ALGEBRA. 



J 6 ! 7 or 5 . either of which, Ex - 6 Lct V*+ T 

 substituted for .< in the original equation, 

 answers the condition, that is, makes the y 6 +r 



whole equal to nothing. 



Ex. 3. Let x + ^/ (5 x + 10) = 8 ; to 

 lind x. 



Jiv transposition, ^/ (5 x -\- 10) = 8 a- 

 squar. both sides 5 x -\- 10 = 64 16x 



z _ 21 x = 10 64 = 54 



441-216 . 01 , 441 225 



= 'A , or ^21x4-- = - 



4 44 



21 . 15 



extracting the sq. root, x = _ 



, = H1 5 = 3 or 18. 



By this, process two values of x are 

 found, but on trial it appears, that 18 does 

 not answer the condition of the equation, 

 if we suppose that ^/ (5 x 4- 10) repre- 

 sents the positive square root of 5 x 4- 

 10. The reason is, that 5 x 4- 10 is the 

 square of ^/ (5 x +- 10) as well as of 

 4- v^ (5 x 4- 10); thus by squaring 

 both sides of the equation ^/ (.5 x -f- 10) 

 = 8 x, a new condition is introduced, 

 and a new value of the unknown quanti- 

 ty corresponding to it, which had no 

 place before. Here, 18 is the value which 

 corresponds to the supposition that x 

 ^/ (5 x + 10) = 8. 



Every equation, where the unknown 

 quantity is found in two terms, and its in- 

 dex in one is twice as great as in the other, 

 may be resolved in the same manner. 



Ex. 4. Let z+4 z*=21 



z+4 2^4-4=214-4 = 25 



z*4-2=5 

 $=5 2 = 3, or 7 



therefore z=9, or 49. 



Ex. 5. Let y* 6y* 27=0. 

 y* 6y>=27 

 y* 6i,'+9=27-f 9 = 36 



.21 

 27 



When there are more equations and 

 unknown quantities than one, a single 

 equation, involving only one of the un- 

 known quantities, may sometimes be ob- 

 tained by the rules laid down for the so- 

 lution of simple equations ; and one of 

 the unknown quantities being discovered, 

 the others may be obtained by substituting 

 its value in the preceding equations. 



Ex.7. Let $**+** = 65 { Tofindx andy. 



c x y *>** j 

 From the second equation, 2 x /=56 



sub. it from the same, a-2 2.rt/4-j/2=9 

 by extracting the sq. roots, x-\-y=. 11 



and x y = 3 

 therefore, 2 x= 14 

 x=7, or 7 

 and /=4, or- 4 



PROBLEMS PRODUCING Q.AUDBATIC 

 EQ.UATrOIfS. 



Prob. 1. To divide a line of 20 inches 

 into two such parts, that the rectangle 

 under the whole and one part may be 

 equal to the square of the other. 



Let x be the greater part, then will 20 



- _j- \)Q the less 



and j 2 = (20 x) . 20 = 400 20 x by 

 the question. 

 a-2 4-20 x=400 

 *z 4-20 x4-100=4004-100=500 



* 3 6=9, or 3 

 3. 



^=4^/500 10, or v/ 500 10. 



Prob. 2. To find two numbers, whose 

 sum, product, and the sum of whose 

 squares, are equal to each other. 



Let x-\-y and x y be the numbers ; 



their sum is 2 x 

 their product r2 y% 

 the sum of their sqs. 2 or 1 =2 y* 

 and by the question 2 o-=2 x 1 +2 y 1 



or j"=x j 4- y^ 

 also, 2x=.r i y* 

 therefore, 3x=2*z 



