The first member is now a complete square. Taking the square root of both members, we have x+ar+Va+b; whence by transposition, x = -a Va+b. Thus the equation has two roots, which are unequal in all cases except when a'+b=0; in this case we shall have x = a+0, and the equation is said to have two equal roots. Thus take the equation, * - 10x = -25. Adding 25 to both members, -102+25 = 0; whence, by evolution, –5= 0; that is, = 5+0 = 5 or 5. 3+1 13 1. Given + to find the values of x. 3+1 clearing of fractions, 6x® +6x*+12x+6= 13x*+13x; reducing terms, 2+x=6; adding (1)to both members, 2+x+1=; by evolution, *+=+; whence, X = 2 or 3. Hence, for the solution of an affected quadratic equation we have the following RULE. I. Reduce the given equation to the form of x'+2ax b. II. Add to both sides of the equation the square of one half the coefficient of x, and the first member will be a complete square. III. Extract the square root of both members, and reduce the resulting equation. 290. When the equation has been reduced to the form of #*+2ax=b, its roots may be obtained directly by the following obvious rule : Write one half the coefficient of x with its contrary sign, plus or minus the square root of the second member increased by the square of one half this coefficient. 1. Given x*_6x = 55, to find the values of x. OPERATION. x=3+V 55+9; = 3+8=11 or-5, Ans. or, EXAMPLES FOR PRACTICE. Find the values of the unknown quantity in each of the following equations : 1. x+2x = 15. Ans. x = 3 or -5. 2. x'-6x = 16. Ans. X = 8 or — 2. 3. a 20:0 = -96. Ans. x = 12 or 8. 4. 2-6x-7= 33. Ans. x = 10 or -4. 5. x-28x+80=-115. Ans. x= 15 or 13. 6. x*+6x+1='92. Ans. X=7 or -13. 7. 2+12x=589. Ans. x = -31. 8. X-6x+10=65. Ans. x=11 or -5. 9. 30'+12+2=110. Ans. x = 6 or -18. 10. c* _ 14x = 51. Ans. x = 17 or —3. 11. x+20x+19=0. Ans, x = -l or -19. 12. a'-6x+6=9. Ans. x=3+23. . 13. x+8x=12. Ans. x=-4+277. 14. 2+12x = 10. Ans. X= -6+1 46. 15. 3x*—15x =-12. Ans. x = 4 or -1. 16. 4x* +12x = 40. Ans. x=2 or -5. 17 2x*+28 = 18x. Ans. x = 7 or 2. 18. (3x—5) (2x—2)=2(a*+1.5). Ans. x=5 or -1. 19. (2x+2)(5x—8) = (2+1)(5x+4). Ans. x = 4 or -1. 20. (3.0+4)' = 54x. Ans. x = or . 2x 7 21. x? Ans. x= or -1o. 15 12* 291. It frequently happens in reducing a quadratic to the form of x' +2ax = b, that 2a, the coefficient of x, becomes fractional, thus rendering the solution a little complicated. In such cases it will be sufficient to reduce the first member to the simplest entire terms. The equation will then be in the form ax'+bx = 0, (1) in which a and b are integral in form, and prime to each other, and c is entire or fractional. To render the first term of equation (1) a perfect square, multiply both members by a; thus, a'x' +abx = ac. (2) bo 39 (3) 39 where the first member is a complete square. Now if b is even, 4 62 will be entire; but if b is odd, will be fractional, a result which 4 Adding = acta we wish to avoid. To modify the rule to suit the latter case, suppose equation (3) to be multiplied by 4; thus, 4aRx+4abx+6 = 4ac+b'. The first member is now a complete square, and its terms are entire. Moreover, we observe that (4) may be obtained directly from (1) by multiplying (1) by 4a, and adding Go to both sides of the result. Hence, for the second method of completing the square in the first member, we have the following RULE. I. Reduce the equation to the form of ax'+bx = 0, where a and b are prime to each other. II. If b is even, multiply the equation by the coefficient of x', and add the square of one half the coefficient of x to both members. III. If bis odd, multiply the equation by 4 times the coefficient of x', and add the square of the coefficient of x to both members. The above rule may be considered as more general than the first; for if applied to equations in the form of x*+2ax=b, the operation will be the same as by the first rule, with the simple modification of avoiding fractions in the first member, when 2a is fractional. 1. Given 5x*—6x = 8, to find the values of x. OPERATION. 5x'-6x = 8. Multiplying by 5, and adding 3', or 9, to both mem 250"-30x+9=49; bers, by evolution, 5x—3= +7; whence, 5x = 10 or-4; or, X = 2 or 2. Given 15x'—55x = 350, to find x. OPERATION. 15x*_55x = 350. Dividing the given equation by 5, } 36x?_132x+121 =961; 6–11=481; We will now apply this rule to an equation in the form of 2+2ax=b, where 2a is odd. 3. Given x*—72 = 44, to find x. OPERATION. 29 _7* = 44. Multiplying by 4 times 1, 4x*_283+49=225; and adding 72 to both sides, by evolution, 23–7=+15; whence, X = 11 or 4. Thus we may always operate in such a manner as to avoid fractions in the first member; and indeed in the second member, if we first reduce both members of the equation to entire terms. EXAMPLES FOR PRACTICE. -3. Solve each of the following equations : 1. 5x*+4x = 204. Ans. = 6 or — 34. 2. 5.0'+43 = 273. Ans. =7 or 3. 70*_20x = 32. Ans. x = 4 or . 4. 6x +15x = 9. Ans. x = or 5. 2xc _5X 117. Ans. x = 9 or . 6. 21x?_292x = --500. Ans. w = 11ļi or 2. 7. 6x'-13.+6= 0. Ans. x = or . 8. 73_3x = 160. Ans. x = 5 or -32. 9. 3x_53x =-34. Ans. x = 17 or . 10. x+13x-140=0. Ans. X= 7 or 20. 11. 3x'—8x = 5+413. Ans. x = 2+1/3 or ģ-V3. 12. * +11x480 = 0. Ans. 3 = 5 -16. or |