taken from the difference of their ages, the remainder will be equal to the age of the younger. What then were their ages? Ans. 30 and 12. 24. To find four numbers such, that the sum of the 1st, 2d, and 3d shall be 13 ; the sum of the 1st, 2d, and 4th, 15; the sum of the 1st, 3d, and 4th, 18; and lastly the sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 25. To divide 48 into 4 such parts, that the first increased by 3, the second diminished by 3, the third multiplied by 3, and the 4th divided by 3, may be all equal to each other. Ans. 6, 12, 3, 27. QUADRATIC EQUATIONS. QUADRATIC Equations are either simple or compound. A simple quadratic equation, is that which involves the square of the unknown quantity only. As ax? = b. And the solution of such quadratics' has been already given in simple equations. A compound quadratic equation, is that which contains the square of the unknown quantity in one term, and the first power in another terni, As ax? + bx = c. All compound quadratic equations, after being properly reduced, fall under the three following forms, to which they must always be reduced by preparing them for solution. 1. x2 + ar = b b The general method of solving quadratic equations, is by what is called completing the square, which is as follows: 1. Reduce the proposed equation to a proper simple form, as usual, such as the forms above ; namely, by transposing all the terms which contain the unknown quantity to one side of the equation, and the known terms to the other; placing the square term first, and the single power second; dividing the equation by the co-efficient of the square or first term, if it has one, and changing the signs of all the terms, when that term happens to be negative, as that term must always be made positive before the solution. Then the proper solution is by completing the square as follows, viz. 2. Complete a.r = 2. Complete the unknown side to a square, in this manner,' viz. Take half the co-efficient of the second term, and square it; which square add to both sides of the equation, then that side which contains the unknown quantity will be a complete square. 3. Then extract the square root on both sides of the equation*, and the value of the unknown quantity will be determined, As the square root of any quantity may be either + or -, therefore all quadratic equations admit of two solutions. Thus, the square root of tn? is either + nor en; for tnx tn and - nx n are each equal to + n°. But the square root of - no, or ✓ –no, is imaginary or impossible, as neither + n nor - n, when squared, gives - no. So, in the first form, x2 + ax=b, where x + ja is found = vb + a2, the root may be either + vo + tu?, or -7b+fa”, since either of them being multiplied by itself produces b + ja'. And this ambiguity is expressed by writing the uncertain or double sign I before 7b+ Lao; thus ==Vb+ a? — {a. In this form, where x = vb.+ da? - ju, the first value of X, viz. r = + vb + ja - ža, is always affirmative; for since fa? + b is greater than ja”, the greater square must necessarily have the greater root; therefore vb+ba? will always be greater than va', or its equal ja; and consequently + Vb+ca - ja will always be affirmative. The second value, viz. x = Nb + ju? - ja will always be negative, because it is composed of two negative terms. Therefore when x2 + ax = b, we shall have x = + vota? - ja for the affirmative value of x, and x abt ta' the negative value of x. In the second form, where x = + vb tas + fa the first value, viz. x = + b + a + La is always affirmative, since it is composed of two affirmative terms. But the second value, viz. vb + a2 + a, will always be negative ; for since b + la is greater than La', therefore vb + jak will be greater than visa, or its equal a; and consequently - 76 +12° + ja is always a negative quantity. • Therefore, ža for determined, making the root of the known side either + or - , which will give two roots of the equation, or two values of the unknown quantity. Note, 1. The root of the first side of the equation, is always equal to the root of the first term, with half the co-efficient of the second term joined to it, with its sign, whether'+ or -. 2. All equations, in which there are two terms including the unknown quantity, and which have the index of the one just double that of the other, are resolved like quadratics, by completing the square, as above. Thus, x4 + ax? = ab, or 22n + axh = b, or r + ax? 3b, are the same as quadratics, and the value of the unknown quantity may be determined accordingly. Therefore, when x2 ax. =b, we shall have x = t.Vb+ 9 + ja for the affirmative value of x; and x = vbitu?+ a for the negative value of x; so that in both the first and second forms, the unknown quantity has always two values, one of which is positive, and the other negative. But, in the third form, where x = # b. + a, both the values of x will be positive, when {a? is greater than b. For the first value, viz, =+vab + a will then be affirmative, being composed of two affirmative terms. The second value, viz. x - VIQ? - b + ja is affirmae tive also; 'for since {ais greater than 12% – b, therefore vaor ja is greater than via - b; and consequently - vla - 6+ La will always be an affirmative quantity. So that, when x2 – ux =- b, we shall have r = + vad 6 + ja, and also x = Vlu -6+ Zu; for the values of r, both positive. But in this third form, if b be greater than 1 a?, the solution of the proposed question will be impossible. For since the square of any quantity (whether that quantity be affirmative or negative) is always affirmative, the square root of a negative quantity is impossible, and cannot be assigned. But when b is greater than Lu”, then ja?'- b is a negative quantity; and therefore its root va -b is impossible, or imaginary; consequently, in that case, x = a=N— b, or the two roots values of t, are both impossible, or imaginary, quantities. VOL. I. R EXAMPLES EXAMPLES. 1. Given era + 4x = 60; to find x. First, by completing the square, r' + 4x + 4 = 64; Then, by extracting the root, x + 2 = $ 8; Then, transpos. 2, gives x = 6 or – 10, the two roots. 2. Given r? – 6x + 10 = 65; to find z. First, trans. 10, gives ro - 6.r = 55; Then trans. 3, gives x = 11 or -5. First by transpos. 20, it is 2.1? + 8.5 = 90; And by compl. the sq. it is 2.2 + 4.3 + 4 = 49; And transp. 2, gives x = 5 or — 9. 4. Given 3x – 3x + 9 8; to find .r. And transp. *, gives x = g or . First by transpos. 30, it is 12 - r = 225; And transp. , gives x = 7 or -63. 6. Given ax? -- bx = c; to find x. b 6 62 Then compl.the sq. gives x? X + + 4a? 6 4ac + bi And extrac. the root, gives x = 2a 4a? b. 4ac + 62 b + 2a 7. Given x4 2ax = b; to find r. First by compl. the sq. gives r4 - 2ax? t a = a +b; And a a с a a Then transp And extract. the root, gives x -a = Iva + b; And extract. the root, gives x = + Vad + b. And thus, by always using similar words at each line, the pupil will resolve the following examples. EXAMPLES FOR PRACTICE. 1. Given r? – 6.3 - 7 = 33; to find x. Ans. = 10. 2. Given x2 5x - 10 = 14; to find x. Ans. I = = 8. 3. Given 5.x2 + 4.2 - 90 = 114; to find r. Ans. X = 6. 4. Given {x?-12 + 2 = 9; to find r. Ans. x = 4. 5. Given 3.r4 _ 2x? = 40; to find x. Ans. .X = 2. 6. Given x - *V x = 1*; to find x. Ans. r = 9. 7. Given 1.x2 + çox = ; to find x. Ans. x = 727766. 8. Given 200 + 4x3 = 12; to find z. Ans. t = 32 = 1*259921. 9. Given x2 + 4x = a + 2; to find x. Ans. * = va+6- 2. QUESTIONS PRODUCING QUADRATIC EQUATIONS. 1. To find two numbers whose difference is 2, and product 80. Then the first condition gives x-y = 2, * These questions, like those in simple equations, are also solved by using as many unknown letters, as are the numbers required, for the better exercise in reducing equations; not aim, ing at the shortest modes of solution, which would not afford so much useful practice. R 2 2. To |