inscribe two chords, AB = a, AD = b

c, both from any common assumed point A. Then, produce AD to F so that DF = c, and about the centre c of the former circle, with the radius CF, describe another circle, cutting the chords AD, AB, produced, in F, E, G, H: so shall AG be the affirmative and AH the negative root of the equation x2 + ax-be = 0; and contrariwise AG will be the negative and AH the affirmative root of the equation x2 bc0.

ax

For, AF or AD + DF = b, and DF or AE = c; and, making AG or BH = x, we shall have AH = a + x: and by the property of the circle EGFH (theor. 61 Geom.) the rectangle EA AF GA. AH, or bc (a + x)x, or again by transpo

sition x2+ ax bc

=

= 0. Also if AH be =-x, we shall have

AB =

AG or BH or AH
x2+ax, as before.

AH =

a: and conseq. GA . AH = So that, whether AG be = x, or

x, we shall always have x2 + ax − bc = 0. And by an exactly similar process it may be proved that AG is the negative, and AH the positive root of r2 -ax bc = 0.. Cor. In quadratics of the form x2 + ax − bc = 0, the positive root is always less than the negative root; and in those of the form r2 0, the positive root is always greater than the negative one.

2. The third and fourth cases also are comprehended under one method of construction, with two concentric circles. Let x2 + ax + bc = 0. Here describe any circle ABD, whose diameter is not less than either of the given quantities a and b + c; and within that circle inscribe two chords AB α, AD = b+c, both from the same point A. Then in AD assume DF = c, and about c the centre of the circle ABD, with the radius CF describe a circle, cutting the chords AD, AB, in the points F, E, G, H: so shall AG, AH, be the two positive roots of the equation rax + bc = 0, and the two negative roots of the equation x2 + ax + bc=0. The demonstration of this also is similar to that of the first

case.

Cor. 1. If the circle whose radius is CF just touches the chord AB, the quadratic, will have two equal roots; which can only happen when 4a2 = bc.

Cor. 2. If that circle neither cut nor touch the chord AB, the roots of the equation will be imaginary; and this will always happen, in these two forms, when be is greater than a2.

PROBLEM

PROBLEM III.

To Find the Roots of Cubic and Biquadratic Equations, by Construction.

1. In finding the roots of any equation, containing only one unknown quantity, by construction, the contrivance consists chiefly in bringing a new unknown quantity into that equation; so that various equations may be had, each containing the two unknown quantities; and further, such that any two of them contain together all the known quantities of the proposed equation. Then from among these equations two of the most simple are selected, and their corresponding loci constructed; the intersection of those loci will give the roots sought.

Thus it will be found that cubics may be constructed by two parabolas, or by a circle and a parabola, or by a circle and an equilateral hyperbola, or by a circle and an ellipse, &c and biquadratics by a circle and a parabola, or by a circle and an ellipse, or by a circle and an hyperbola, &c. Now, since a parabola of given parameter may be easily constructed by the rule in cor, 2 th. 4 Parabola, we select the circle and the parabola, for the construction of both biquadratic and cubic equations. The general method applicable to both, will be evident from the following description.

ΜΙ

M

D

B

M

2. Let M'AM'M be a parabola whose axis is AP, M'M'GM a circle whose centre is c and radius CM, cutting the parabola in the points M, M, M", M": from these points draw the ordinates to the axis MP, M'P', M"P", M"P"; and from c let fall cp perpendicularly to the axis; also draw CN parallel to the axis, meeting PM in N. Let AD DCb, Cмn, the parameter of the parabola P, ARX, PM = y. Then (pa. 31) px = y2: also CM2 CNNM2, or n= (x = a)2 + (y b); that is, x2 ± 2ax + a2 + y2 ± 2by + b2 = n2. Substituting in this equation for x, its value and arranging the terms according to the dimensions of y, there will arise

૫૩

p

2

a,

M

y ± (2pa + p2)y2 ± 2bp2y + (a2 + b2 — n2)p2 = 0, a biquadratic equation, whose roots will be expressed by the ordinates PM, PM', P"M", "M", at the points of intersection of the given parabola and circle.

3. To make this coincide with any proposed biquadratie whose second term is taken away (by cor. theor. 3); assume

यु

y1 — q32 + ry — s = 0.

Assume also p = 1; then comparing the terms of the two equations, it will be, 2a-1=q,

9+1

or a = -2b=r, or b == ; a2 + b2 — n2 = − s, or

, 2

9

n2 = a2 + b2 + s, and consequently n = √(a2 + b2 + s). Therefore describe a parabola whose parameter is 1, and in at right angles to it draw DC and

the axis take AD =

---

9+1
2

r; from the centre C, with the radius √(a2+b2 + s), describe the circle M'M'GM, cutting the parabola in the points M, M', M", M""; then the ordinates PM, P'M', P"M", P"M", will be the roots required.

Note. This method, of making p= 1, has the obvious advantage of requiring only one parabola for any number of biquadratics, the necessary variation being made in the radius of the circle.

Cor. 1. When DC represents a negative quantity, the ordinates on the same side of the axis with c represent the negative roots of the equation; and the contrary.

