SECTION III. On the Solution of Pure Quadratics, and others which may be solved without completing the Square. (27.) WHEN the terms of an equation involve the square of the unknown quantity only, the value of the square will be found by the preceding articles; and extracting the root on each side of the equation, the unknown quantity itself will be determined. In the same way any pure equation may be solved; for the power of the unknown quantity standing alone on one side of the equation, the known quantities being transposed to the other, the simple unknown quantity will be determined by extracting the root. And by the same process, any equation containing the powers of a function of the unknown quantity, or containing the powers of two unknown quantities, may frequently be reduced to lower dimensions. EXAMPLES. 1. Given x2 - 17 = 130 2x2, to find the values of x. By transposition, 3x2 = 147; .. x2 = 49, and x = 7.* * The square root of a quantity may be either + or, and consequently all quadratic equations admit of two solutions. Thus, +7 × +7, and —7 × —7, are both equal to 49; and both, when substituted for a in the original equation, answer the conditions required. = 2. Given x2+ ab 5x2, to find the values of x. By transposition, ab = 4x2; From the second equation, x = by. Substituting this value in the first equation, therefore, extracting the square root, x = ± 3, 2x whence y = = ± 2. E 2x: 2y :: 4 : 2; . (Alg. 184.) x : y :: 2 : 1, Substituting this value of x in the second equation, To the second equation, adding twice the first, x2 + 2xy + y2 = s2 + 2a2 ; .. extracting the square root, x + y = ± √ s2 + 2a2; and from the second, subtracting twice the first, x2 - 2xy + y2 = s2 — 2a2; .. extracting the square root, x - y = ±√82 - 2a2 but x + y = ± s2 + 2α2; .. by addition, 2x = ± 82 + 2a2± √ 82 and x = 2 by subtraction, 2y+82 + 2a2± √ s2 — 2a2, = ± - ✓ s2 - 2a2). to find the values of x and y. |