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Example. If u is a real quantity, prove that the expression 22 +230 - 11

can have all numerical values except such as lie between 2 (x2 – 3) 2 and 6. Let the given expression be represented by y, so that

2? + 2x – 11

=y;

2 (x2 – 3) then multiplying across and transposing, we have

x2 + 2x (1 y)+by - 11=0. This is a quadratic equation, and if « is to have real values 4 (1 - y)2 – 4 (6y - 11) must be positive; or simplifying and dividing by 4, ya - 8y +12 must be positive; that is, (y-6) (y 2) must be positive. Hence the factors of this product must be both positive, or both negative. In the former case y is greater than 6; in the latter y is less than 2. Therefore y cannot lie between 2 and 6, but may have any other value.

In this example it will be noticed that the expression ye - 8y+12 is positive so long as y does not lie between the roots of the corresponding quadratic equation ya — 84+12=0.

This is a particular case of the general proposition investigated in the next article.

308. For all rcal values of x the expression ax2 + bx+c has the same sign as an except when the roots of the equation ax2+bx+c=0 are real and unequal, and x lies between them. CASE I. Suppose that the roots of the equation

axl+bx+c=0 are real; denote them by a and B, and let a be the greater.

6 Then ax: +bx+c=a ( 203 +

x+

(

a

=a {x?- (a+)x+aß} [Art. 304.]

=a (x a) (x B). Now if x is greater than a or less than B, the factors x-a, 2 B are either both positive or both negative; therefore the expression (x a) (x –B) is positive, and ax2 +bx+c has the same sign as a. But if x lies between a and B, the expression

(x-a) (-8) is negative, and the sign of ax2 +bx+c is opposite to that of a.

CASE II. If a and ß are equal, then

ar+bx+c=a (x-a)?, and (r. – a)” is positive for all real values of x; hence a.ro +bx+c has the same sign as a.

CASE III. Suppose that the equation a.ro+bx+c=0 has imaginary roots; then

b ax2 +bx+c=a22+

- +
a

+

b\2 4ac - 6.27
+ +
2a

4a2 ji but since 12 – 4ac is negative [Art. 303], the expression

b )2 4ac- 62 x + +

2a) 4a2 is positive for all real values of x; therefore ax? +bx+c has the same sign as a.

EXAMPLES XXXIV. a. Find (without actual solution) the nature of the roots of the following equations:

1 1. x2 +0 - 870=0. 2. 8+6x=522. 3. x2=14 - 3x2

sa,

a.

12.

0, 8

4. x2 +7=4x. 5. 2x= x2 +5. 6. (x+2)=4c+15.

Form the equations whose roots are 7. 5, – 3.

8. -9, -11,

9. a+b, a -b.
3 5
2 4

7 10.

11. 2' 6'

3 5 13. If the equation x2 +2 (1+k) x+ka=0

has equal roots, what is the value of k? 14. Prove that the equation

3mx2 (2m +3n) x + 2n=0 has rational roots. 15. Without solving the equation 3.22 — 4x – 1=0, find the sum,

the difference, and the sum of the squares of the roots. 16. Show that the roots of a (22–1)=(6 - c) x are always real. Torm the equations whose roots are

6 17. 3+1/5, 3-5. 18. – 2+13, -2- 3. 19.

a

[blocks in formation]

If
a,

Bare the roots of the equation p.x2+2x+r=0 find the

values of 23. a? +ß2. 24. (a-3)?

25. a ß+aß?.

a 26. a4 +84 27. a582 +aß5.

28.

B

[ocr errors]

+

a

a

29. If a, 3 are the roots of 22 – px+9=0, and a', B3 the roots of

22 - Px+Q=0, find P and Q in terms of p and q. 30. If a, ß are the roots of 22 — ax+b=0, find the equation whose roots are

B

B?' a? 31. Find the condition that one root of the equation

axl+bx+c=0 may be double the other. 32. Form an equation whose roots shall be the cubes of the

roots of the equation 2x (x a)=a”. 33. Prove that the roots of the equation

(a+b) 22 (a+b+c)x+;=0 are always real. 34. Show that (a+b+c).x2 – 2 (a+b)x+(a+b-c)=0)

has rational roots. 35. Form an equation whose roots shall be the arithmetic and

harmonic means between the roots of x2 px+q=0. 36. In the equation px2+2x+r=0 the roots are in the ratio of l to in, prove that (1? + m2) pr+lm (2pr -2)=0.

x? – 15 37. Show that if x is real the expression

cannot lie

2X – 8 between 3 and 5.

