Imágenes de páginas
PDF
EPUB

When besides c positive, b is negative, the roots are both plus.

From (1) When besides c positive, b is positive, the roots

are both minus.

When c is negative, the roots are necessarily real and of unlike signs. We shall illustrate the application of these inferences by a few examples.

[merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors][ocr errors][merged small][merged small][ocr errors][merged small][merged small][ocr errors][ocr errors][merged small]

82. It appears from the properties (1) and (2) above, that when any quadratic equation ax2+bx+c=0 is divided by the

b

leading coefficient a, and thus reduced to x2+ -x+

α

с

α

a=0, the coefficient of the second term is always equal to the sum of the roots with their signs changed; and that the third term is always equal to the product of the roots; we know also that the product of two simple factors, (-), (x-'), is a quadratic expression x2—(r+r')x+rr', of which the coefficient of the second term is the sum of and r' with their signs changed, and the third term equal to their product. Consequently if r, r', represent the roots of the quadratic equation x2+-+--0, then it must

b

b

α

C

necessarily follow that a2+ 2x + 2 = (x—r') (x—r'); so that

a

α

the significant side of every quadratic equation, of which the leading coefficient is unity, is compounded of two simple factors, which, when separately equated to zero, furnish the two roots of that quadratic, the factors being x minus one root, and x minus the other. If the quadratic be not divided by the leading coefficient a, then ax2+bx+c=a(x-—r)(x—r'); and hence, to decompose any quadratic into its simple factors, all we have to do is to find the roots r, r' of the equation, for the required factors will be (x-r), (x-2), and a. And conversely, to construct a quadratic, with prescribed roots r, ', we have

I

[ocr errors]

only to multiply (x-r) and (x-') together; or rather to write down 2 for the first term, the sum of r and r', with changed sign, for the coefficient of the second term, and the product rr' for the third term, and then to equate the expression to zero; thus the quadratic whose roots are 2 and 3 is x2-5x+6=0; that whose roots are 2 and 3, is x2+x-6=0; that whose roots are 5 and -7, is x2+12x+35=0, and so on.

If the roots are imaginary, or r=p+q√−1, r'=p—q√—1, then (x—r) (x—r') = (x—p)2 + q2, the sum of two squares; hence, in this case, whatever real number be put for x, the first member of the equation will be positive.

83. QUESTIONS TO BE SOLVED BY QUADRATIC EQUATIONS WITH ONE UNKNOWN QUANTITY.

(1) A person bought a certain number of sheep for £72; and found that if he had bought six more for the same money, he would have paid £1 less for each: how many did he buy, and what did he pay for each?

72

Let a represent the number bought, then is the price of

[ocr errors]

each in pounds; but if he had bought 6 more, the price of each 72 would have been

x+6

72

.. by the question. =

72

-1..72x=72x+432-x2-6x;

x+6 Ꮳ

.. transposing, x+6x=432.. completing, x2+6x+9=441; Extracting the root, x+3=±21 .. x=18 or —24;

Hence the number of sheep purchased was 18, and the price of

[blocks in formation]

This is an instance in which a particular restriction in the question is not introduced into the algebraical statement of it; the question requires not only that we discover a number fulfilling certain conditions, but that the number may be positive. In representing this number by x we impose no such limitation, and we thus arrive at the widest interpretation of it possible. The algebraic statement is equally confirmed, whether be 18 or —24; but from the peculiar nature of the question we see that the negative value of x must be rejected; the only admissible answer being the positive one.

(2) A company at a tavern had £8 15s. to pay: but before the bill was paid, two of them left, and those who remained had in consequence 10s. each more to pay: how many were in the company at first?

175

Suppose were the number, then shillings is the fair

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors]

.. 10x2-20x350, or x2-2x=35; Completing the square, a-2x+1=36.. x-1=6; .. x=7, or —5;

Consequently, there were seven persons at first.

(3) A vintner sold 7 dozen of sherry and 12 dozen of claret for £50, and finds that he has sold 3 dozen more of sherry for £10 than he has sold of claret for £6. Required the price of each.

Let x be the price in pounds of a dozen of sherry ..

