356 



THE POPULAE EDUCATOR. 



derived. Thus, in the seventh example, half the co-efficient of 



is . and this is the root of the third term . 

 2b' 4b 2 



When the first power of the unknown quantity is in several 

 terms, these should be united in one, if this can be done by the 

 rules for reduction in addition. But if there are literal co- 

 efficients, these may be considered as constituting, together, a 

 compound co-efficient or factor, into which the unknown quantity 

 is multiplied. 



Thus ax -\- bx 4- # = (a 4- b + d) X x. The square of half 

 this compound co-efficient is to be added to both sides of the 

 equation. 



EXAMPLES. 



1. Reduce the equation a; 2 4- Sx -{- 2x + x = d. 

 Uniting terms, se* -\- $x = d. 

 Completing the square, a; 2 4- 6x -\- 9 = 9 4- d. 

 And x = - 3 A/94-d. Ans. 



2. Reduce the equation a: 2 4- ax 4- bx = h. 



a: 2 4- (a + 6) X IB = h. 



Therefore *' + (a + 6) x x + fetf Y = (^ Y + * 

 \ ^ / \ ^ / 



Before completing the square, the known and unknown 

 quantities must be brought on opposite sides of the equation 

 by transposition ; the square of the unknown quantity must 

 also be positive, and it is preferable to make it the first or lead- 

 ing term. 



EXAMPLE. 



Reduce the equation a + 5x - 3b = 3x - x 2 . 



Transposing and uniting 



terms, a; 2 + 2x = 36 - a. 



Completing the square, a 2 -f 2x + 1 = 1 + 3b - a. 

 And * = 1 4- </r+ 3b - a. Ans. 



If the highest power of the unknown quantity has a co-efficient, 

 or divisor, before completing the square it must be freed from 

 these by multiplication or division. 



EXAMPLE. 



Reduce the equation a: 2 -f 24a Gh = 12 5x*. 



Transposing and uniting 



terms, 6* 12x = 6h 24o. 



Dividing by 6, x* 2x=h 4a. 



Completing the square, x 2 2x-i-l = l + h 4a. 

 Extracting and trans- _ 



posing, x = +1 -v/1 + h 4a. Ans. 



If the square of the unknown quantity is in several terms, 

 the equation must be divided by all the co-efficients of this 

 square. 



EXAMPLE. 



Reduce the equation bx* + d 2 - 4x = b - h. 



4-P b _ fa 



Dividing by I + d, x* = . 



o -f- d 6 + a 



Given ax- -\-bx = d, to find x. 



If this equation is multiplied by 4a, and if I 2 is added to both 

 sides, it will become 



4aV + 4ab + b 2 = 4,ad + I 2 ; 

 the first member of which is a complete square of the binomial 



EXERCISE 63. 



1. Beduce the equation - = 

 2 



nf* 



4. 





2. Beduce the equation h + 2x i= d -- . 



a 



3. Beduce the equation ax s + a; = h + 3z a;*. 



From the foregoing principle is deduced 



A SECOND METHOD OF COMPLETING THE SQUARE, 



called the Hindoo method. 



Multiply the equation by 4 times the co-efficient of the highest 

 power of the unknown quantity, and add to both sides the square 

 of the co-efficient of the loivest power. 



The peculiar advantage of the Hindoo method is, that it 

 avoids the introduction of fractions in completing the square. 



DEMONSTRATION. 



1. The object of multiplying the equation by the co-efficient 

 of the highest power, is to render the first term a perfect square 

 without removing its co-efficient, and at the same time to 

 obtain the middle term of the square of a binomial. But we 

 must multiply all the terms of the equation by this quantity 

 to preserve the equality of its members. Thus the equa- 

 tion a* 4 + bx = d, when multiplied by a, becomes a z j? + abx 

 = ad. 



That the first term will, in all cases, be rendered a complete 

 square when multiplied by its co-efficient, is evident from the 

 fact, that it will then consist of two factors, each of which is a 

 square, viz., x 2 , and the square of its co-efficient. But the pro- 

 duct of the squares of two or more factors is equal to the square 

 of their product. 



2. It will be seen that one term is still wanting in the first 

 member, in order to make it the square of a binomial, viz., the 

 square of the last term. 



This deficiency may be supplied by adding to both sides the 

 square of half the co-efficient of the lowest power, as in the first 

 method of completing the square. But in taking half of this 

 co-efficient, the learner will often be encumbered with fractions 

 which it is desirable to avoid. Thus, in the equation above, 



half of the co-efficient of the lowest power is ^, the square of 



6 2 2 



which is Adding this to both sides, the equation will 



become aV 2 4- abx -f = ad 4- , the first member of which 



4 4 b 



is a complete square of the binomial, ax -f- . 



2 



Now it is obvious to the student that multiplying the 

 equation by 4 has precisely the same effect as removing the 

 denominator 4 from the third term. Hence, if we multiply 

 the equation by 4, we not only avoid the introduction of frac- 

 tions, but also leave the square of the whole of the co-efficient 

 of the lowest power to be added to both sides according to 

 the rule. 



The first term evidently continues to be a square after it is 

 multiplied by 4, for it is still the product of the powers of 

 certain factors. 



3. It will be perceived at once, that the second term is com- 

 posed of twice the root of the first term multiplied into the co- 

 efficient of the last term, which constitutes the middle term of a 

 binomial square. 



Observation. It is manifest from the preceding demon- 

 stration, that multiplying by 4 is not a necessary stop in com- 

 pleting the square, but is resorted to on this particular occasion 

 as an expedient to prevent the occurrence of fractions. When, 

 therefore, the co-efficient of the lowest power is an even number, 

 so that half of it can be taken without a remainder, we may 

 simplify the operation by multiplying by the co-efficient of the 

 highest power alone, and adding to both sides the square 

 of half the co-efficient of the lowest power of the unknown 

 quantity. 



EXAMPLE. 



Take the equation 7x- + 40z 715. 



Multiplying by 7 it becomes 49z 2 -f- 280x = 500. 



Adding the square of half 



the co-efficient, 49a; 2 + 280z + 400 = 900. 



By evolution and trans- 

 position, 7a; = 10, or x = 1. 