3x2+2 38. If x is real, prove that

can have all values except 22 - 2x -1

3 such as lie between 2 and

2

MISCELLANEOUS THEOREMS. 309. We shall now give general proofs of the statements made in Art. 55. We suppose n to be positive and integral.

1. To prove that x" — y" is always divisible by x-y.

Divide an – yn by x - y till a remainder is obtained which does not involve x. Let Q be the quotient, and R the remainder; then

21-y=2(x - y) + R. Since R does not contain x, it will remain unaltered whatever value we give to x.

Put- x=y, then y" - y"=Qx0+R; .: R=0; that is, since there is no remainder, 21- yn is divisible by x-y.

II. To prove that x"+yn is divisible by x+y when n is odd, but not when n is even. With the same notation as before,

2"+y"=2(x+y)+R. Now since R does not involve x, it will remain unaltered whatever value we give to x. Put x=

(-y)"+y"=Qx0+R; that is,

R=(-y)"+y".
(1) if n is odd, (-y)"+y=-y"+y"=0;

(2) if n is even, (-y)n + yn= yn +yu =2y" ; hence there is a remainder when n is even, but none when n is odd; which proves the proposition.

In like manner it may be proved that xn - yn is divisible by x+y when n is even ; and xn Fyr is never divisible by x - y.

The results of the present article may be conveniently stated as follows:

(i) For all values of n,

x" yn=(x - y) (22-1+am-2y+2n3y2 + ...... tyn-1). (ii) When n is odd,

21 +yn =(x+y) (2n-1 – 2n-2y+wn–3y2 . (iii) When n is

even,
qin yn = (x+y)(xn-1 – 2n-2y +.xn-3y2

-y, then

+yn-).

- yn2).

310. The Remainder Theorem. If any algebraical expression

x"+P23"-1+P2x9-2+Pzx"-8 + ...... +Pn-1X+Pn be divided by x-a, the remainder will be

a" +pan-1+pga"-9 +Pga"-3+.... + Pn-ja + Pn Divide the given expression by_x – a till a remainder is obtained which does not involve x. Let Q be the quotient, and R the remainder; then

200+P12-1+PqX-?+ ...... +P:-1&+Px=(2-a)+ R. Since R does not contain x, it will remain unaltered whatever value we give to X. Put x=a, then

an +Pan-1+pan-2+ +Pr-1a+Pn=Qx0+R,

R=an + Pan-1+p,an-2+. +Pn-1a + Pni which proves the proposition.

From this it appears that when an algebraical expression is divided by x – a, the remainder can be obtained at once by writing a in the place of x in the given expression.

311. The remainder is zero when the given expression is exactly divisible by x- a; hence we deduce :

The Factor Theorem. If any algebraical expression involving x become equal to 0 when a is written for x, it will contain x-a as a factor.

Example 1. The remainder when x4 – 2x3 + x - 7 is divided by X + 2 is

(-2)* -2(-2)+(-2) - 7; that is,

16+16 – 2 – 7, or 23. Or the remainder may be found more shortly by substituting x= -2 in [{(x - 2)x} x + 1] x – 7.

Example 2. Resolve into factors x3 + 3x2 – 13.x – 15.

Dy trial we find that this expression vanishes when x=3; hence 2-3 is a factor. .. 23 + 3x2 – 13x – 15= x2 (x – 3) + 6x (x – 3) +5 (x – 3)

=(x – 3) (x2 + 6x+5)

= (x - 3) (t+1) (-x + 5). Note. The only numerical values that need be substituted for a are the factors of the last term of the expression. Thus, in the present case, by making trial of - 5, we should have detected the factor x+5.

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