Cor. 2. If the circle touch the parabola, two roots of the equation are equal; if it cut it only in two points, or touch it in one, two roots are impossible; and if the circle fall wholly within the parabola, all the roots are impossible.

-

Cor:3. If a + b2 n2, or the circle pass through the point A, the last term of the equation, i. e. (a2 + b2— n2)p2=0; and therefore y1 ± (2pa + p2)y2± 2bp3y = 0, or y3± (2pa + p2)y ± 2bp2 = 0. This cubic equation may be made to coincide with any proposed cubic, wanting its second term, and the ordinates PM, P"M", P"M' are its roots. Thus, if the cubic be expressed generally by y3±qy±s=0. By comparing the terms of this and the preceding equation, we shall have 2pa + p2 = ± 9, and ± 26p2 = ± s, or Fap F and b So that, to construct a 2p' cubic equation, with any given parabola, whose half parameter is AB (see the preceding figure): from the point B take, in the axis, (forward if the equation have -q, but backward if be positive) the line BD ⇒ ; then raise the perpendicular

9

±

9

2p

DC = 2p2, and from ç describe a circle passing through the vertex A of the parabola; the ordinates PM, &c, drawn from the points of intersection of the circle and parabola, will be the roots required.

PROBLEM

PROBLEM IV.

To Construct an Equation of any Order by means of a Locus of the same Degree as the Equation proposed, and a Right Line.

-As the general method is the same in all equations, let it be one of the 5th degree, as x3-bx++ acx3 — á2dx2 + a3ex -af0. Let the last term afbe transposed; and, taking one of the linear divisors, f, of the last term, make it equal

to z, for example, and divide the equation by at; then will

z=

x5 bx4 + acx3- a2dx4 + a3ex

a4

On the indefinite line вQ describe the curve of this equation, BMDRLFC, by the method taught in prob. 2, sect. 1, of this chapter, taking the values of x from the fixed point B. The ordinates PM, SR, &c, will be equal to z; and therefore, from the point в draw the right line BA =f, parallel to the ordinates PM, SR, and through the point A draw the indefinite right line KC both ways, and parallel to BQ. From the points in which it cuts the curve, let fall the perpendiculars MP, RS, ca; they will determine the abscissas BP, BS, BQ, which are the roots of the equation proposed. Those from A towards a are positive, and those lying the contrary way are negative.

If the right line AC touch the curve in any point, the corresponding abscissa x will denote two equal roots; and if it do not meet the curve at all, all the roots will be imaginary. If the sign of the last term, af, had been positive, then we must have made z = -f, and therefore must have taken BA=-f, that is, below the point è, or on the negative side.

EXERCISES.

Ex. 1. Let it be proposed to divide a given arc of a circle into three equal parts.

Suppose the radius of the circle to be represented by r, the sine of the given arc by a, the unknown sine of its third part by x, and let the known arc be 3u, and of course the required arc be u. Then, by equa. VIII, IX, chap. iii, we shall have

་

Putting, in the first of these equations, for sin 3u its given value a, and for sin 2u, cos 2u, their values given in the two other equations, there will arise

3 sin u

cos2 u. sin3 u

r

Then substituting for sin u its value x, and for cos2 u its value r2x2, and arranging all the terms according to the powers of x, we shall have

x3 — r2x + žar2 = 0,

a cubic equation of the form x3- px + q = 0, with the condition that p3 > 4q2; that is to say, it is a cubic equation falling under the irreducible case, and its three roots are represented by the sines of the three arcs u, u + 120°, and u + 240°.

Now, this cubic may evidently be constructed by the rule in prob. 3 cor. 3. But the trisection of an arc may also be effected by means of an equilateral hyperbola, in the following manner.

Let the arc to be trisected be AB. In the circle ABC draw the semidiameter AD, and to AD as a diameter, and to the vertex A, draw the equilateral hyperbola AE to which the right line AB (the chord of the

H

E

arc to be trisected) shall be a tangent in the point A; then the arc AF, included within this hyperbola, is one third of the arc AB.

For, draw the chord of the arc AF, bisect AD at G, so that G will be the centre of the hyperbola, join DF, and draw GH parallel to it, cutting the chords AB, AF, in I and K. Then, the hyperbola being equilateral, or having its transverse and conjugate equal to one another, it follows from Def. 16 Conic Sections, that every diameter is equal to its parameter, and from cor. theor. 2 Hyperbola, that GK. KI = AK2, or that GK: AK:: AK: KI; therefore the triangles GKA, AKI are similar, and the angle KAI=AGK, which is manifestly Now the angle ADF at the centre of the circle being equal to KAI or FAB; and the former angle at the centre being measured by the arc AF, while the latter at the circumference is measured by half FB; it follows that AF = FB, or = AB, as it ought to be.

Ex. 2.

= ADF.

Given the side of a cube, to find the side of another of double capacity.

Let the side of the given cube be a, and that of a double oney, then 2a3=y3, or, by putting 2a=b, it will be a2b=y3: there are therefore to be found two mean proportionals be

tween