This equation shows that to obtain the true quotient, we must divide one number by the other regardless of the root, and then write the root over the quotient. (ART. 86.) When the roots are different, we proceed on the same principle, which will be sufficient for every possible case. For example, divide 7 by 7; the quotient must be some number which we can represent by Q, and from the equation SECTION IV. EQUATIONS. (ART. 87.) WE have thus far been able to resolve only simple equations, or equations of the first degree; but many problems and many philosophical investigations present equations of the second, third, and higher degrees, which may demand a solution, as, for instance, the first example of Art. 86 incidentally demanded the solution of the equation Q6=7 which is an equation of the sixth degree; but it appears in so simple a form that there is no mistaking the principle on which its solution depends, and thus generally, When an unknown quantity is involved to any power, we find the first power (that is, the quantity itself) by extracting the corresponding root of both members. As in that equation. Q=(7) * 1 In the same manner, if x"=a, then x=a". The converse of these equations may often occur, that is, the unknown quantity may appear under the form of a root, as in the following equation: Here, it is obvious that the value of a must be found by cubing; but if we cube the first member of the equation, we must cube the second to preserve equality, (Ax. 9). That is x=(a+c)3 A (ART. 88.) From the foregoing observations, we draw the following general rule of operation: RULE. To free a quantity from a power, extract the corresponding root. To free it from a root, involve to the correspond ing power. When the unknown quantity is connected to a known quantity, and the whole number a power or root, the power or root, as the case may be, is removed in the same manner as before. The value of x is found, by first taking the cube root of both numbers and afterward transposing a. Here, after squaring both members, we have 2x+c=a2, a simple equation. (ART. 89.) The equations that appear in Articles 87 and 88, and all other equations of like kind where the unknown quantity is raised to a complete power, or is under some one particular root, are called PURE EQUATIONS. Thus, the equation ax2=b, or, which is the same thing, b 3 is a pure equation, because the power of the unknown = а quantity is complete; but the equation a2+bx=c is not a pure equation, because it contains different powers of the unknown quantity. 1 The equation (x+a)=c is a pure equation; but the equation x3+x=c is an impure equation, because it contains no complete power of the unknown quantity. Again, x2+2ax+a2=c+b is a pure equation, because the first member is a complete power of (x+a), and (x+a) may be represented by y, then y2=c+b, obviously a pure equation. There is no difficulty in resolving pure equations as we have already seen, for all we have to do is to apply the rule expressed in Art. 88; but impure equations in the higher degrees, present serious difficulties; and even equations of the second degree, when impure, compel us to complete the power before we can solve the equations. Equations of the second degree can be represented by a geometrical square; and when the equation is pure, the square corresponding to the first member is complete, and when impure, it must be completed, and the necessary operation is very appropriately called COMPLETING THE SQUARE. But before we go into the investigation of completing a square, we will give some examples to exercise the learner in resolving pure equations. EXAMPLES. 1. Given √4+(x—2)*=3, to find x. . . Ans. x=27. To remove the first radical sign, we square both members, Dropping 4 from both members, and then squaring, we Transpose x for the purpose of having the quantity under the radical stand alone as one member of the equation, Now, by squaring, the radical sign will disappear; but if any other quantity were joined to this by + or, the radical could not disappear in the square. The square is x2+6=4-4x+x2 3. Given x+√x2—7—7, to find x. 4. Given +12=2+x, to find x. N. B. No rules can be given that will meet every case, for the combination of quantities is too various. The pupil must depend mainly on general principles and his own practical experience. 5. Given 2+(3x)=√5x+4, to find x. . Ans. x=12. 8. Given x+2=√4+x√√64+x2, to find x. Ans. x= 6. 9. Given x- ≥√x=√x2—x, to find x. Ans. x=. 10. Given xa2+x2=a2x2, to find x. 11. Given √x-32=16-√x, to find æ. Ans. x=a√. Ans. x 81. For the sake of brevity, put a=16, then the last equation will be x-2a=a—↓x. At the conclusion resubstitute the PROBLEMS PRODUCING PURE EQUATIONS. (ART. 90.) In solving problems, it often depends on the manner or means of notation we employ, whether the equation comes out simple or complex, or whether it is a pure equation or a common quadratic. For example, 1. Find two numbers, whose difference is 6, and their product 40. Ans. 4 and 10. If we represent the least number by x, x +6 But, by the problem this product is 40. |