We shall in the first place briefly investigate the value of x in this equation, and then point out the connection of the process with the foregoing problem.

Dividing the above equation by c, the coefficient of the first term, we obtain

Let a be a quantity, such that being substituted for x in this expression, it shall fulfil the conditions of the equation by making the first member equal 0; then we shall have

transposing

a2 + pa + q = 0,

substituting this value of q in equation (B), we obtain

and again transposing

but x2

x2 + px · a2 — pa = 0,

a2 + px pa = 0;

· a2 = (x + a). (x − a) [II. 5, cor. 1], and papa = p (x − a), which values being substituted, we have

(x + a + p). (x − a) = 0.

Now it is evident that the first member of this equation can only become equal to 0 by one of its factors becoming equal to 0; that is, either when xα= = 0, or when x+a+p=0; and we thus see that there are two values of x, which will fulfil the conditions of the original equation (A); namely, either x = a, or x = -a - p. These values (a and -a-p) are termed the roots of the quadratic equation, and the symbols being perfectly general, we derive the following proposition:

THEOREM 1. Every quadratic equation has two roots.

If x and x, be put for the two values of x, and they be added together, we have

but p is the coefficient of the second term px in equation (B), therefore:

THEOREM 2. The sum of the roots of a quadratic equation is equal to the coefficient of the second term with its sign changed.

Again, if the two roots be multiplied together, we have

x1 x2

= a. (

p)

ap, and q is the third term of equation (B), therefore:

THEOREM 3. The product of the roots of a quadratic equation is equal

to the third term.

=

substituting this value of x in the equation x + x2 =

P, we have

Then from the equation x1 + x2:

=

p, we have, by transposition, 2 = - P 1, and substituting the value of x, in equation (C), we obtain

Now examining the equation x2 + bx b2 already derived (page 73) from the algebraical expression of proposition xi., we immediately perceive that it is a quadratic equation, and may be reduced to the general form (B) by transposing b2, when we have

in which p b, and q

-b2. In order then to obtain the two roots of this equation, or the two values of a which will fulfil the conditions of the same, we have only to substitute the above values of p and q in the expressions (C) and (D); doing which, the first of these becomes

a result identical with that already obtained at page 73 by a different process. And in a similar manner equation (D) becomes

also identical with the second value of x, previously obtained.

Now recollecting that b AB, and x = AH, in the first case of the geometrical problem, when the point H is taken in the line AB itself,

and in the second case, when taken in the production of AB,

And this problem affords, geometrically, an illustration of the first theorem, enunciated at page 74, that every quadratic equation has two roots.

Now AH, (or AH in the second figure) is, from the construction, evidently equal to the sum of AB and AH, (or AH in the first figure), that is

AH, + AB = AH2;

but AH, is negative, because measured in the opposite direction to AH, (as already explained), therefore

and by transposition,

AH,

AH, + AB

=

and therefore a geometrical proof of the second theorem that the sum of the roots is equal to the coefficient of the second term.

Again, in the geometrical demonstration of the second case of the problem, it is shown that the rectangle (CG) under CK and CF is equal in area to the square (AD) on AB; but CF is equal to the value of AH in the first case (or AH,), and CK is equal to its value in the second case (or AH,); also AB is equal to b, and the square on AB to b2, which is the third term; therefore the product of the two roots (AH, and AH) is equal to the third term (AB2); and we hence derive a geometrical proof of the truth of the third theorem.

2. When a line is divided as in the eleventh proposition, it is said to be cut "in extreme and mean ratio." For the segment BH is to the segment AH as AH is to the whole line AB; and in such a proportion the product of the extremes equals the product of the means, or in this instance the rectangle under AB and BH is equal to the square on AH, which is propo

sition xi. itself.

COROLLARY 1. If a line (AB) be cut in extreme and mean ratio (in C), the greater segment will be cut in the same manner by taking on it a part equal to the less.

