CHAPTER XXVI. QUADRATIC EQUATIONS. 206. SUPPOSE the following problem were proposed for solution: A dealer bought a number of horses for $280. If he had bought four less each would have cost $8 more: how many did he buy? We should proceed thus: 280 Let x=the number of horses; then the number of dollars each cost. If he had bought 4 less he would have had x- 4 horses, and 280 each would have cost dollars. X-4 280 280 .. 8+ -4 whence X (2 — 4) +35 (2 — 4)=35.0 ; .. 22 — 4x +35x – 140=35x ; .:. x2 - 4x=140. Here we have an equation which involves the square of the unknown quantity; and in order to complete the solution of the problem we must discover a method of solving such equations. 207. DEFINITION. An equation which contains the square of the unknown quantity, but no higher power, is called a quadratic equation, or an equation of the second degree. If the equation contains both the square and the first power of the unknown it is called an affected quadratic; if it contains only the square of the unknown it is said to be a pure quadratic. Thus 2.22 – 5x=3 is an adfected quadratic, and 5x2 =20 is a pure quadratic. 208. A pure quadratic may be considered as a simple equation in which the square of the unknown quantity is to be found. 9 25 Example. Solve 22-27 32_11 Multiplying across, 922—99=25x2 – 675; .. 16x2=576 ; .. 22=36; and taking the square root of these equals, we have X= +6. Note. We prefix the double sign to the number on the righthand side for the reason given in Art. 178. 209. In extracting the square root of the two sides of the equation x2=36, it might seem that we ought to prefix the double sign to the quantities on both sides, and write Ex=+6. But an examination of the various cases shows this to be unnecessary. For Ex= +6 gives the four cases : +x= +6, +x= -6, - x = = +6, =-6, and these are all included in the two already given, namely, x= +6, x=-6. Hence, when we extract the square root of the two sides of an equation, it is sufficient to put the double sign before the square root of one side. -X= -+24. EXAMPLES XXVI. a. Solve the following equations: 1. 4x2 +5=22 +17. 2002 3 232-6 x2 – 4 522-10 =0. 7 =5. (x - 1)3 — x+24 3+2 -2. or AFFECTED QUADRATIC EQUATIONS. 210. The equation x2=36 is an instance of the simplest form of quadratic equations. The equation (x – 3)2 = 25 may be solved in a similar way; for taking the square root of both sides, we have two simple equations, X-3= +5. Taking the upper sign, X—3= +5, whence x=8; taking the lower sign, x - 3= -5, whence x=... the solution is X=8, or – 2. Now the given equation (0 – 3)2=25 may be written x2 - 6x +(3)2 =25, x2 - 6x=16. Hence, by retracing our steps, we learn that the equation x2 - 6x=16 can be solved by first adding (3)2 or 9 to each side, and then extracting the square root; and the reason why we add 9 to each side is that this quantity added to the left side makes it a perfect square. Now whatever the quantity a may be, x2 + 2ax+a-=(x+a)?, and x2 – 2ax+a?=(x-a); so that, if a trinomial is a perfect square, and its highest power, x?, has unity for a coefficient, the term without x must be equal to the square of half the coefficient of x. Example 1. Solve 7x=22–8. Transpose so as to have the terms involving & on one side, and the square term positive. Thus 2c2 – 7x=8. (7\2 49 Completing the square, 22 – 73+ (2)" =8+4 : Example 2. Solve 32 - 3x2=10x. 3x2 + 10x = 32. Divide throughout by 3, so as to make the coefficient of a unity. 10 32 Thus 3 3' 10 32 25 completing the square, 2+ 3 ; 5\2 121 that is, 2+ 9 ; + 100 5 11 } =2, or -51. 3 3 Example 3. Solve 7 (c +2a)+ 3a2=5a (726+ 23a). Simplifying, 7cco +28ax + 28a? + 3a2=35ax +115a”, that is, 70c2 – 7ax=84a2. Whence 22 – ax=12a”; 2 ax+ 2 C a completing the square, 22 = 12aa + 4 49a2 that is, ; 4 7a 2 2 ; i's x=4a, or – 3a, 211. We see then that the following are the steps required for solving an affected quadratic equation. (1) If necessary, simplify the equation so that the terms in x2 and x are on one side of the equation, and the term without x on the other. (2) Make the coefficient of x2 unity and positive by dividing throughout by the coefficient of x2. (3) Add to each side of the equation the square of half the coefficient of x. (4) Take the square root of each side. 212. In all the instances considered hitherto the quadratic equations have had two roots. Sometimes, however, there is only one solution. Thus if x2 – 2x+1=0, then (x-1)2=0, whence x=1 is the only solution. Nevertheless, in this and similar cases we find it convenient to say that the quadratic has two equal roots. EXAMPLES XXVI. b. 1. 5.x2 + 14x=55. 2. 3x2 + 121=44.x. 3. 25x=6x2 +21. 4. 8x2+x=30. 5. 3x2 +35=22x. 6. x+22 - 6x2=0. 7. 15=17x +4.ro. 8. 21+x=2x2. 9. 9x2 - 143 – 6x=0. 10. 12.x2=29.x – 14. 11. 20.x2=12 – x. 12. 19x=15 - 8x2. 13. 21x2 + 22x+5=0. 14. 50.x2 – 15x=27. 15. 18x2 – 27x – 26=0. 16. 5.x2=8x+21. 17: 15x2 – 2ax=a2. 18. 21.x2=2ax +3a2. 19. 6x2=11kx + 742. 20. 12x2 +23kx + 10k2=0. 21. 12.x2 - cx – 20c2=0. 22. 2(x-3)=3(x+2)(x-3). 23. (x+1) (2x +3)=4x2 - 22. 24. (3.c-5) (2.6-5)=.co+2x - 3. 5x+7 5x - 1 32 - 8 5x – 2 25. = 3x + 2. 26. 27. X-1 x+1 2 X – 2 x+5 3.x - 1 6 5.X – 7 28. 1 29. 4x +7 x+7 72 - 5 2x – 13" x +3 2x – 1 1 1 6 31. 3 X 1 4 1 32. + =6) 3 5 9 - 2x * 4 5 3 4 3 34. 35. 1 x + 2 2 X+6° X 2 13 + 3.3 X – 5 30. 1+x 35 * 33. OT 20 SOLUTION BY FORMULA. 213. After suitable reduction and transposition every quadratic equation can be written in the form axo+bx+c=0 where a, b, c may have any numerical values whatever. If therefore we can solve this quadratic we can solve any. |