By Charles Denlinger and Elaine Jacobson (Auth.)

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**Extra info for Algebra Review**

**Sample text**

LI. 12. L3. 14. L5. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 50. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. i . [i - ( i . (i 51. 1 INTRODUCTION If 2 is substituted in the equation 3x- 2 = 4 the two sides become equal. We say that x = 2 is a solution of the equation. The following operations are helpful for finding the solu tions of an equation: I. II. The same quantity may be added to or subtracted from both sides of an equation. Both sides of an equation may be multiplied (or divided) by the same non-zero quantity.

To explain the meaning of x , where a is an irrational number, is beyond the scope of this text. But it is worth noting that it is possible to use arbitrary real numbers as exponents, while pre serving all the laws of exponents. a EXAMPLE 4 16 8 1 / 2 2/3 =Vl6 = 4 . ( V5 V5 = 5 3 3 ^ , 2 . 2 5 = 5 1 / 2 1 / 3 2 = 4 9 o r g 2/3 . V? = V5I25 EXAMPLE 5 (2 xV) 2 1 / 2 -2 2 / V / 2 y 6 / 2 - 2x y 2 3 28. 1/4 16 does not equal 4. Beware: EXERCISE SET In fact, 16 1/4 2,3 Simplify each of the following: 1.

Thus, x=0 or x+ 7 = 0 , which yields the solutions x=0 (b) or x = -7 or x - 3 = 0, or x=3 . x - 5x+6 = 0 2 (x- 2)(x- 3) = 0. Thus, x- 2 = 0 which yields the solutions x=2 (c) x - 2x+l = 0 2 (x- l)(x- 1) = 0. Thus, x- 1 = 0 , which yields the solution x= 1 . EXAMPLE 3 S O L V E : 2y = 30- 4y 2 62. SOLUTION W e begin by bringing all terms to the left side 2 y + 4 y - 30 = 0. 2 Dividing by 2, we have y +2y-15 = 0 2 (y+5)(y-3) = 0. Thus, y+ 5 = 0 or y- 3 = 0 , or y=3. which yields the solutions y = -5 The solutions of a quadratic equation 2 ax + bx + c = 0 can always be obtained by the quadratic formula: (3) This formula is not "magic", but has a logical derivation from still another method of solving a quadratic equation called "completing the square".