for the roots or values of x in the proposed 30. Given, ax2 + bxc, to find x. If we multiply the terms of the equation by 4a (or four times the coefficient of the square of the unknown letter), and add (or the square of the coefficient of the simple power of the unknown letter) to both sides, we shall get 4a2x2+4abx+b2 = b2 + 4ac or (2ax + b) = b2+4ac; consequently, by extracting the square root in both members, we - b± √b2 + 4ac 2a shall get 2ax + b = ± √b2 + 4ac or x = b ± √b2 + 4ac then (a Remarks.-1. Resuming x= being positive) if c is positive, since b+4ac is not less than b2, it follows that b24ac is not less than b; consequently, when is greater than 0, the values of x must be real and have contrary signs; and when c = 0, one of the values of b a is 0, and the other equals. a 2. If c is negative, we have x = − b ± √ b2 4ac ; which shows if 4ac is not greater than 62, that the values of a must have the same sign as when 4acb, and that If 4ac is greater than 2, since b2 — 4ac is negative, it follows that the values of a will be impossible or imaginary. с 3. Resuming the equation a+b= c, then if c is finite and a an infinitesimal, by supposing a∞, or to be infinitely great, we must reject e on account of its comparative minuteb ness, and we shall have a2 + bx =0 or x= an unlimitedly great result. α ∞= Again, still supposing a to be an infinitesimal, if we regard x as being finite, then on account of the minuteness of a, we must reject ax3, and shall thence get bx = c or x b 31. Given, a2aab, to find x. Adding a to both members of the equation, we get an 2αx2 + a2 = (x” — a)2= a2+b; consequently, by extracting the square roots, etc., we get x" a± √a2 + bor x = Remark. If the given equation had been "- n we should have found x = (a ± √a2+b)2. 32. Given, 122 + 7 = a2 — 4, to find æ. By transposition, the equation can be put under the form x2 - √2x2 + 7 = 4, then multiplying the members by 2 and adding 7 to the products, we get 2x2 +7-2√2x2 + 7 = 15; consequently, putting y = √2x2 + 7, the equation is reduced to the quadratic y 2y = 15. Solving the quadratic, we get y 14 or y = 5 and y = 3; consequently, by restoring the value of y, we shall get 12+ 7 = 5 and √2a +7=3. If we suppose the radical in the given equation to be positive, it is clear that the second of the preceding equations must be rejected, and that the values of a must be found from the equation 12+7=5. Squaring the members of this equation, we get 2a2+7= 25 or a2 = 9, whose square root gives a = 3; consequently, if a is required to be positive, we must reject the equation = -3 and put x = 3. 33. Given, - 4x3 + 8x2 — 8x = 1365, to find the positive value of x. By extracting the square root of the first member of the equation, it is easy to put it under the form (x2 — 2x)2 + 4(x2 — 2x) = 1365; consequently, putting - 2x = y, the equation is reduced to the quadratic y2 + 4y = 1365. Solving the quadratic, we get y = x2 — 2x = 2x 35, and thence x = 7. The equation being equivalent to x-3=√5 — x2, is easily reduced to 5-x2+15-a2= 2; consequently, 2. QUADRATIC EQUATIONS. 35. Given, 13x-5+ √2x-1=4+ √5x — x, regarding the radicals as being positive. - 1 x 1 x 22, to find Ans. x=7. =(x+4), to find the real values 1 +1 x Ans. x+15. = x, to find x.” and squaring, we get (Va 2 X x+1, which is easily |, or x = x2 — 2 √x2 − x + 12 reduced to 2-x-2 √x2-x+1=0, whose square root gives Vax-1= 0, or x2-x=1; solving this, we get Otherwise. If we put y=x- and 22 = 1 — x equation will be reduced to y + z = x. Subtracting the second of the assumed equations from the first, we get y3 - z2 = (y + z) (y — 2) = x(y — z) = x − 1, and from 22=1 = 1 − (1 + 2 − y)=y-z, the equation 1 is reduced to 2y-y=1, or y2 - 2y + 1 = 0, or y Hence, the equation y + z+2y= 1 reduces to 2+2 38. Given, Va + x + √ b + x = Ve+ 2x, to find x. Put ya+x, z2=b+x, and v2=c+2x; then the equation becomes y + z = v, or y2 + 2yz + ≈2 = v2. From the assumed equations we have y2+22= a + b + 2x=a+b-c+v2, which reduces the preceding equation to 2yz = c(a + b). = Also, from the assumed equations we have y2- 2a— a — b, whose square gives y1- 2y2 + 2 = (a - b), and adding 4y2z2 = [c − (a + b)] to this, we have y + 2y2z2 + z1 (a − b)2 + [c − (a + b)]2, and denoting the square root of this by A, we have y2+ 22 = A. y2 + 22 — (a + b) 2 we get x = A-(a + b) Hence, from x = as required. 39. Given, x+y=a and xy=b, to find x and y. 40. Given, ax + by = c and xy=d, to find x and y. Multiplying the terms of the second by xy, and dividing by the corresponding members of the first, we get = x + y Because the first equation gives x2 + 2xy + y2 = a2, we 3a + b easily reduce the preceding equation to _xy = a2, or xy α Hence, the solution of the given equations is reduced to that of the equations x + y = a and xy: = a3 3a + b If a = consequently, from x+y= 12 and ay and y = 4, or x = 4 and 12, and b 18, we shall have = 1728 32 we get x=8 43. Given, a+aya and xy+2y=b, to find x and y. + a − x2 = b, or x1 — (3a + b)x2 : - -2a; whose solution gives x=± = (3a+b± √a2+6ab+b2 3 :(3a+ 2 ; consequently, having found x, we get y from the equation y x If a = 56 and b = 60, we get √a2 + 6ab + b2 = 164 and x = ± √114 ± 82 = ± 14, x = ± 4 √2. Hence, if we take the rational values, we have x = 14, and z." Dividing the numerators and denominators of the first members of these equations by xyz, the equations become 1 1 b, =c, which are easily re 1 = a, 1 1 1 1 + + + yz x2 xy 1 duced to = + + = + = yz xy If from the half sum of these equations we subtract the equations severally, we shall get Dividing the second of these by the product of the first and third, and extracting the square root of the result, we 1. To find two numbers, such that the ratio of the greater to the less shall be 5, and their product 845. The numbers may be represented by a and 5, since the |