10

[ocr errors]

is

the number of dozens of sherry for £10; and by the question, 10-3x

10

-3, or.

is the number of dozens of claret for £6.

[ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors]

6x

7x+12

10-3x

10-3x

50, or 70x-21x+72x-500-150x;

Transposing, 21-292x=-500; hence by Rule II. p. 178, 42x-292-(-42000+2923)=√43264208;

[merged small][merged small][ocr errors][merged small][merged small][merged small]

Hence the price of a dozen of sherry was £2, and of a dozen of

[blocks in formation]

£3. If the other value of x be taken, the ex

pression for the number of dozens of claret will be negative; this value is, therefore, inadmissible.

(4) Divide a given number a into two parts, such that their product may be a given number b; and find the parts when their product is the greatest possible.

Let x be one part and .. a—x the other; then by the question, x(α-x)-b2, or x2-ax——b2;

.. Rule II. 2x—a=√ (a2 —4b2) .. x=

part; and a-x:

a±√(a2—4b3), one

2

[merged small][merged small][ocr errors]

If 462 > a2, these values are imaginary; in this case, therefore, the proposed division is impossible. The greatest possible value of b2 is that which makes a2-4b2=0; that is, it is

a2

4

since a greater value gives imaginary results. For this

value of b2 the parts are a and a; so that the product is the greatest when the two parts are equal.

(5) Find two numbers whose difference is 9, and such, that their sum multiplied by the greater produces 266.

(6) Find two numbers whose sum is 41, and the sum of whose squares is 901.

(7) Divide 20 into two parts, such that the product of the whole and one of the parts may be equal to the square of the other part.

(8) A person bought a certain number of sheep for £60; and having reserved 15, sold the remainder for £54, gaining two shillings a head by them: how many sheep did he buy, and what was the cost of each ?

(9) A draper bought two pieces of cloth of different textures; the finer cost 4 shillings a yard more than the other. The sum paid for the finer was £18, and the sum paid for the coarser, which was two yards longer than the other, was £16: how many yards were there in each piece, and how much did a yard of each cost?

(10) Find a number such, that if three times its square root be taken from five times its fourth root, the remainder may be . (11) Twenty persons, men and women, receive £48: the men £24 and the women £24; but the men receive £1 each more than the women. How many men were there?

(12) Into what two parts must 39 be divided, so that the sum of their cubes may be 17199?

185

CHAPTER VIII.

ON CERTAIN EQUATIONS OF THE THIRD AND FOURTH DEGREES WHICH ARE REDUCIBLE TO QUADRATICS.

84. As already noticed, equations of the third degree, or those in which the third, but no higher power of the unknown quantity enters, are called cubic equations. Equations of the fourth degree are called biquadratic equations. General formulæ for the solution of these equations, analogous to the formula already given for quadratics, and which would serve for the determination of the roots in every particular case, have been hitherto sought for in vain. Algebraists have met with but partial success in this department of research; the formulæ that they have discovered leading to the accurate determination of the roots only in particular circumstances. But this defect is not looked at at present with the same concern as it used to be. The stream of science, when thus checked by some unyielding impediment, usually finds, at length, a new channel; and the efforts exercised on the obstacle, withdrawn from their unprofitable expenditure, generally give impetus to the diverted current, and so accelerate its onward progress. In practice the equations proposed to us for solution have numerical values for the coefficients of their terms, and although the solution of equations with letters for the coefficients may be impossible by algebraic formulæ, yet it is quite possible that the solution of numerical equations may be within the powers of common arithmetic. This idea has been very successfully followed up of late years. It was first fully developed by Mr. Horner, of Bath, in 1819, and his researches, amplified a little by the contributions of others, have given such a degree of completeness to this department of inquiry, that little more is either to be expected or desired; we shall give a brief sketch of the arithmetical process hereafter, confining ourselves at present to purely algebraical methods of reducing equations of the third and fourth degrees to quadratics; a subject well worthy of attention though not hitherto fully examined.

85. At page 71 an easy method was proposed for dividing a polynomial of an even degree into two factors each consisting of a rational part and of an irrational part, and differing from one another only in the sign of the latter. The theory of this

« AnteriorContinuar »