negative root.

EXAMPLES L. e. Find the nature of the roots of the following equations : 1. 24+2x3 – 13x2 – 14x +24=0. 2. 24 — 10x3 +35x2 – 50x +24=0.

3. 3x4 + 12x2 +5x—4=0.

4. Show that the equation 2.7 — 24+4x3–5=0 has at least four imaginary roots.

5. What may be inferred respecting the roots of the equation 210 — 4x6 +244-28-3=0 ?

6. Find the least possible number of imaginary roots of the equation 29 — 206+ 24 +22+1=0.

DERIVED FUNCTIONS. 565. To find the value of f(x+h), when f(x) is a rational integral function of x.

Let f(x)=P02" +2,02-1+pyQ"-2+ +Pn-1x+Pn; then f(x+h) =Po(a+h)n +P1(x+h)n-1+P2(x+h)n-2+

+Pn--1(x+1)+P. Expanding each term and arranging the result in ascending powers of h, we have Porn +2222-1+PqXn-2+ ... +Pn--1x+pn

+h{npoxn-1+(n-1)P,2n^2+(n— 2)p22n–3+ +Pn-1}

h2
+2{n(n − 1)Poem-2+(n-1)(n2)P,27-3+ ... +2pn-2}

+

hin
+ {(n-1)(n-2)...2.1pok:
This result is usually written in the form

h2
h3

hin
f(x+h)=f(x) +hf' (x) + f'(x) + f''(x) +
12

f'x) 134

Ino and the functions f'(x), f'(x), f'(x),... are called the first, second, third, ... derived functions of f(x).

ha

we see that to obtain f'(x) from f(x) we multiply each term in f(x) by the index of x in that term and then diminish the index by unity.

Similarly we obtain f"(), f'(),...

Examining the coefficients of h, ***

EQUAL Roots. 566. If the equation f(x)=0 has r roots equal to a, then the equation f'(x)=0 will have r– 1 roots equal to a.

[ocr errors]

Let $(x) be the quotient when f(x) is divided by (x - a)"; then f(x)=(x-a)'(x). Write x+h in the place of x; thus f(a+h)=(x-a+h)$(x+h);

h2
:: f(x) +hf'(x)+5f"(x) + .

2

h2 -a)

12 In this identity, by equating the coefficients of h, we have

f'(x)=r(ic-a)-1(x)+(2-a)'*'(x). Thus f'(x) contains the factor x—a repeated r-1 times; that is, the equation f'(x)=0 has r-1 roots equal to a.

Similarly we may show that if the equation f(x)=0 has s roots equal to b, the equation f'(x)=0 has s-1 roots equal to b; and so on.

From the foregoing proof we see that if f(x) contains a factor (x-a)", then f'(x) contains a factor (2-a)?-?; and thus f(x) and f'(x) have a common factor (x-a)-1. Therefore if f(x) and f'(x) have no common factor, no factor in f(x) will be repeated; hence the equation f(x)=0 has or has not equal roots, according as f(x) and f'(x) have or have not a common factor involving x.

567. It follows that in order to obtain the equal roots of the equation f(x)=(, we must first find the highest common factor of f(x) and f(x), and then placing it equal to zero, solve the resulting equation.

Example. Solve the equation 24 — 1128 +44x2 — 76x+48=0, which has equal roots. Here

f(x)=x+ - 1103 +44x2 - 76x +48,

f" (2)=423 — 33x2 +884— 76 ; and by the ordinary rule we find that the highest common factor of f(x) and f'(x) is 3 —2; hence (x - 2)2 is a factor of f(x); and

f(2)=(x-2)(302–7+12)

=(x-2)(x-3)(3-4); thus the roots are 2, 2, 3, 4.

« PreviousContinue »