SECTION 11. Affected Equations. As a pure equation of the second degree contains the unknown quantity only in the form of its second power, all the terms in which it appears can be united in one term, whose root, as we have seen, can be readily found. An affected equation of the second degree contains not only the square of the unknown quantity in one term, but also the unknown quantity itself in another term. Thus, x2 + 4 x = 77 is an affected equation of the second degree, in which the unknown quantity appears in two terms; for x2 and x cannot be actually added together so as to make but one term. When an equation of this sort is formed, it may contain the unknown quantity in any number of terms, provided it be only in the first and second powers; for, in this case, the terms may all be reduced to two. Thus, if we have the equation 24 + x2 - 3 x = 4 x2 + x + 48, by transposition we obtain 5 x2 + 8 x and, by dividing all the terms by 2, x2 + 2 x = 36. Affected and pure equations of the second degree are solved in the same manner: in both, we find the value of the unknown quantity by extracting the square root of each member of the equation. When the member containing the unknown quantity is a complete power, the process is as direct and simple in an affected as in a pure equation. 1. Given x2 + 2 a x + a2 = b2, What is the value of x? х a. ANS. b In this question, a and b represent known quantities. Now, we know by inspection, that the first member of the equation, x2 + 2 a x + a2, is the complete second power of the binomial quantity x+a; and the other member, 62, is the second power of b. By extracting the square root of each member, therefore, we obtain the equation, x + a = b. And xba, by transposition. If we suppose a = 4 and b = 9, by raising x +4 and 9 to their second powers, instead of the above equation, we shall have x2+8x + 16 = 81, or x2 + 8 x + 16 = 81. and That is, x+4= 9, 2. Given x2 + 2 a x = b. What is the value of x? ANS. /ba2 a. In this question, the member of the equation containing the unknown quantity, is not a complete power; and, of course, while it remains in its present form, its root, and consequently the value of the unknown quantity, cannot be found. But it is possible to add such a quantity to the first member as shall make it a perfect square. The necessary quantity must be added to both members of the equation, to preserve its equality. But how shall we find the quantity which must be added to complete the square? If we examine the equation given above, x2 + 2 ax = b, we shall be satisfied that the root of the first member, whatever it may be, is not a single term; for any power of one term consists of but one term. But if its root consist of two terms, one term is wanting to complete the square; for the second power of a binomial quantity contains three terms, [See Chap. VII. Sec. III.] whereas the given quantity contains but two. The terms given, x2 + 2 a x, are the first two terms of the second power of the binomial x + a; and the third term of the same quantity is a2. [See Chap. IX. Ques. 18.] By adding this quantity, therefore, to both of the given members, we make the first member a complete square; and the equation becomes x2 + 2 a x + a2 = b + a2. 2 b And x2 + 2 a x + a2 = √ b + a2, or x + ab + a2, by evolution, and x = Observe that the coefficient of the second term given is 2 a, one half of which is I a or a; and that the quantity added to complete the square, a2, is the second power of a. If, in the last example, we suppose a = 4, and b = 84, the given equation will be x28x84. Now, if we add, as before, the second power of the value of a, (4 X 4 = 16,) to both sides of the equax4 tion, the first member will be a complete square; and the equation will be solved thus: x2 + 8 x + 16 = 84 + 16, or 100. And2+8x+16=100, Observe that 16, which is added to complete the square, is the second power of 4, or half of the coefficient of the second term of the given equation. Hence, to render the first member a complete second power, we add the square of half the coefficient of the second term to both members of the equation, 3. Given the equation 87 +7 x2 123 + 3 x = 5x21185 x. What is the value of x? 7x2-5x2+3x+5x=118+123-87, by transposition. 2x2 + 8 x = 154, by uniting terms, and x2 + 4 x = 77, by division. x2+4x+4=77+ 4, or 81, by completing the square. x+29, by evolution, and x = 9 -- 2, or 7. ANS. 7. 3 ...4. Given the equation 3x2+89++ 20 x = 224 + 3x2+8x; to find the value of x. x2 3 -3x2+20x—8x=224-89,by transposition. + 12 x = 135, by uniting terms. x2 + 36 x = 405, by removing the denominator. x2+36x+324=405+324, or 729, by completing the x+18=27, by evolution. x=27-18, or 9. [square. 5. What is the value of x in the equation 3 x2 + 2 x = 161? 2x 3x2 + 2 x = 161. 22 +2.00 = 161, by removing the coefficient of x2. 3 x2+2+3=181+3=184, by completing the square. In this equation, the second term has a fractional coefficient; for 2 is the same as 3 x is, the second power of which is x. The half of . The fractions of the second member, 161 and 4, are reduced to a common denominator, and added together in the usual way. 6. Given 5 x2 + 3 x 344. What is the value of x? 5 x2 + 3 x = 344. 244, by removing the coefficient of x2. x2 + 3x + 180 = 344 + 180, or 5889. Half of is; and the second power of is 180. |