SlNCrl^E KqUATIONvS. 83 



2. Solve V-^-l- V-iH^6=3. /,; 



Squaring each side of the equation, obtain 



.r-f 2 \/^'-f-6i+x+6=9. ■ (2) 



Transposing all but the radical to the right-hand side this 

 becomes 



2\/r'+6.t — 3 — 2:r. (2,) 



Squaring, we obtain 



^x" + 24.r = 9 — 1 2Ji: + 4.r' ^4 j 



or i^=i- 



What are the questionable steps ? What is their effect ? 

 The above solution is really equivalent to the following : 

 sf x-\- \/ X + 6=3. 

 Transpose the 3 to left member, obtaining 



V -^ "+ V -1^ + 6— 3=0 ; 

 Multiplying through by the rationalizing factor of left member, 

 (III, Art. 26) we obtain 



(V-r+ V^rT6-3)( V-^+ V.r + 6+3) 



x(V-t— \/r-f-6— 3)(\/-^— V-r-j-6-f 3)=o, 

 which reduces to 



.4.r— 1=0, (5) 



or t = 1 . 



The introduced equations arc 



s/ x-^ 1^ X -\-b-\-2,—o \ 

 \/.r— V-^ -I- 6— 3=0 

 V -^'— V -^ ■ + 6 + 3 =0. 

 Here, then, is an apparent paradox : three solutions seem to 

 have been introduced, yet there is only one in all ! 



This can be explained in the following manner. If we regard 

 the radical signs as calling for any one of the two roots of the ex- 

 pression underneath, then the introduced equations are all iden- 

 tical with the original equation ana hence could not give rise to 

 a differejit solution. If we restrict ourselves to using that square 

 root in each case which has the sign given before the radical, 

 then none of the introduced equations have any solution ivhatever^ 

 and hence no solution is introduced in this ca.se. 



s/ x-\-sf a~^x 

 3. Solve -^ ^^^^=A-. 

 V X— V ^ + '^' 



