54 Algebra. 



Each of the first two of these is equivalent to x= \^ q , and each 

 of the last two is the same as jr= — v/,^, and, on the whole, we 

 merely have 



Whence it is seen to be sufficient to write the sign ± on but 

 one side of the equation. 



7. ExAMPi^ES OF Incomplete Quadratics. 

 Solve the following equations : 



2. (ax~.b)(ax-\-b)=c. 



J. (x+2r-{-(x-2r=24. 



4- (x^br^(x-bT=c. 



5. (ax-\-b)''^-(ax—bf=c. 



6. (x+T)(x-g) + (x-'j)(x-^g)r='^6. 



7. (x^-a)(x-b) + (x-a)(x+b) = c. 

 S. (x-^a)^=q. 



Show that examples i , 3 and 6 may be solved by proper substi- 

 tution in the results to examples 2, 5 and 7 respectively. 



8. We have solved the equation x"=q, and also the equation 

 (x-\-a)^—q (Ex. 8) in a similar manner. Now it is evident that 

 the equation 



x'-\-px=q 

 can be solved if it can be put in either of the above forms. It can 

 be placed in the form (x-\-a)'^=q if the first member can be made 

 the square of a binomial. On inspection it is seen that x^-\-px are 

 the first two terms of the square of a binomial, the third term of 

 which must be \p\ Hence, if we add \p'" to both sides of the 

 equation 



it takes the form 



x^+px-^\P^=q + \P% 



or ^^ + i/>/=^ + i/^ 



which is of the form (x-\-a)-=q. 



The process of putting a quadratic equation in this form is 

 called complctimr the square. 



