Algebra Review by Charles Denlinger and Elaine Jacobson (Auth.)

By Charles Denlinger and Elaine Jacobson (Auth.)

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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".

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