ALGEBRA. 



x 6=1 



x = 6 1 = 7 or 5; either of which, 

 subsutuied for .- in the original equation, 

 answers the condition, that is, makes the 

 whole equal to nothing. 



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

 find jc. 



By transposition, v/ (5 .r -f- 10) = 8 x 

 squar. both sides 5 x + 10 = 64 I6x 



+ * 1 



x i _ 21 x = 10 64 = 54 



-*s 



3 or 18. 



Ex. 6. Let y*+r */3+ = 0, 



21 

 extracting the sq. root, .r 



21 15 



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 + 10) repre- 

 sents the positive square root of 5 x + 

 10. The reason is, that 5 x + 10 is the 

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

 + v/ (5 a? + 10) ; thus by squaring 

 both sides of the equation ^/ (5 x + 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 

 V/(5;r+10)=8. 



Every equation, where the unknown 

 quantity is found in two terms, and its in- 

 dex in one is twice as great as|inthe other, 

 may be resolved in the same manner. 



Ex. 4. Let 2+4 



+4=21+4=25 



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 X *+J 65 To find x and y. 



z^=5 2=3, or 7 

 therefore z=9, or 49. 



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

 y4_6y-27 



^ 6^+9=27+9=36 

 F 1 3#=6 

 y 1 36=9, or 3 

 y=3, or ^/ 3. 



From the second equation, 2 x y=56 

 & adding this to the lst,o: 1 +2^+i/ l =121 

 sub. it from the same, a; 1 2.n/+i/ 2 ==9 

 by extracting the sq. roots, a+y= 11 



and x y = 3 



therefore, 2 x = 14 



x=7, or 7 



and y=4, or 4 



PROBLEMS PRODUCING QUADRATIC 

 EQ.UATIOXS. 



Prob. 1. To divide aline 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 

 x be the less. 



and* 1 = (20 or). 20 = 400 20 x by 

 the question. 

 X* +20 or=400 

 # +20 .r+ 100=400+ 100=500 



x= + V/500 10, or ^/ 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 x 1 r/ s 

 the sum of their sqs. 2 cc s =2 y 2 

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



or ar=x 1 + y 1 

 also, 2 x=x* y 1 

 therefore, 3 x=2 x 1 



