5+2✓5 5+2✓5 3+√5 = √9+4/5 2√(5+2√/5)' 2√(5+2/5) 2√(5+2√✅√5) 2+✓5 = 2/(5+2/5) 2√(5+25) (multiplying num. and 2. Required the sum of 5/2-1, and 3/2+2. Ans. 8/2+3. 1 33 3. What is the sum of 12 and sy? 1 32 6. Required the product of 4+2√2 by 2—2? Ans. 2ab(a-b) yb. Ans. 4. Product, a+bo. 7. Multiply a+b√−1 by a-b√✓—1. 9. Required the square root of 51-10/2? Ans. 52-1. 10. Required the 3d power of a-b✅—1. Ans, a3-3a2b−1+3ab2+b3✓—1. 11. Required the square root of 7—24/—1 ? 15. Divide 18+26/-1 by 3+√-1. Quot. 8+6-1. 16. Required the square root of 13—20✓/—3 ? Ans. 5-2-3. Imaginary surds are of great importance, in the investigation of several valuable formula, in the arithmetic of sines. SECTION XI. EQUATIONS IN GENERAL. 114. It has been remarked, page 77, that quadratic equations sometimes admit of more answers than one; the principles on which the ambiguity of quadratic equations depends, are productive of similar results in equations of the higher orders. In the general equation "-p"-1+qxn−2......+r=0, where p, q, r, are supposed to be given, the different values, of which x is susceptible, are called the roots of the equation. 115. If two binomials, as x-a, x-b, be multiplied together, the product x-ax-bx+ab, or x2-(a+b) x+ab, is manifestly a quadratic; if now, x-a=0, or x-b-0, we have evidently x2-(a+b)x+ab=0. On the other hand, the given quadratic equation x2—px+ q=0, may be resolved by making a+b=p, ab=q, and determining a and b. For, on these assumptions being made, the equations become identical, and their conditions are fulfilled by taking x=a, or x=b; the equation x2-px+q=0, has therefore, two positive roots, which are both possible when ip is greater than q, and impossible when q exceeds ip. This expression corresponds Չ to form 3d, A, (art. 62.) 116. Again, x-a×x+b=x2—ax+bx-ab, which being supposed =0, corresponds to x+px-q=0, or to x2-px-q=0, according as b is greater or less than a. The conditions of this equation are answered by taking x-a=0, or x+b=0; the equation has, therefore, two roots, a, and b. These expressions correspond to forms 1, 2, A, (art. 62.). Those equations consequently admit of a negative and a positive root, which are both possible, because when the square is completed, the second member consists of positive quantities. 117. Moreover, (x+a).(x+b)=x2+ax+bx+ab,which being supposed =0, agrees with x2+px+q=0. Hence an equation of this form is resolved by making a+b=p, ab=q, and determining a and b. The roots of this equation are manifestly both negative; and as in (art. 115,) both possible or both impossible. 118. It is worthy of remark, that in the equation x2-px+q=0, the signs are alternately + and, that is, they are twice changed; and the equation has two positive roots. But in the equations (art. 116,) the same sign is once continued, and the sign once changed; and these equations have one affirmative and one negative root. In the equation (art. 117,) the sign is twice continued; and this equation has two negative roots. It also appears, from what is above shown, that every quadratic equation has two roots, which are both possible, or both impossible.* 119. Assuming three factors, (-a).(x—b).(x—c)= x®—(a+b+c)x2+(ab+ac+bc)x-abc, which being supposed =0, will be identical with x-px2+qx-r-0, if p=a+b+c, q=ab+ac+bc, and r=abc. But the con-ditions of the former are answered by making x—a=0, x―b=0, or x-c=0; therefore, the latter has three positive roots, a, b, and c. It has likewise three changes of the signs. 120. If instead of x-c we take x+c, our equation (x-a).(x-b)(x+c)=x3—(a+b—c) x2+(ab-ac-bc)x+ abc=0, will, manifestly, have two positive roots, a and b, and one negative root, C. In this case, if a+b>c, the sign of the second term is, the equation may, therefore, be expressed x-pa+qx+r=0, in which there are two changes of the signs, and one continuation of the same sign. If'c> a+b, the second term will have the sign, but the third, because in that case (a+b). c>(a+b), and therefore, >ab. The equation then becomes x+px2-qx+r=0; having, as before, two changes of the signs, and one continuation of the same sign. 121. A cubic equation, composed of the factors (-a). (x+b).(x+c)=x3 +(b+c—a) x2-(ab-ac+bc)x-abc=0 has plainly one positive root, a, and two negative roots, -b, and -c. But in, this case, if b+c>a, the sign of the second term is +; and the equation may be expressed x+px2+qx―r=0, having one change of the signs, and *In form 3, A, (art. 62,) when ba2, the roots are equal to each other. In forms 1 and 2, if 2a-0, the equation becomes a simple quadratic, and the roots are equal quantities with contrary signs. Every simple quadratic has, therefore, a positive and negative root. two continuations of the same sign. If a>b+c, the second and third terms are both negative, because (b+c) a>(b+c)>bc; the equation, therefore, may be expressed x-px-qx-r=0, having, as before, one change of the signs, and two continuations of the same sign. 122. If we use the factors x+a,x+b,x+c, their product 3+(a+b+c) x2+(ab+ac+be)x+abc, being put =0, the equation may be expressed, x2+px2+qx+r=0, in which there are three continuations of the same sign. The equation has likewise three negative roots, -a,b, and --C. By pursuing this inquiry it will be found, that any equation of this kind, admits of as many roots as there are units in the index of the highest power of the unknown quantity; that the number of positive and negative roots, will be, respectively, equal to the number of changes in the signs, and the number of continuations of the same sign. It likewise appears, that the last term, or absolute number, is the continued product of all the roots with their signs changed. 123. It may be observed, that as every cubic equation is composed of three factors, and every quadratic of two; a cubic equation may always be considered as the product of a simple and a quadratic equation. But (art. 118,) every quadratic has, either two possible, or two impossible roots; hence, a cubic equation, the terms of which are possible, having one impossible root, has two; and, as in the multiplication of compound quantities, containing impossible parts, those impossible parts can disappear only when two like roots are multiplied together, it follows, that every cubic equation, consisting of possible quantities, has, at least, one possible root. 124. In like manner, it appears that every equation consisting of possible quantities, having an odd number of roots, has, at least, one of those roots possible. And that every equation which is made up of possible quantities, has all its roots possible, or an even number of impossible roots. |