Example. If x is a real quantity, prove that the expression x2+2x-11 can have all numerical values except such as lie between 2(x-3) 2 and 6. Let the given expression be represented by y, so that then multiplying across and transposing, we have x2+2x (1− y) +6y-11=0. This is a quadratic equation, and if x is to have real values 4(1-y)2-4 (6y-11) must be positive; or simplifying and dividing by 4, y2-8y+12 must be positive; that is, (y-6) (y-2) must be positive. Hence the factors of this product must be both positive, or both negative. In the former case y is greater than 6; in the latter y is less than 2. Therefore y cannot lie between 2 and 6, but may have any other value. In this example it will be noticed that the expression y2-8y+12 is positive so long as y does not lie between the roots of the corresponding quadratic equation y2 — 8y+12=0. This is a particular case of the general proposition investigated in the next article. 308. For all real values of x the expression ax2+bx+c has the same sign as a, except when the roots of the equation ax2+ bx+c=0 are real and unequal, and x lies between them. CASE I. Suppose that the roots of the equation ax2+bx+c=0 are real; denote them by a and B, and let a be the greater. Now if x is greater than a or less than ẞ, the factors x-a, x-B are either both positive or both negative; therefore the expression (x - a) (x – B) is positive, and ax2 + bx+c has the same sign as a. But if x lies between a and B, the expression is negative, and the sign of ax2+bx+c is opposite to that of a. CASE II. If a and B are equal, then ax2+bx+c=a (x − a)2, and (-a) is positive for all real values of a; hence ax2+bx+c has the same sign as a. CASE III. Suppose that the equation ax2+bx+c=0 has imaginary roots; then but since b2-4ac is negative [Art. 303], the expression is positive for all real values of x; therefore ax2+bx+c has the same sign as a. EXAMPLES XXXIV. a. Find (without actual solution) the nature of the roots of the following equations: 13. If the equation 2+2 (1+k)x+k2=0 has equal roots, what is the value of k? 14. Prove that the equation 3mx2-(2m+3n) x+2n=0 has rational roots. 15. Without solving the equation 3.2-4x-1=0, find the sum, the difference, and the sum of the squares of the roots. 16. Show that the roots of a (x2 — 1)=(b − c) x are always real. Form the equations whose roots are 17. 3+√5, 3-√√5. 18. −2+√3, −2−√3. 19. α b 5' 6' If a, ẞ are the roots of the equation pr2+7x+r=0 find the values of 29. If a, 3 are the roots of x2-px+q=0, and a3, ß3 the roots of x2 - Px+Q=0, find P and Q in terms of p and q. 30. If a, ẞ are the roots of x2-ax+b=0, find the equation a B B2' a2' whose roots are 31. Find the condition that one root of the equation ax2+bx+c=0 may be double the other. 32. Form an equation whose roots shall be the cubes of the roots of the equation 2x (x - a) = a2. 33. Prove that the roots of the equation 34. Show that (a+b+c) x2-2 (a+b)x+(a+b-c)=0 has rational roots. 35. Form an equation whose roots shall be the arithmetic and harmonic means between the roots of x2-px+q=0. 36. In the equation px2+qx+r=0 the roots are in the ratio of I to m, prove that (12+m2) pr+lm (2pr — q2)=0. 37. Show that if x is real the expression can have all values except MISCELLANEOUS THEOREMS. 309. We shall now give general proofs of the statements made in Art. 55. We suppose n to be positive and integral. I. To prove that x" -y" is always divisible by x- Divide any by x-y till a remainder is obtained which does not involve x. Let be the quotient, and R the remainder; then x-y"=Q(x-y)+R. Since R does not contain x, it will remain unaltered whatever value we give to x. .. R=0; that is, since there is no remainder, a" — y" is divisible by x-y. II. To prove that x+y" is divisible by x+y when n is odd, but not when n is even. With the same notation as before, x+y=Q(x+y)+R. Now since R does not involve x, it will remain unaltered whatever value we give to x. Put xy, then that is, (y)"+y"=QxO+R; (1) if n is odd, (-y)"+y"-y"+y"=0; (2) if n is even, (-)"+y" y"+y"=2y"; hence there is a remainder when n is even, but none when n is odd; which proves the proposition. In like manner it may be proved that "-y" is divisible by x+y when n is even; and "+y" is never divisible by x-y. The results of the present article may be conveniently stated as follows: (i) For all values of n, xn — yn = (x − y) (xn−1+xn−2y+xm¬3y2+......+yn−1). (ii) When n is odd, x2+yn=(x+y) (xn−1 — xn−2y+xn−3y3 — . (iii) When n is even, xn−yn=(x+y) (x2-1 — xn−2y+xn−3y2 — ...... +yn−1). — yn−1). 310. The Remainder Theorem. If any algebraical expression x"+P1x"¬1+P2x"−2+P3x"¬3+.............. +Pn-1x+Pn be divided by x-a, the remainder will be Divide the given expression by x-a till a remainder is obtained which does not involve x. Let Q be the quotient, and R the remainder; then Since R does not contain x, it will remain unaltered whatever value we give to x. Put xa, then a"+P1a"-1+P2a"-2+...+P2-1a+P2=Q×0+R, which proves the proposition. From this it appears that when an algebraical expression is divided by x- -a, the remainder can be obtained at once by writing a in the place of x in the given expression. 311. The remainder is zero when the given expression is exactly divisible by x-a; hence we deduce: The Factor Theorem. If any algebraical expression involving x become equal to 0 when a is written for x, it will contain x-a as a factor. Example 1. The remainder when x4 - 2x3+x-7 is divided by x + 2 is that is, (-2)-2(-2)+(-2)-7; Or the remainder may be found more shortly by substituting x = - - 2 in [{(x-2)x}x+1] x -7. Example 2. Resolve into factors x3+3x2 - 13x-15. =3; hence Dy trial we find that this expression vanishes when x= 2-3 is a factor. .. x3+3x2 – 13x-15=x2 (x-3) + 6x (x − 3) + 5 (x − 3) NOTE. The only numerical values that need be substituted for x are the factors of the last term of the expression. Thus, in the present by making trial of 5, we should have detected the factor x + 5. case, |