DEMONSTRATION. On AC and CB, the two segments, construct the squares ADEC and CFGB (a), and produce GF to H. Because BG is equal to CB (b), the rectangle AG is equal in area to the square AE (c), taking away from both the common rectangle AF, and the rectangle HE is equal in area to the square CG; but DE is equal to EC, therefore HE is the rectangle under EF and EC;

also EC is equal to the greater segment AC, and FC is equal to the less CB; therefore the rectangle under EC and EF is equal in area to the square on CF, and the line CE is cut in extreme and mean ratio.

PROPOSITION XII.

THEOREM.-If a perpendicular be drawn from any of the acute angles of an obtuse-angled triangle (ABC) to the opposite side (BC) produced, the square on the side (AB) subtending the obtuse angle is greater than the sum of the squares on the two sides (BC and CA), which contain the obtuse angle, by double the rectangle under the side (BC), which is produced, and the external segment (CD) between the obtuse angle and the perpendicular.

CONSTRUCTION. Produce BC, and from the acute angle A draw AD perpendicular to BC produced (a).

B

(a) I. 12.

(b) II. 4.

(c) I. 47.

DEMONSTRATION. Because the straight line BD is divided into two parts in C, the square on BD is equal in area to the sum of the squares on BC and CD, together with double the rectangle under BC and CD (b); to each of these equals add the square on AD, and the sum of the squares on BD and AD is equal in area to the sum of the squares on BC, CD, and AD, together with double the rectangle under BC and CD. But because D is a right angle, the square on AB is equal in area to the sum of the squares on BD and AD (c), and the square on CA is equal in area to the sum of the squares on CD and AD (c); therefore the square on AB is equal in area to the sum of the squares on BC and CA together with double the rectangle under BC and CD; that is, the square on AB is greater than the sum of the squares on BC and CA by double the rectangle under BC and CD.

COROLLARY. If in any obtuse-angled triangle (ABC) the sides (AC and CB) which contain the obtuse angle be produced, and perpendiculars be drawn to the acute angles, the rectangle under one of those sides (BC) and the produced part (CD) between the obtuse angle and the perpendicular, is equal in area to the rectangle under the other side (AC) and its produced part (CE).

For it may be proved by the foregoing proposition that double the rectangle under AC and CE is also equal in area to the excess of the square on AB above the sum of the squares on BC and CA; therefore the rectangle under AC and CE is equal in area to the rectangle under BC and CD.

PROPOSITION XIII.

THEOREM. If in any triangle (ABC) a perpendicular be drawn to one of the sides (BC) which contains an acute angle, from the opposite angle, the square on the side (AC) subtending that acute angle is less than the sum of the squares on the sides (AB and BC) which contain that angle, by double the rectangle under the side (BC) to which the perpendicular is drawn, and the segment (BD) between the perpendicular and the acute angle.

CONSTRUCTION. From A draw AD perpendicular to BC, produced if necessary (a).

DEMONSTRATION. Because when a straight line is divided, the sum of the squares on the whole line and one of the segments is equal in B area to double the rectangle under the whole line and that segment together with the square on the other seg

ment (6); therefore the sum of the squares on BD and CB is equal in area to double the rectangle under BD and CB, together with the square on CD; to each of these equals add the square on AD, and the sum of the squares on CB, BD, and AD is equal in area to the sum of the squares on CD and AD, together with double the rectangle under BD and CB. But, because the angles at D are right angles, the square on AB is equal in area to the sum of the squares on BD and AD (c), and the square on AC is equal in area to the sum of the squares on CD and AD (c); therefore the sum of the squares on AB and CB is equal in area to the square on AC together with double the rectangle under BD and CB; that is, the square on AC is less than the sum of the squares on AB and CB by double the rectangle under BD and CB.

SCHOLIUM. Euclid has separated this proposition into three cases, depending upon whether the perpendicular falls within or without the triangle, and employs the twelfth proposition to prove the second case. This division is not, however, necessary, as all the cases may be demonstrated from the seventh proposition, as is done above. By comparing the demonstrations of this and the preceding propositions, it will be seen how nearly identical they are, and they may be combined in one general proposition in the following terms:-"The difference between the square on one side of a triangle and the sum of the squares on the other two sides, is equal in area to double the rectangle under either of these two sides and the segment between the perpendicular on it from the opposite angle and the angle included by thre sides."