If we add together the two values of x in the general equa tion, the radical parts having opposite signs disappear, and we obtain p Ρ = 2 p. Thus, in the equation x2-10x=-16, the two roots are 8 and 2, whose sum is +10, the coefficient of x taken with the contrary sign. If the two roots are equal numerically, but have opposite signs, their sum is zero, and the second term of the equation vanishes. Thus the two roots of the equation x2-16 are +4 and -4, whose sum is zero. This equation may be written x2+0x=16. = 274. The product of the two roots is equal to the second member of the equation taken with the contrary sign. If we multiply together the two values of x (observing that the product of the sum and difference of two quantities is equal to the difference of their squares), we obtain Thus, in the equation a2-10x=-16, the product of the two roots 8 and 2 is +16, which is equal to the second member of the equation taken with the contrary sign. 275. The last two principles enable us to form an equation whose roots shall be any given quantities. Ex. 1. Find the equation whose roots are 3 and 5. According to Art. 273, the coefficient of the second term of the equation must be −8; and, according to Art. 274, the second member of the equation must be -15. Hence the equa Ex. 2. Find the equation whose roots are -4 and -7. Ex. 7. Find the equation whose roots are and +1. Ex. 8. Find the equation whose roots are 1±√5. 276. Every equation of the second degree whose roots are a and b, may be reduced to the form (x-a) (x—b)=0. Thus the equation x2-10x=-16, whose roots are 8 and 2, may be resolved into the factors x-8-0 and x-2=0. It is also obvious that if a is a root of an equation of the second degree, the equation must be divisible by x-a. Thus the preceding equation is divisible by x-8, giving the quotient x-2. Ex. 1. The roots of the equation x2+6x+8=0 are -2 and -4. Resolve the first member into its factors. Ex. 2. The roots of the equation x2+6x-27=0 are +3 and 9. Resolve the first member into its factors. Ex. 3. The roots of the equation x2-2x-24=0 are +6 and 4. Resolve the first member into its factors. Ex. 4. Resolve the equation a2+73x+780-0 into simple factors. Ans. (x+60) (x+13)=0. Ex. 5. Resolve the equation x2-88x+1612=0 into simple factors. Ans. (x-62) (x−26)=0. Ex. 6. Resolve the equation 2x2+x-6=0 into simple factors. Ans. 2(x+2)(x-3)=0. Ex. 7. Resolve the equation 3x2-10x-25=0 into simple factors. Ans. 3(x−5)(x+3)=0. Discussion of the General Equation of the Second Degree. 277. In the general equation of the second degree x2+px=q, the coefficient of x, as well as the absolute term, may be either positive or negative. We may therefore have the four following forms: We will now consider what conditions will render these roots positive or negative, equal or unequal, real or imaginary. Therefore, in the first and second forms, the sign of the roots will correspond to the sign of the radicals; but in the third and fourth forms the sign of the roots will correspond to the sign of the rational parts. Hence, in the first form, one root is positive and the other negative, and the negative root is numerically the greatest; as in the equation x2+x=6, whose roots are +2 and -3. In the second form one root is positive and the other negative, and the positive root is numerically the greatest, as in the equation x2-x=6, whose roots are -2 and +3. In the third form both roots are negative, as in the equation x2+5x=-6, whose roots are -2 and -3. In the fourth form both roots are positive, as in the equation x2-5x=-6, whose roots are +2 and +3. 279. Equal and unequal roots. In the first and second forms the two roots are always unequal; but in the third and fourth forms, when q is numerically equal to 4' x 22, the radical part of both values of a becomes zero, and the two roots are then said to be equal. In this case Thus, in the equation x2+6x=-9, the two roots are -3 and 3. We say that in this case the equation has two roots, because it is the product of the two factors x+3=0 and x+3=0. So, also, in the equation x2-6x=-9, the two roots are +3 and +3. 280. Real and imaginary roots. 4 Since 22, being a square, is positive for all real values of p, it follows that the expression 2+ can only be rendered neg 4 ative by the sign of q; that is, the quantity under the radical sign can only be negative when q is negative and numerically greater than 22. Hence, in the first and second forms, both roots 4 are always real; but in the third and fourth forms both roots are imaginary when q is numerically greater than p2 d Thus, in the equation x2+4x=-6, the two roots are and in the equation x2-4x=-6, the two roots are +2±√2. It will be observed that when one of the roots is imaginary, the other is imaginary also. 281. Imaginary roots indicate impossible conditions in the proposed question which furnished the equation. The demonstration of this principle depends upon the following proposition: the greatest product which can be obtained by dividing a number into two parts and multiplying them together is the square of half that number. Let p represent the given number, and d the difference of the parts. Then, from page 89, and and Ρ d + 22 the greater part, Now, since p is a given quantity, it is plain that the prod uct will be the greatest possible when d=0; that is, the great est product is the square of, half the given number. For example, let 10 be the number to be divided. We have 10=1+9; and 9×1= 9. 10=2+8; and 8x2=16. 10=3+7; and 7x3=21. 10 4+6; and 6x4=24. Thus we see that the smaller the difference of the two parts, the greater is their product; and this product is greatest when the two parts are equal. Now, in the equation x2-px=-9, p is the sum of the two roots, and 7 is their product. Therefore q can never be greater than p2 |