QUADRATIC EQUATIONS.] 



MATHEMATICS. ALGEBRA. 



473 



solution an additional principle ; what this is, is now to 

 be explained. 



And first, you are to observe that the square of a 

 quantity, consisting of but one term ; that is, the square 

 of a monomial is itself a monomial : thus the square of 

 1 z is z 2 ; the square of ax is a-x 2 , and so on. But the 

 square of a binomial consists of three terms ; that is, it 

 is a trinomial: thus, (z + a) 2 = x- + 2ax + a 3 , (x a) 2 

 = z 2 2ax -\- a 2 , and so on. 



The next thing to observe is, that if only the first and 

 second terms of the square of a binomial, x + a, or 

 x o, be given, the third term of it can always be found ; 

 for you see that this third term is nothing more than the 

 square of half the coefficient in the second term. Thus, if 

 the first and second terms, viz., z 2 -\-2ax, or x 2 2ax, 

 be written down, we can complete the square to which 

 these two terms belong, by dividing the coefficient of x, 

 namely 2a or 2a, by 2 ; and then adding the square of 

 the result. Thus, half the coefficient of x, in x 2 + 2ax, 

 is a ; the square of this is a 2 ; adding this .'. to the two 

 terms proposed, we have x- -f- 2ax + a 2 , for the complete 

 square of x -f- a. In like manner, knowing the two terms 

 x* 2ax, we have only to add to them the square of 

 half 2a, that is, ( a) 2 , or a 2 , to get z 2 2ax + a-, 

 the complete square of (x a) 2 . It follows from this, that 

 whenever we have an expression of this form namely, 

 a; 2 + pz, whether p be positive or negative, we can 

 always add to it a term, easily found, that will make 

 the expression a complete square ; this term being no 

 othi-r than the square of ip ; that is ^p 2 ; and moreover, 

 that the root of the square may at once be pronounced 

 to rxs x +J p. In fact, common multiplication shows 

 that z 2 + px + Jp 2 = (x + If)*, whether p be + or . 



We shall give you an example or two of completing 

 the square, as it is called 



). Having the two terms z 2 + 6z, complete the square 

 to which these two terms belong, and express the root of 

 that square. Here half the coefficient of x is 3, /. the 

 wanting term is 3 2 = 9, .*. the square is z 2 + 6z + 9, 

 and the root being z, with the half coefficient taken with 

 its own sign added, we have z 2 + 6z -f- 9=(z + 3) z . 



2. Given z 2 8-r to complete the square. Here the 

 half coefficient is 4, of which the square is 16, /. the 

 complete square is z 2 8z + 1C, the root of which being 

 x, together with the half coefficient, we have z 2 8z + 

 16 = (z 4) 2 . 



3. Given z 2 Zx to complete the square. Here the 

 half coefficient is |, /. its square is j, /. the complete 

 square is z 2 3z + j = (z ij) 3 . 



You can now find no difficjilty in rendering an ex- 

 pression of the form z 2 + px, in which the coefficient of 

 z 2 is unity, and that of z anything, a complete square. 

 It is the accomplishment of this single matter that con- 

 stitutes the whole mystery of quadratic equations. 



To solve a Quadratic Equation. 



RTOE 1. As in a simple equation dispose all the 

 unknown terms on one side of the equation, and all the 

 known on the other ; the unknown side, when the terms 

 are collected, will then consist of but two terms the 

 first containing z 2 , and the second z simply. 



2. If z 2 , in the first term, have a coefficient other 

 than unity, divide both sides by that coefficient; the 

 equation will thus be reduced to the form x 2 + pr. = a, 

 where a is the known side, and p a known coefficient. 

 +3. Add the square of half this coefficient to both sides ; 

 the unknown side will then be a complete square, viz., the 

 square of z + ip, the sign of p being the same as that in 

 the above reduced form; and the known side will be 

 o + J;> 2 . 



4. Extract the square root of each side, and the result 

 will be a simple equation, viz., x + lp = /( + tp 2 )- 



You will observe, that although we extract the square 

 root of each side, yet as far as the unknown side is con- 

 cerned, there is, in reality, no actual extraction per- 

 fi/rmed ; you have nothing to do but to write down z 

 together with half the coefficient of z, in the reduced 

 formx 2 +pz; but, as respects the known side, when 

 this is a number, the square root is actually found by 



VOL. I. 



common arithmetic, unless it is seen to be a surd, when 

 we may leave it under the sign */, the extraction being 

 merely indicated. You must remember, however, that 

 when the square root of a number is actually determined, 

 the sign to be prefixed to that root is ambiguous ;* it may 

 be either + or . 



1. Find the values of x in the equation z 2 +6x=55. 

 Completing the square, z' 2 + 6x + 9 = 64. 

 Extracting the square root, z + 3 = +8. 



.-. z = 3+8 = 5, or 11. 

 If 5 be put for x, the proposed equation is 

 25 + 30 = 55. 



If 11 be put for x, the equation is 121 66 = 55. 



2. 3x 2 + 2.c 9 = 76. Transposing, and then dividing 

 by 3, the coefficient of z 2 , z 2 + $x = *$. 



Completing the square, z 2 + gx + J r = - 3 - + J=" a -' 

 Extracting the root, z + J = ^/ f* = + l .^. 



3. 3x 5 9x 4 = 80. Transposing and dividing by 

 3, z 2 3x = 28. 



Completing the square, z 2 3z + J = 23 + J = "- 



Extracting the root, z $=\/~4 ~ == + T- 



..*=f+y-=7,or-4. " 



4. 7 -\ = 5J. Multiplying by 5x, to clear fractions, 



J 3 -f" 23 = 2G.T, .- . ;r* 2G^ = 25. 

 Completing the square, x* 26-r + 1 3 1 = 1 G9 25 = 144. 

 Extracting the root, x 13= V 144 = +12. 



.-. 1 = 13 + 12 = 25, or 1. 



7 , 2 



5. . . + - = 5. Clearing fractions, 



4 2 

 Transposing, 5a 4* = 2 ..** x 5 ' 



4 4 2 4 _ 14 



Completing the square, *' j x + ^ ~ 5 + 25 25' 



2 14 1 



Eitracting the root, x g=^^j = j 



The square root of 14 is found, by arithmetic, to be 



1 2 



3742; and - of this is '748; also since ;= '4, wo 



have z = -4 + -748 = 1-148, or -348. 



6. J^LL + ?.+ [ 4j =7. Multiplying by x', to clear 



fractions, x* + 11* + 9 + 4x= 7*' .-. 6*' 15* = 9. 



5 3 



Or dividing by 6, ** ^ x = ^. 



5 25 3 25 _ 49 



Completing the square, ** 2 x ~t~ To = 2 "" 10 ~ 16' 



5 49 ,7 



Extracting the root, x j=v-jg=:j7j* 



5,7 1 



.^=44 = 3,or 2- 



7. . i ; = 



-T 



Clearing fractions, and remem- 



bering that sum X dif. = dif. sqrs. 



2r + 2 A/x=16 .# .-.2 Vx = \6 3x. 

 Squaring, to remove the radical, 



.-.9*' 100* = 20fl. 



. .,. , 100 256 



Dividing by 9, ** -- g~' =: ~ef 



Completing the square, 



? 100 2500_^500 J36_._19G 



' 9 -r + 81 "" 81 " ' 9 '" HI 

 50 196 _ , 14 



Extracting the root, x --= ^ ~ai ~ ~f 



g-= ~ai 



14 _ 64 



- 



_ M) 



Sec ante, p. 4C6. 



3 P 



