Page:Encyclopædia Britannica, Ninth Edition, v. 1.djvu/580

Rh 542 ALGEBRA [QUADKATIC EQUATIONS, x y ax z by z c If these be added, and their sum divided by 2, we find M+---+-+-. x y n 2a 26 2c From this equation let each of the three preceding be subtracted in its turn : thus we get 26 26 2c Zctbo ab + ac + be Hence 26 2c 26c 2abc 7200 x + 6 + ac-6c 120 2abc _7200 ac + bc~ 360 2abc 7200 -ab + ac + bc 240 = 60, = 20. = 30. SECT. IX. SOLUTION OP QUADKATIC EQUATIONS. Quadratic 71. &quot;We are next to explain the resolution of equations Equations, of the second degree, or quadratic equations. These in volve the second power of the unknown quantity, and may be divided into two kinds, pure and adfected. I. Pure quadratic equations are such as after proper reduction have the square of the unknown quantity in one term, while the remaining terms contain only known quantities. Thus, x 1 = 64, and ax 2 + b = c, are examples of pure quadratics. II. Adfected quadratic equations contain the square of the unknown quantity in one term, and its first or simple power in another; the remaining terms consisting entirely of known quantities. Such are the following, x 2 + 3x = 28, 2x 2 = 33 - 5x, ax 2 + bx-c = d. The manner of resolving a pure quadratic equation is sufficiently evident. If the unknown quantity be made to stand alone on one side, with unity as a coefficient, while the other side consists entirely of known quantities, and the square root of each side be taken, we immediately ob tain the value of the simple power of the unknown quan tity as directed by rule 5th of Sect. VI. In extracting the square root of any quantity, it is necessary to observe, that the sign of the root may be cither + or, and that consequently a quadratic must always have two solutions. 72. When an adfected quadratic equation is to be re solved, it may always, by proper reduction, be brought to the followin form : where p and q are numerical quantities, + or. Let us compare the side of it which involves the un known quantity x with the square of a binomial x + a ; that is, let us compare x 2 +px with x 2 + 2ax + a 2 = (x + a) 2, and fQ it will presently appear, that if we suppose p 2a, or ^ = a, 4i the quantities x 2 +px and x 2 + 2ax will be equal; and as x 2 + 2ax is rendered a complete square, by adding to it a 2 , so also may x 2 +px be completed into a square by adding 2 p 2 to it, which is equal to a 2 ; therefore, let ~ be added to 4 4 both sides of the equation x 2 +px = q, and we have and, extracting the square root of each side, p /m 2 p a? + ^= /^ + 2; hence x= - 73. From these observations we derive the following general rules for resolving adfected quadratic equations. 1. Bring all the terms involving the unknown quantity to one side, and the known quantities to the other side, and so that the term involving the square of the unknown quantity may be positive. 2. If the square of the unknoAvn quantity be multiplied by a coefficient, let the other terms be divided by it, so that the coefficient of the square of the unknown quantity may be 1. 3. Add to both sides the square of half the coefficient of the unknown quantity itself, and the side of the equa tion involving the unknown quantity will now be a com plete square. 4. Extract the square root of both sides of the equation, by which it becomes simple with respect to the unknown quantity; and, by transposition, that quantity may be made to stand alone on one side of the equation, while the other side consists of known quantities; and therefore the equa tion is resolved. Ex. 1. Given a? 2 + 2x= 35, to determine x. Here the coefficient of the second term is 2 ; therefore, adding the square of its half to each side, we have x 2 + 2x + 1 = 35 + 1 = 36, And, extracting the square root, x+l = /3G= 0. Hence x = 6-1, that is, x = + 5, or x = - 7, and either of these numbers will be found to satisfy the equation, for 5x5 + 2x5 = 35, also - 7 x -7 + 2x -7 = 35. x 2 Ex. 2. Given - 12 = x, to find x. This equation, when reduced, becomes x 2 - Gx 72 , And, by completing the square, x 2 - Gx + 9 = 72 + 9 = 81. Hence, by extracting the square root, x - 3 = 9, and x= 9 + 3; Therefore x= +12, or # = - G ; and upon trial we find that each of these values satisfies the original equation, for 12X12 -12 = 12, also 6 * 6 -12= -6. 6 6 Ex. 3. Given x- + 28 = 1 Ix, to find a;. Then x 2 - llar= -28. And, by completing the square, 121 121 9 4 ~ 4 4 11 3 Therefore, by extracting the root, x- - 11 3 Hence a; = - ; that is, x = + 7, or x = + 4. In the first two examples, we found one positive value for x in each, and also one negative value; but in this example both the values of x are positive, and, upon trial, each of them is found to satisfy the equation; for 7x7 + 28 = 11x7, also 4x4 + 28 = 11x4. 74. As at first sight it appears remarkable, that in every quadratic equation the unknown quantity admits always of two distinct values or roots, it will be proper to consider a little further the circumstances upon which this peculiarity depends. To do this, let us re-examine the equation x 2 + 2x = 35. By bringing all the terms to one side, the equation may be expressed thus, x 2 + 2x - 35 = ; so that we shall have determined x, when we have found such a number as, when substituted for it in the quantity x 2 + 2x - 35, will render the result equal to 0. But x 2 + 2x - 35 is the product of these two factors x - 5 and x + 7, as may be proved by actual multiplication; therefore, to find x, we have (x - 5) (x + 7) = ; and as a product can only become = when one of its factors is reduced to 0, it follows that either of the two factors x - 5 and &amp;lt;K + 7 may be assumed = 0. If