By Harley Flanders and Justin J. Price (Auth.)

Algebra and Trigonometry

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**Additional info for Algebra and Trigonometry**

**Example text**

Examples: x + 1= 3 solution: 2 4x = 4 solution: 1 x2 = 25 solutions: 5 and —5. For solving harder equations, the usual strategy is to keep modifying the equation until you arrive at some equation you can solve. Take, for example, x2 — x = 25 — x. If x satisfies this equation, it also satisfies x2 = 25, and conversely. The latter equation is simpler; x = 5 or JC = —5. If x satisfies 2x + 1 = 7, then x also satisfies 2x = 6, from which x = 3. In the first example, we added x to both sides of the equation; in the second we subtracted 1 from both sides.

The product is 0 only if x — 4 = 0 or x + 1 = 0 , that is, only if x = 4 or x = — 1. Answer (a) 0,2 (b) 4 , - 1 . Completing the Square The quadratic equation (JC - 2) 2 = 9 is easy to solve because the left-hand side is a perfect square. By inspection, either x — 2 = 3 or x — 2 = —3. Hence x = 5 and x = — 1 are solutions. Now look at the equation x2 + 6x + 4 = 0. The left side is not a perfect square, but something similar is: x2 + 6x + 9. So add 3. Quadratic Equations 5 to both sides: x2 + 6x + 9 = 5, (x + 3)2 = 5.

8. 11. 14. 17. 20. 23. 26. 29. 32. 35. i). x2 + 11* + 30 x2 _ 9X + H 3. 6. X3 _ 4*2 _μ 3 χ 9. 2x2 + llx + 12 12. x6 + 7JC3 + 12 15. 4x2 - 81 18. 9a2 - 4 + (3a - 2)(fl + 1) 21. („4 _ 8 1 ) + v3 + 9 i ; 24. x4-8* 27. x5- 1 30. x6-64 33. 8x376 + 1000 x2 + x - 6 JC2 + 5x - 36 x4 + x3 - 20x2 10x2 - 17x + 3 x 4 _ 5x2 + 6 9y2 - 100 a4 - 1662 b3 - 2 7 8(χ + yf + 27x x5-32 x8- 1 37. 5 using formula (6) and show that your answer agrees with the answer in the text. 38. Write out formula (6) for n = 5 and verify it by multiplication.