47* 



MATHEMATICS. ALGEBRA. 



[QUADRATIC EQUATIONS. 



night hare completed the solution without clearing 

 the radical from tho equation LV + 'J N 'x-10 x, 

 thu* : by transposing. 



2 16 



Now, since x in Uio first term U the sgium of Jx'm tho 

 second, it U plain that wo mar treat this as a quadratic 

 equation, regarding ,/ * M tuo unknown quantity . . 

 completing toe square, 



Extracting tho root, 



8 G^ 



" 3 ' ''"* ' r 9 

 NOTE. And in a similar way may any equation of tho 

 -e*" -\-pjf + o, in which the exponent of x in tho 

 first term is double that of x in tho second, be solved 

 as a quadratic ; the equation just considered Li x + 



* Tr *here & ex P- 1 of x, in the first term, is 

 double J, tho exp. of x in the second. 



EXAMPLES FOB EXERCISE. 

 3. *' 8* = 9. 



7. 5** + 4* = 273. 



9. 



2. 

 4. 

 6. j' x 170 = 40. 



4x' x 

 8. 3 = 11- 



10. 5** + 3=4x + 159. 



H. 



13. + = l 



14. * 40x + 39 = 0. 



20. ^r-r4 + 



15. 



19. 15*' 48x + 45=0. 



Tho preceding method of solving a quadratic equation 

 is that which is usually employed for the purpose ; in 

 certain cases, however, it is attended with inconvenience ; 

 in those cases, namely, where the completion of the 

 square introduces numerical fractions into the work. 

 We shall now show you an improvement upon tho com- 

 mon rule ; by adopting which, in practice, you will 

 prevent the introduction of fractions. 



The most general form of the quadratic equation, after 

 tho proper preliminary reductions, is ax* + bx = e. Let 

 each side be multiplied by 4a ; tho equation will then bo 

 40*2* + 4ake 4*io, or. which U tho same thing, 

 = 



The first member of this, you will observe, consists of 

 two terms namely, the $quare of 2ax and "1> times 2ax ; 

 ao that, regarding 2ox as tho unknown quantity, we 

 ball complete the square by adding b 1 ; and tho root of 

 that square will bo 2ox + 6 : we therefore proceed thus 

 Completing the square, (2ox)' + 26(2ax) + fc 1 = 4nc + 1-. 

 Extracting tho root, 2ax + 6 = J (4<M + 6 1 ). 



Hence any quadratic may bo reduced to a simple 

 equation, of which this is tho general model, without the 

 introduction of any fractions. Tho steps from tho pro- 

 posed quadratic ax* + fee c, to the simple equation 

 Zoz + 6 J (4ac + 6*), need not be gone through in 

 each particular example ; tho reduced equation may bo 

 derived at once as follows : 



SM mlr. p. 4*3. 



TH roou oT tkU rqiuttao. nd aba thoM of ranaUnn 10, ar imagin. 



Tt ; JkM to, Uw nlM of * la neb will b. found to InrolT. tho tqiun 



of MtuiT* n,im Vr ; w* nur conrtudf , thrrcfoir, tht It li not 



P-rtbU to *!* 4UMT o( UMM ^mtOam b, *nj ml r .lu. of * (m p. 4W) . 



Itulo II. Take twice tho coefficient of x* in the pro- 

 posed quadratic ; this will bo tho coefficient of x in tho 

 derived simple equation. 



tho first term thus found, connect with its own 

 sign tho coefficient of x in that proposed : and the 

 member of tho derived simple equation will be obtained. 



To form the second member, multiply tho second 

 member of tho proposed, by four times the coefficient of 

 x*, add the square of tho next coefl'u irnt (that of f) to 

 the result, and cover the whole with tho radical sign. 



This is the rule for tho solution of any quadratic ; it is 

 a mere translation in words of the formula for the 

 solution ; namely, 



Proposed quadratic. Derived simple equation. 



ox s + ox c. Zax + (/=/ (4ac + b"). 



You may either commit tho rti/c to memory, or keep 

 tho formula before your mind's cyo ; remembering that, 

 as a square is always plus, b- is to take tho positire sign, 

 whether b, in tho quadratic, is + or . Thus (see Ex. 6, 

 P- 473), 



Quadratic. Simple equation. 



6x' 15x = 9 



or 



2x 3 5.r=3. . . . 4* 5=^(24 + 25)= V49=+ 7. 

 Again, Ex. 5, p. 473, 

 5*' 4x=2. . . 

 Also Ex. 3, p. 473, 



And from each of these simple equations you may 

 deduce tho value of x, at a single step : thus 



T, 



Ex. 3. x 



4=^(40+16)= V50= 



,_=k__7,or-4; 



5 + 7 

 Ex. 6. = ^ = 3, or ^, 



as at the pages referred to. All the examples worked by 

 the common method you may re-sol ro in this way, which 

 will be much the shorter and easier, whenever fractions are 

 unavoidable by the former rule. Wo refrain from giving 

 additional examples here, solely because we expect you 

 to act upon this recommendation. 



To solve apair of Equations with Two Vnknoien Quan ! 

 when one is a Simple Equation and the other a Quadratic. 

 Tho mode of proceeding in this case scarcely requires 

 any formal directions. From the simple equation an 

 expression for one of tho unknowns may be easily found 

 in terms of the other and known quantities, as at pago 

 471 ; and this expression substituted in the quadratic 

 will, of course, give an equation with only one unknown 

 quantity ; the other being eliminated in consequence of 

 the substitution. Thus 



From the first equation x = 7 

 f 2y ', Substituting this in tho 



second, it becomes 

 (7 2y)S + 3 (7 2>/)i/ y 3 = 23 ; that is, 49 28u 

 + 4y + 21i/ Ci/- y"- = 23, or 3y= + 7y 

 .-. Rule II., Oy+7= /(312 + 49)= >/361 = 



1 3 



so that tho two values of x and ;/ which satisfy tho pro- 

 posed pair of simultaneous filiations arc either 



2 *~^f = J = 2o} From the first equation *= ; 

 Substituting this in the second, it becomes 







5y a = 20, or, multiplying by 2, and transposing, 



