408 



THE POPULAR EDUCATOR. 



preference to any other square number. For multiplying the 

 equation by 4 times the co-efficient of the highest power, will 

 produce the middle term of a binomial square, the third term 

 of which is the square of the co-efficient of the lowest power. 



In the square of a binomial, the first and last terms are 

 always positive. For each is the square of one of the terms of 

 the root, arid all even powers are positive. 



If, then, - a; 2 occurs in an equation, it cannot with this sign 

 form a part of the square of a binomial. But if all the signs in 

 the equation be changed, whilst the equality of the sides will be 

 preserved, the term x~ will become positive, and the square 

 may then be completed. 



EXAMPLE. 



Reduce the equation 

 Changing all the signs, 



x* + 2x = d h. 

 x- 2x = h d. 



In a quadratic equation the first term a; 5 is the square of a 

 single letter. But a binomial quantity may consist of terms, 

 one or both of which are already powers. 



Thus x 3 + a is a binomial, and its square is x 6 + 2ax s -f- a 2 , 

 where the index of x in the first term is twice as great as in the 

 second. When the third term is deficient, the square may be 

 completed in the same manner as that of any other binomial. 

 For the middle term is twice the product of the roots of the 

 two others. 



So the square of a n + a, is x* a + 2az n + a-. 



1 2 i 



And the square of x u + , is a; a + 2aa; u + a 2 . Therefore 



Any equation ivhich contains only two different powers or 

 roots of the unknown quantity, the index of one of which is twice 

 that of the other, may be solved in the same manner as a 

 quadratic equation, by completing the square. 



N.B. It must be observed, that in the binomial root, the 

 letter expressing the unknown quantity may still have a frac- 

 tional or integral index, so that a further operation may be 

 necessary. 



EXAMPLE. 



Reduce the equation x* - x~ = b a. 



Completing the square, x 4 - x 1 + J = J + & - o. 



Extracting and transposing, x~ = J 4; / + Z> a. 



Extracting again, x = + A/ + A/( + 6 a). 



EXERCISE 64. 



1. Eeduce the equation or' + dx = h. 



2. Reduce the equation 3x 2 + 5x = 42. 



3. Reduce the equation x* 15* = 54. 



4. Reduce the equation 4x z* = 32. 



5. Reduce the equation z ?a 4bx = a. 



6. Reduce the equation x 4- 4 *Jx = h n. 







7. Reduce the equation x" + 8" = a + b. 



The solution of a quadratic equation, whether pure or ad- 

 fected, gives two results. For after the equation is reduced, it 

 contains an ambiguous root. In a pure quadratic, this root is 

 the whole value of the unknown quantity. 



Thus the equation x- = 64. 



Becomes, when reduced, x = + A/64 ; 



that is, the value of x is either + 8 or 8, for each of these 

 is a root of 64. Here both the values of x are the same, except 

 that they have contrary signs. This will be the case in every 

 pure quadratic equation, because the whole of the second 

 member is under the radical sign. The two values of the un- 

 known quantity will be alike, except that one will be positive 

 and the other negative. 



But in adfected quadratics, a part only of one side of the 

 reduced equation is under the radical sign. When this pa.rt is 

 added to, or subtracted from, that which is without the radical 

 sign, the two results will differ in quantity, and will have their 

 signs in some cases alike, and in others unlike. 



EXAMPLES. 

 Thus the equation x* + 8x = 20, 



Becomes, when reduced, x = - 4 + A/16 + 20. 

 That is, x = - 4 + 6. 



Here the first value of a; is -4 + 6 = + 2") one positive, and 

 And the second is -4-6 = -10J the other negative. 



Also the equation 



Becomes, when reduced, x = 4 Hh A/16 - 15. 



That is, a: = 4+1. 



Here the first value ofa;is4 + l = + 5}, ,, 

 And the second is 4 - 1 = + 3 j botli P sitl . 



That these two values of x are correctly found, may be 

 proved by substituting first one and then the other, for x itself, 

 in the original equation. 



Thus 5 2 - 8 X 5 = 25 - 40 = 15, 

 And 3 Z - 8 X 3 = 9 - 24 = 15. 



In the reduction of an adfected quadratic equation, the value 

 of the unknown quantity is frequently found to be imaginary. 



Thus the equation a: 2 - 8x = - 20, 



Becomes, when reduced, x = 4 + A/16 20. 

 That is, x = 4 + A/ 4. 



Here the root of the negative quantity - 4 cannot be 

 assigned, and therefore the value of x cannot be found. There> 

 will be the same impossibility in every instance in which the 

 negative part of the quantities under the radical sign is greater 

 than the positive part. 



When one of the values of the unknown quantity in a 

 quadratic equation is imaginary, the other is so also. For both 

 are equally affected by the imaginary root. 



Thus, in the example above, 



The first value of a; is 4 + A/ - 4, 



And the second is 4 V - 4 ; each of which 



contains the imaginary quantity A/ - 4. 



An equation which, when reduced, contains an imaginary 

 root, is often of use to enable us to determine whether a pro- 

 posed question admits of an answer, or involves an absurdity. 



EXAMPLE. 



Suppose it is required to divide 8 into two such parts that the 

 product will be 20. 



If x is one of the parts, the other will bo 8 x. 



By the conditions proposed (8 x) X x 20. 



This becomes, when reduced, x = 4 +_ A/ 4. 



Here the imaginary expression A/ 4 shows that an answer 

 is impossible ; and that there is an absurdity in supposing that 

 8 may be divided into two such parts that their product shall 

 be 20. 



Although a quadratic equation gives two results, yet both 

 these may not always be applicable to the subject proposed. 

 The quantity under the radical sign may bo produced either 

 from a positive or a negative root. But both these roots may 

 not, in every instance, belong to the problem to be solved. 



EXAMPLE. 



Divide the number 30 into two such parts that their product 

 may be equal to 8 times their difference. 



If x = the less, then 30 - x = the greater part. 



By the supposition, x X (30 - a:) = 8 X (30 - 2x). 



This reduced, gives x = 23 + 17 = 40, or 6, the less part. 



But as 40 cannot be part of 30, the problem can have but one 

 real solution, making the less part 6, and the greater part 24. 



The preceding principles in quadratic equations may bo 

 summed up in the following 



GENERAL RULE. 



1. Transpose all the unknown quantities to one side of the 

 equation, and the known quantities to the other. 



2. Make the square of the unknown quantity positive (if it is 

 not already) by changing the signs of all the terms on both sides ; 

 and place it for the first or leading term. 



3. To complete the square, 



(1.) Remove the co-efficient of the second power of the unknown 

 quantity, and add the square of half of the co-efficient of the first 

 power of the unknown quantity to both sides of the equation. 



(2.) Or multiply the equation by four times the co-efficient of 

 ihe highest power of the unknown quantity, anil add to both 

 sides the square of the co-efficient of the first power of the un- 

 known quantity. 



4. Reduce the equation by extracting the square root of both 

 sides ; and transpose the known part of the binomial root thus 

 obtained to the opposite tide, 



