second term. For, by the binomial formula, we have a2 = 2 (y — A)"= y" — Ay"—1 + = A(y - A)"1 (4)- etc., A-1 n n-1 = Ayn-1- etc., and so on. Hence, by addition, we get " + Aan-1+ Ba"-2+ etc. y" + B'y"-2 + C'yn-etc.0, since the coefficient of y1 is 0. 2. It is clear if n = 2, that the preceding transformation 3. It may be noticed, that, in the reduction of a question to a quadratic equation, we can often so represent whatever is to be found, as immediately to obtain a pure quadratic. Thus, if we wish to divide 100 into two such parts, that the sum of their square roots may be 14; then if we represent the roots by 7 + x and 7 x, their sum will equal 14, and we shall get (7 + x)2 + (7x)= 100, or 98 + 2x2 = 100 or a2 = 1 and x = 1. Hence, 7 + x and 7 x become 8 and 6, which are the roots of the sought squares, for 8 + 6 14, and 82 + 6 = 64 + 36 = 100. = (4.) Any equation of the form an+aab, which is such that the index of a in one of the terms is twice as great as it is in the other term, is of such a form that a" may be found by the solution of a quadratic. For if we put "y, we get any; consequently, the proposed equation is reduced to the quadratic y + ay = b. Hence, if r and r' denote the roots of the quadratic, we shall have "r and "r'; consequently, if we extract the nth roots of these equations, we shall have the required values of x, or those values which will satisfy the equation can + ac = = b. Thus, if n = 2, a = 2, and b = 24, the equation in a becomes a1+2=24, and thence the equation in y becomes y+2y=24, whose roots are 4 and 6. Hence, the equations ar and "r' become 2 4 and 2- 6; consequently, extracting the square roots of these, we get x=±2 and x = ± √— 6 for the roots of the proposed equation; the = first two being real and the last two impossible or imaginary. Remark.-An equation which is not given in the form a2 + ax = b, can often be reduced to it. Thus, if we take the equation + √2x+9= 13; then if we multiply its members by 2 and add 9 to the products, we shall get 2x + 9+212x+9= 35, which is of the requisite form. For, by putting 12x+9=y, we have y2+2y= 35; whose roots give y= 5 and y=-7. Hence we have √295 and 2x+9=7; which, by squaring and reducing, give a 8 and x = 20; noticing, that a 20 requires the radical in the proposed equation to be taken with the sign x = (5.) When two or more separate equations, containing as many unknown letters as there are equations, are to be solved, then the general method of proceeding (whatever may be the degree of one or more of the equations) consists in reducing their solution by the methods of elimination given in Simple Equations, to that of a single equation containing one unknown letter. Hence, if the single equation can be solved by the rule for the solution of quadratic equations, the solution of the given equations is reduced to that of a quadratic. Thus, if we take the equations + xya and y-xy= b for an example, then, since the second gives x = y3 — b the first, by the substitution of this value of x, is easily re duced to (32 —1)2+ y2 — b = a, or y a + 3b = ; 2 which, by putting yz, becomes the complete quadratic 22 a + 3b 2 2= ; consequently, the solution of the b2 given equations is reduced to that of a quadratic. When two equations containing two unknown letters are proposed for solution, there are particular methods of elimination (such as those given in Simple Equations, which result from the forms of expressions) that are often preferable to b the general methods. Since the solution of a2 + x = Chas in (2) been reduced to that of the equations x + y = b and xy = с - it is clear that one of the most important of these methods consists in reducing (when it is possible) the solution of the given equations to that of two equations of the forms x + y =a and xy=p; in which a and p are given, while x and y are the unknown letters. For an example of this method, let the equations x + y = a and x3 + y3 = b be given, to find x and y. 3xy = 2, or a2 - 3p = 2; consequently, we have p = a a and the solution of the given equations is reduced to that of the equations x + y = a and xy=p= as - b For another example, take the equations x + y = a and x + y2 = b. From the second equation we get x*+ y* — 2x2y2 = b— 2p3, or (x2 — y2)2 = (x + y)2(x − y)2 = a2(x2 — 2xy + y2) = a2[(x + y)2 - 4xy]=b-2p3, or a[a2 - 4p]=b-2p, which is easily b- at reduced to the quadratic p2 - 2a3p 2 = Completing the square, etc., we have p2a2p + a* = a* + a2 + b 2 or p― a2 ticing that for both positive. b- at 2 we must use -; supposing x and y are Hence, the solution of the given equations is reduced to that of x+y=a and xy = p = a2 — a + b 2 To illustrate what has been done, take the following EXAMPLES. 1. Given, 4x2 - 30 = 33 — 3x2, to find x. 2 By transposition, etc., we get = 9; consequently, the roots of the given equation are expressed by ∞ = ±3, or by x=3 and x = 3. Freeing the equation from fractions, etc., we have x2 = 3600; consequently, by extracting the square roots, we have x=60. 3. Given, 3(x2- 8) + 2(4x2 — 5) = 8(x2 − 1) + 22, to find æ. Ans. x= ± 4. 4. Given, Va-25 12, to find a. = x. Squaring the members of the equation so as to remove the surd sign, we have 25 = 144; consequently, ∞ = 13. 2ax + b2x2 5. Given, √ a2 + x2 =Va2x2 to find x. Freeing the equation from fractions, etc., we have b22 = (a); consequently, extracting the square root, we get bx = ±(a− x), or bx = a − x and be a +x, which -- Remarks. If b = 1, the first value of a becomes x = α Now, since the given equation is equivalent to 2ax + b2x2 = a + a2, if we put ± 1 for b, it will become 2ax + x2 = a2 + 22. Regarding as finite, this equation, by erasing α from both members, reduces to 2ax a2, or x= which 2 = accounts for one of the values found for a when b = ± 1. If equals infinity, we must reject the terms 2ax and a2 (since a is supposed to be finite) from the equation 2ax + x2 = a2 + 2, on account of their comparative smallness; consequently, the equation is satisfied by a = ±∞, since it becomes x2 = x2. The equation is easily reduced to aa", which gives v x = ± √ a2 = a -1, which are imaginary values, since a is supposed to be real. 8. Given, Va+ x + √ a − x = √2b, to find x. Squaring the members of the equation, etc., we easily get √ a2x2=b-a; then squaring the members of this, etc., we get a√2ab — b2. -. = Otherwise, by putting a + xy' and a-x=22, the equation becomes y + z = √2b; and by taking the half sum and half difference of the equations a+ay' and a-x= y-2. Squaring the mem22, we get y2 + 22. =2a and x = 2y2 2 22 = bers of y + z = √2b, we get y2 + 2yz + z2 = 2b, which, subtracted from 2y+22=4a, gives y 2yz +22 = 4a2b, or, extracting the square root, we have y-2= ± √4a — 2b. Taking half the products of the corresponding members of y2 — 22 = y + z = √2b and y − z = ± √4a 2b, we get x= 2 ±√2ab-b2; consequently, we have the same values of æ as before. Remarks. If a and b have like signs, and a is numerically not less than, it is clear that the values of x will be b real, and that they will be 0 when a = have unlike signs, or if they have the same signs, and a is b numerically less than the values of a will always be 2' imaginary, since the expression under the surd will be negative. |