30. Determine the values of k in order that the equation 31. Determine the value of m in order that the equation may have equal roots. (mx + b)2 = 4ax 32. Find the value of m in order that the equation and determine the value of m in order that the resulting quadratic equation in x may have equal roots. 34. Solve the equation x2+1 = x into a product of two factors of the first degree. 37. Between what limits must x lie, if the expression is to be negative? x2-12x+27 38. Between what limits must the values of the fraction 5x 21 x2 + 16 lie, if is restricted to taking real values? CHAPTER IV THE RESOLUTION OF A TRINOMIAL OF THE SECOND DEGREE INTO FACTORS OF THE FIRST DEGREE. 413. Consider the trinomial x2 + px + q, in which p and q are given constants and x an arbitrary quantity of any magnitude. The value of this trinomial expression will not be changed if the same quantity, 22, is added and subtracted; the fol-. lowing identities result: 4 x2 + px+9=x2 + px + 22 + 9 Since the difference of two squares is equal to the product of the sum of the quantities by their difference (94), then Hence, the trinomial x2 + px + q may be written (6) x2 + px+q= (x − x ̧) (x — x„). This product can be made zero by putting x = = x19 when the second factor is zero, factor is zero, or by putting x = x2, when the first Thus by a second method the two roots of the equation x2 + px + q = 0, have been found, as well as the factors of its first member. This decomposition of a trinomial of the second degree into factors holds in all cases, whether the roots are real or imaginary and whatever is the value of x. 414. The most general trinomial of the second degree is which can be written ax2 + bx + c, b ax2 + bx + c = a(x2 + 1/14 x + 1 ) = a(x2 + px + q), where p = and q By 413, (6), this polynomial in paren b a = с a thesis can be decomposed into the factors x -x, and x therefore, (7) X1 ax2 + bx + c = a(x - x) (x − x2) whatever is the value of x. Here x1 and x2 are the two roots of the equation x2 + px + q = 0, or, what is equivalent to the same thing, the roots of Hence, x2 — x — 6 — ( x − x ̧) (x − x ̧) = (x − 3) (x + 2). - — Therefore, according to formula (7), 4.x2 4.c 15 + (x − 1 ) (x+3)=(2.x — 5) (2x + 3). 415. Factor ar2+2bxy + cy2+ 2dx + 2ey+f. Arranging the terms with respect to x and x, and equating to zero, (1) ax2 + 2(by + d)x + (cy2 + 2ey + ƒ) = 0. Solving by formula (4), 2408, then NOTE. It has been assumed that a is not zero. In case a is zero and b not zero, solve for y as we have above for x and proceed in a similar manner. EXAMPLE.-Factor 2 x2 xy-y2+3x+3y 2. On comparing this equation with the general equation, it is seen 1 (ax+by+d+ R) (ax + by + d − R) -- ! (2 x − b y +} + ↓ (3 y − 5)) × a 2 In case the quantity under the radical in R is a perfect square (8411, 2), In this case the factors in (3) above are rational. This is the case in the preceding example; for (bd — ae)2 — (d2 — ac) (d® — uƒ ) : : ( − 15)* − ()}) () = (25) 0. EXERCISE LXXI Investigate whether the following expressions can be separated into factors or not, and if this is possible, whether the factors are rational or irrational. If the factors are rational, they can be found directly by 413, or indirectly by solving the given equation. 1. x2 -7x+12. 2. x2 + 13x + 30. 3. x2 9 x 15. 4. x2 + 12 x + 27. |