The way in which this principle may be used in solving equations is illustrated below: EXAMPLE 1. Solve the equation x2-8x+15=0. SOLUTION. Since the trinomial x2-8 x+15 may be factored (see § 58) into (x-3)(x−5), the given equation may be written in the form (x-3)(x-5)=0. Thus, we are to have the product of two numbers equal to zero, and it will be, according to the Principle above, whenever either of the factors is equal to zero. That is, when either x-3=0, or x-5=0. But these last two equations give us x=3 and x=5. Therefore, either x=3 or x=5 is a solution (root) of our equation. Ans. CHECK. When x=3, the left side of the given equation becomes 32-8.3+15, which reduces to 9-24+15, and this reduces (as the equation demands) to 0. When x=5, the left side of the given equation becomes 52-8.5+15=25-40+15=0. EXAMPLE 2. Solve the equation x2-4 x-21=0. SOLUTION. Factoring, we have (x−7)(x+3)=0. Whence, by the Principle above, we have either x-7=0, or x+3=0. Therefore, the roots are 7 and -3. Ans. CHECK. 72-4.7-21-49-28-21=0; (−3)2—4(−3)−21=9+12-21=0. NOTE. Each of the equations just considered has two roots. This is true of every equation which contains the second (but no higher) power of the unknown letter. Such equations are called quadratic equations and will be more fully considered in Chapter XVI. WRITTEN EXERCISES Solve the following equations by factoring. Check the first ten. 1. x2-7x+10=0. 2. x2-5x+6=0. 3. x2+8x+15=0. 4. x2+7x-30=0. For further exercises on this topic, see the review exercises below, and Appendix, p. 301. 26. — x2−x+25=0. 27. r2+r=0. EXERCISES - REVIEW OF CHAPTER VI 1. Using Formulas I and II (§§ 48 and 50) state (orally) a different form for each of the following expressions. 2. Using Formula III (§ 51) state (orally) a different form for each of the following expressions. (a) 3(x+y). (b) m(a+b). (d) m2n2(a2+b2+c2). (e) ab(c+d)-pq(r+s). (c) mn(a-b). (f) xy(x+y)+yz(z2+y). 3. Factor each of the following expressions. (See §§ 52 and 54.) 4. By use of Formula IV (§ 57) state (orally) the result of each 6. By use of Formulas V and VI (§§ 59 and 60) expand each of the following expressions by inspection. 7. Show that each of the following trinomials is a perfect square and find its square root. (See §§ 63 and 64.) 8. By use of Formula VII (§ 65) state the result of each of the following multiplications.. (a) (x-1)(x+1). (b) (x2+1)(x2−1). (c) (x2+y2)(x2-y2). (d) (2x+3 y) (2 x −3 y). (e) (2 ab+3) (2 ab-3). (f) (xy+z2)(xy-z2). (g) (3 m2n−q) (3 m2n+q). . (h) (3 x3-4 y2) (3 x3+4 y2). (i) (-5x-y)(-5x+y). (j) (-3 a3b+c2d2) (-3 a3b-c2d2. (k) (xm-1+yn+1)(xm−1 — yn+1). 9. Factor each of the following expressions. (See § 67.) (a) a2-16. (b) 25 a2-36 b2. (c) 400 x2-81 y2. (d) x2y2-256. (e) x2n-y2m. (ƒ) x2-(y+z)2. (g) 9 m2-(p+q)2. (h) x2-(2x-3y)2. (i) 81 x2-(3 m-2 n)2. (j) (2a+3b)2- (a+b)2. (1) (a+b+c)2− (a−b—c)2. 10. Solve (by factoring) each of the following equations. (See EXERCISES-APPLIED PROBLEMS 1. A certain rectangle is 2 feet longer than wide, and its area is 8 square feet. What are its dimensions? HINT. Let x = the width. Then x+2= the length. Now form an equation for x, transpose all its terms to the left side and solve as in § 68. Keep only the + root, since the - root has no meaning in this problem. 2. The perimeter (distance around) a certain rectangle is 10 feet, while the area is 6 feet. What are the dimensions? 3. The figure represents a rectangular plot of ground, within which are two equal square flower beds. The border (shaded in the figure) is everywhere 3 feet wide and contains 201 square feet. side of each flower bed, and what are What is the length of O FIG. 35. FIG. 34. the dimensions of the rectangle? 4. In the figure are two circles so placed that the inner one just touches the outer one. Show that the area of the crescent thus formed (shaded in the figure) may be expressed in the form T(R+r) (R-r), where R is the radius of the large circle, and r that of the small circle. |