REDUCTION OF AFFECTED QUADRATIC EQUATIONS. k 96. An affected quadratic equation is that which contains both the first and second powers of the unknown quantity; every equation of this kind is comprehended under one of the three following forms, viz. First form, x2+ax= b. Where is the unknown Second form, x2. b. quantity to be found, and a Third form, x2-ax=—b.) and b known quantities. The value of the unknown quantity x, in each of these three forms, is found by one general method, as follows. 97. To find the roots of affected quadratic equations. RULE I. Range the terms of the given equation according to the dimensions of the unknown quantity; namely, let the term containing the square stand in the first place, (to the left,) and that containing the first power in the second, on that side of the equation in which the square will be affirmative. II. Transpose all the known quantities to the other side of the equation; and if the square of the unknown quantity have a multiplier or a divisor, it must be taken away by the methods employed in simple equations. III. Take half the coefficient of the unknown quantity in the second term, square it, and add this square to both sides of the equation, then will that side which contains the unknown quantity be a complete square '. * Quadratic is derived from the Latin quadratus, squared. The term adfected, or affected, (from affecto to pester or trouble,) was introduced by Vieta, the great improver of Algebra, about the year 1600: it is used to distinguish equations which involve, or are affected with different powers of the unknown quantity, from those which contain one power only, which are therefore called pure. Dr. Hutton sometimes calls the former compound equations: this term the venerable and learned Baron Maseres highly approves of, observing that it is less obscure, and therefore more proper, than that of affected or adfected equations. 1 Since the square of every simple quantity is a simple quantity, and the square of every binomial is a trinomial, it follows that no quantity in the form of a binomial can be a complete square; but that, in order to make it such,` another terin must be added to it, which term may in every case be found, from IV. Extract the square root from both sides of the equation, prefixing to that of the known side the double sign ± m. V. Transpose the known part of the root; this incorporated (according to the import of its sign) with the root of the known side, will be the value required. Note. The root of the unknown side is readily found by taking the roots of the first and third terms, and connecting them by the sign of the second". the two given ones: to make this appear, let a + b be any binomial, the square of which is a2 +2ab+b2; now suppose a2+2 ab to be given, if half the coefficient (2b) of a in the second term, viz. b, be taken, and squared, this square, viz. b2, will be the remaining term; in like manner if b2+2ab be given, if half the coefficient (2 a) of b in the second term be squared, its square a2 will be the remaining term; wherefore, since this quantity being added to both sides of the equation does not affect their equality, this part of the rule is manifest. TM The reason for the double sign is this-every square has both an affirmative and a negative root; thus a2 may arise either from the multiplication of + a into + a, or -a into -a; therefore a2 has both + a and -a, or+a for its roots, and the same is evidently true in general. ■ Because the square root of a2 +2ab+b2 is a+b, and that of a 2-2 ab +b2 is nb, it follows, that in every complete square, the root of the first term (a) connected with the root of the third term (b2) by the sign of the middle term (2 ab), will be the root of the square: thus in the above example, a+b is the root in the first instance, and a-b is the root in the second. In the first form of quadratics, where x2 +ax=b, x will always have two values, one affirmative, and one negative, and the negative value will be the greatest; for from the solution of the above equation, we have a= 土 √b+aaa; now the former of these values, namely, + √/b+aa-a, will be affirmative, since baa is always greater than aa, or its equal a. The second value, namely, —√b+aa-a, being composed of two negative terms, will evidently be negative; moreover, since the affirmative root is the difference of the two quantities, (√b+aa and a,) and the negative root their sum, it follows that the negative root will be the greatest. In the second form, where x2-ax=b, x will likewise have two values, one affirmative, and one negative, and the affirmative value will be the greatest; for from the solution = ·±√b + ÷ aa+a, the first value of x, namely, + √b+aa+a, being composed of two affirmative terms, will evidently be affirmative. The second value of x, namely, ➡/b++ aa +±a, will always be negative, for since b+aa is greater than (aa, or than its qual)a, it follows that~√/b+aa+a will be negative; and because EXAMPLES. 1. Given x2+6x=40, to find the values of x. Half the coefficient of x in the second term, is (=) 3, this squared is (32) 9, which being added to both sides, the equation becomes x2+6x+9=(40+9=) 49. The square root of which is x+3= (±√49) ±7. the former root is the sum, and the latter the difference, of two affirmative quantities, it follows that the affirmative root is the greatest. In the first and second forms, the quantity under the radical sign is always affirmative, and therefore it can never happen that either of the roots in these two forms is impossible. In the third form, where x2—ax=—b, or == ±√aa-ba, there are three cases; b may be either greater than aa, equal to aa, or less than aa; if b be greater than aa, the quantity under the radical sign will be negative; and since the square root of a negative quantity is impossible, both values of r will evidently in this case be impossible. = nothing, If b be equal to aa, the quantity under the radical sign will be and will have but one value, namely, a. But if b be less than aa, x will have two values both affirmative; for the first value of x, namely, + aa-ba, will be the sum of two affirmative terms, and will therefore be affirmative. 4 The second value of r, namely, √ aa- ba, will likewise be affirmative; for since aa is greater than 6, it is plain that ✓aa, or its equal a, is greater than aa-b, and therefore-aa-b+a will always be affirmative therefore, when x2—ax——b, if a be greater than b, we shall have x= + √1 aa−b+3a, and x=— √ aa—b+a, both affirmative values of a: hence this is sometimes called the ambiguous form. Either of these roots, whether affirmative, negative, or impossible, will answer the algebraic conditions of the equation from whence it is derived; but in the application of quadratics to the solution of problems, impossible roots always imply inconsistency in the conditions, or that the problem, as to any real use, is impossible. The affirmative roots, in the first and second forms, are in most cases the answers to the question proposed; sometimes however the negative roots are to be taken, when the quadratic forms a subordinate part of some more extensive solution: in the application of algebra to geometry, both the affirmative and negative roots have place, each having a distinct and necessary import. Of the two affirmative roots in the third form, one only will for the most part be the answer to a numerical problem, the conditions of which will always point out which it is in geometrical problems both roots have place, as we have already observed, Whence, by transposition x= (±7—3=) 4, or−10, the answer. 2. Given x2-8x=20, to find x. Half the coefficient 8 is 4, this squared is 16, and the square added to both sides, gives x2—8x+16= (20+16=) 36. By evolution x-4=(±√36=) ±6. By transposition x=(+6+4) 10, or -2, the answer. 3. Given x2-4x=-3, to find x. Half the coefficient 4 is 2, this squared is 4, which added to both sides, we have x2-4x+4= (−3+4=) 1. Whence by evolution x−2= (±√l=) ±1. And by transposition x= (2+1=) 3, or 1, the answer. 4. Given 2x2+16 x=40, to find x. Divide by 2, and x2+8x=20. Complete the square, and x2+8x+16= (20+16=) 36. Whence, by transposition x= (±6—4=) 2, or —10, the ans. 5. Given 5 x2—60 x—12=788, to find x. By transposition 5 x2 - 60 x= (788+12=) 800. By division x2—12 x=160. Complete the square, and x2—12x+36= (160+36=) 196. By transposition x= (6+14=) 20, or -8, the answer. 6. Given 8x2+8x-4=12, to find x. By transposition 8 x2+8x= (12+4) 16. By division x2+x=2: here the coefficient of x is 1, half of |