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# 0011 chapter iv

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### 0011 chapter iv

2. 2. x2 – 3x + 2 = 0
3. 3. s6 + 8s3 – 6 = 0
4. 4. n2 – 6n + 10 = 0
5. 5. g8 + 2g4 – g = 0</li></ul>-317500-619941Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Solve the equation that is in quadratic form.<br /><ul><li>a8 + 2a4 – 8 = 0</li></ul> _____________________________________________________<br /><ul><li>l2 + 4l2 – 6 = 0</li></ul> _____________________________________________________<br /><ul><li>e4 – 8e2 – 3 = 0
6. 6. _____________________________________________________
7. 7. l6 – 10l – 5 = 0
8. 8. _____________________________________________________
9. 9. i10 – 8i5 – 4 = 0
10. 10. _____________________________________________________
11. 11. s6 – 5s3 – 25 = 0
12. 12. _____________________________________________________
13. 13. -349954-516441h2/4 + 8h1/4 – 12 = 0
14. 14. _____________________________________________________
15. 15. a6- 5a4 – 15 = 0
16. 16. _____________________________________________________
17. 17. n8 + 12n2 – 8 = 0
18. 18. _____________________________________________________
19. 19. e9 – 3n3 – 10 = 0
20. 20. _____________________________________________________
21. 21. x2/3 – 2x 1/3 = 8
22. 22. _____________________________________________________
24. 24. organize the equation if it is quadratic equation; and
25. 25. solve the equation by factoring or quadratic formula.</li></ul>A variety of equations can be transformed into quadratic equations and solved by methods that we have discussed in the previous section. We will consider fractional equations, equations involving radicals and equation that can be transformed into quadratic equations by appropriate substitutions. Since the transformation process may introduce extraneous roots which are not solutions of the original equation, we must always check the solution in the original equation.<br />Example: Solve 1 1 7<br /> x+2 + x+3 = 12<br />Solution: First note that neither -2 nor -3 can be a solution since at either of these points the equation is meaningless.<br />Multiplying by the LCD, 12(x+2) (x+3), we get<br />12(x+3) + 12(x+2) = 7(x+2) (x+3)<br />24x + 60 = 7(x2 + 5x + 6)<br />or<br />7x2 + 11x – 18 = 0<br />Factoring, we get, (7x + 18)(x – 1) = 0<br />x = 1 or -18 <br /> 7<br />If x = 1, _1_ _1_ _1_ _1_ _7_<br /> 1+2 1+3 = 3 + 4 = 12<br />Therefore x = 1 is a solution.<br />If x = -18, __1__ + __1__<br /> 7 -18/7 + 2 -18/7 + 3<br />= __7__ + __7__<br />-18 + 14 -18 + 21<br />= _-7_ + _7_ = _7_<br /> 4 3 12<br />Therefore, x = _-18_ is a solution.<br /> 7<br />Example 2. 3x + 4 = x + 16 - 2<br />Solution: squaring both sides of the equation, we obtain,<br />(3x+4) 2 = x + 6 + 4<br />2x – 16 = 4 x + 16 <br />Dividing both sides by 2 gives, x – 8 = -2 x + 16 <br />Squaring both sides of the equation we get <br /> (x - 8) = (–2x + 16) 2<br />x2 - 16x + 64 = 4 (x +16)<br />x2 – 20x = 0 <br />x(x – 20) =0<br />x = 0 or x = 20 <br />Check: if x = 0, 30+ 4 = 20+16- 2 64 = 36 - 2 <br />8 = 6 – 2 <br />8 ≠ 4<br />Therefore x = 20 is not a solution of the original equation.<br />Thus the only root of 3x+4 = x+16 - 2 is 0.<br />Many equations are not quadratics equations. However, we can transform them by means of appropriate substitutions into quadratics equations and then solve these by techniques that we know.<br />Example: Solve:<br />a. 2x-2 – 7x-1 + 3 = 0<br /><ul><li>x4 – 2x2 – 2 = 0
26. 26. x4x+12 – x4x+1- 2=0</li></ul>Solutions <br />a. Let u = x-1. Then u2 = (x-1)2 = x-2 and our equation becomes <br />2u2 – 7u + 3 = 0, a quadratic equation in u.<br />To solve the equation, we factor the left-hand side.<br />(2u – 1)( u – 3) = 0<br />U = ½ or u = 3<br />Since u = x-1, x-1= 1x=12 or x-1 = 3, from which<br />x = 2 or x = 13<br />Check: if x = 2, 2(2-2) – 7(2-1) + 3 = 24-72+3=0<br />Thus x = 2 is solution<br />If x = 1/3, 2(1/3)-2 – 7(1/3)-1 + 3 = 2(3)2 – 7(3) + 3<br />So, x = 1/3 is a solution.<br /><ul><li>Let u = x2. Then u2 = x4 and the given equation becomes a quadratic equation in u.
