SOLUTIONS TO THE EXERCISES.] M ATHEM ATICS. AL GEBR A. 



603 



and real; for o j 46, under the radical, is in this case 

 positive. 



3. If a' bo less than 46, the two roots must be ima- 

 ginary ; for then a 2 46, under the radical, is negative. 



Hence, by examining the coefficients of a quadratic, 

 we may ascertain tho character of its roots without 

 actually solving the equation. 



16. - H = 6. Clearing fractions, 



(x + o) a + (x a) 2 = 6(x a 2 ) 



that is, 2x2 + 2a 2 = 6x 2 o 2 6 



Transposing, (5 2)x 3 = a 2 (2 + 6), 



.'. x= +a v &_2' 



17. Put y for Jx, then multiplying by 6, we have 

 3^2 2y = 133, 

 2 133 

 Dividing by 3, y * g y = -3 



..... 2 ,1 133 . 1 400 



Completing the square, y 2 - y + ==-- + g = ~{r> 



_ . . . 



Extracting the root, y - 



400 



20 



. . 1/2 = x = 49, or 40J. 

 Or, by Rule II., page 474, 

 ;;;/- 2y= 133, . . Gy 2= V(15%+4)= ^1600 



6 



. . y* = x 49, or 40J. 

 18. (3* 2)(1 *) = 4; that is, 



3x* + x 2 = 4, .-. 3*2 x = 2. 



1 2 



Dividing by 3, x* 3 x = 3 ' 



Completing the square, 



1 1 1 2_ 23 

 * ~3 Z +30 = 30~3 36 



Extracting the root, x ^ = + g-\/ 23, 



19. 15* 3 48x + 45 = 0, or 15z 48* = 45. 

 Dividing by 15, 



--*. 



Completing tho square, 

 16 64 



64 3= ^ 

 38 ~25 



Extracting the root, 



.'. x 



8 + v/- 

 5 



We might have known, without actually solving either 

 of the two equations in Examples 18 and ]9, that the 

 roots were imaginary, for these equations being 3x a x 

 + 2 = 0, and 15-c" 48* + 45 = 0, we see at once 

 that four times the product of the extreme terms 

 exceeds the square of the middle term. (Sec the prin- 

 ciple 3, above.) 



Squaring, 



.. , 



.< + '2 ^ .r '< 



and this equation has already been solved in Ex- 

 ample 12. 



QUESTIONS BEQUIKING QUADRATIC EQUATIONS. 



PAGE 475. 



1. Divide the number 33 into two such parts, that 

 their product may be 162. 



Let x be one part, then 33 a; is the other ; and by 

 the question, 



33x a;* = 162; or x 3 33* = 162, 

 .-. Rule II., 2x 33= V (1089.- 648) = </441 = +_21, 



33 + 21 

 .-.x= = = 27, or 6. 



.-. 33 x = 6, or 27. 

 Hence the parts are 6, and 27. 



Otherwise. Let the two parts be x and y ; then by tho 

 question 



x _i.y = 33, andry= 162 

 .-. x 3 + 2zi/ + j/ 2 = 1089 

 And 4xy = C48 



441 



Subtracting, x* 



That is, 



z _, / )2 = 441, .-. x y 



And since x + y 



33 



.'. adding and subtracting, 2.e = 33 + 21, .'. x = 27or6 

 2y = 33 + 21, .'. y = 6 or 27. 



2. Find two numbers whoso difference is 9, and which 

 are also such that their sum multiplied by the greater 

 gives 266 for tho product. 



Let x be the greater number, then x 9 IB the less, 

 and by the question, (2x 9)x = 20(3 ; that is, 



2x2 _9x 

 .-.Rule II., 4x 9' 



266 



. 



.-. * 9 = 5, or 185; 

 hence the numbers are either 5 and 14, or 9J and 



Otherwise. Let x and y be thejiumbers ; then by the 

 question, x y = 9, and (z + y)x =.266. 



From the first of these equations, x = y + 9, .. by 

 substitution in the second, (2y + 9)(y + 9) = 266 ; or 



2 y 2 + 27y + 81 = 266, .-. 2y2 + 27y = 185 

 . . Rule II., 4y + 27 = V(1*SO + 729)= J 2209 = +47 



27 + 47 

 .-. y = =- - = 5 or 18| 



.'. x = y + 9 = 14, or 9i 

 Or thus : x y = 9 



Adding, 



9 + 



26G 



2x2 _ 9x = C66, 



and the remaining steps as in tho fir.=t solution. 



3. A company at a tavern had 7 <k to pay ; but two 

 of them having left, the others had each la. more to 

 pay than his fair share : how many persons were theru 

 at first? 



Suppose there were x persons, then the fair share of 



each was -- shillings ; but after two had left, the share 



X 



144 



to be paid by each of the others was g _ g ; and by the 



Jfl "* + 

 x 2* z 



Clearing fractions, 144x = 144x 288 + y? 2x 



.-. x 2 2.B = 288, . . completing the square, 

 3? 2x + 1 = 289, . ' . extracting, x 1 -= + 17, 



.-. x= 18, or 16. 



Consequently there were 18 persons at first. Tho shnro 

 of each was therefore 8s. ; but when two had left, the 

 share of each of the remaining 1 6 was 9s. 



4. A purse contains 24 coins of silver and .copper ; 