27. 27. u2 – 2u – 2 = 0</li></ul>By the quadratic formula;<br />u = 2± -22-41-221 = 2±21+22=1±3<br />u = 1 + 3 or u = 1-3<br />Since u = x2 and u = 1-3 < 0, we have to discard this solution.<br />u = x2 = 1+3 implies<br />x = ± 1+ 3<br />It is simple to verify that both values of x satisfy the original equation. The roots of x4 – x2 – 2 = 0 are 1+ 3 and - 1+ 3.<br />Let u = x4x+1 . This substitution yields a quadratic equation in u.<br />u2 – u – 2 = 0<br />(u – 2)(u + 1) = 0 <br />u = 2 0r u = ˉ1<br />u = x4x+1 = 2 implies x = 2(4x + 1)<br /> or x = ˉ27<br />u = x4x+1 = 1 implies x = ˉ4x – 1<br /> or x = ˉ15<br />Again, it can easily be verified that both solutions check in the original equation.<br /> The roots are ˉ27 and ˉ15.<br />Exercise:<br />Solve the following equation by transforming each into a quadratic equation by a suitable substitution.<br />y - y - 20 = 0<br />r6 + 26 r3 – 27 = 0<br />(x – 4)3 + (x – 4)3/2 – 6 = 0<br />-440223-252096Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br /> <br />Instruction: Solve the following equation.<br />1x+1+ 43x+6= 23<br /><ul><li> _____________________________________________________</li></ul>1x+ 1x + 1= 920<br /> _____________________________________________________<br />x+2x+3= 2x-33x-7<br /><ul><li> _____________________________________________________</li></ul>x-1x+ xx-1=3<br /><ul><li> _____________________________________________________</li></ul>3x-12 15= 16-3xx+6+ x+35<br /><ul><li> _____________________________________________________</li></ul>3x+2= 2 6-x2x<br /><ul><li> _____________________________________________________</li></ul>-443592-202169x-2x-3x-1(x-2)+3x+1x+1 =5<br /><ul><li> _____________________________________________________</li></ul>x-2+1=x+1<br /><ul><li> _____________________________________________________</li></ul>x+4=2x+7<br /><ul><li> _____________________________________________________</li></ul>x+6=2x+6<br /><ul><li> _____________________________________________________</li></ul>x-3x+4=2x+2<br /><ul><li> _____________________________________________________</li></ul>3x+4=x+16-3<br /><ul><li> _____________________________________________________</li></ul>3x+2+6x=4<br /><ul><li> _____________________________________________________</li></ul>5y+19+y=7<br /><ul><li> _____________________________________________________</li></ul>x+3-x-2=1x+3<br /><ul><li> _____________________________________________________</li></ul>Solve for x.<br /><ul><li>x + 4 = 4
28. 28. 3x + 4 - 5 = 0
29. 29. 4x = 45
30. 30. 3x - 5 = x+9
31. 31. x²+15x = x + 2
32. 32. Reduce to quadratic equation.
33. 33. x4 – 5x + 4 = 0
34. 34. 4(x + 3) + 5x+3 = 21
35. 35. x2/3 – 5x1/3 – 6 = 0
36. 36. (x2 + 4x)2 – (x2 + 4x) = 20
37. 37. 2x4 – 9x2 + 7 = 0
38. 38. Solve for x.
39. 39. 6x+1 + 11-x = 2x
40. 40. 1x-2 + 3x-3 + 52-x = 0
41. 41. 2x+13 + 22x+1 = 53
42. 42. x-1x+2 + x+3x-2 = 2
43. 43. x-1x+2 + x+3x²-4 = 3x-15(x-2)