anonymous one year ago PLEASE HELP I WILL GIVE MEDAL Part B: Solve 4x2 -12x + 5 = 0 using an appropriate method. Show the steps of your work, and explain why you chose the method used. (4 points)

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1. pooja195

Have you learned factoring?

2. anonymous

This is an example of factoring trinomials :)

3. anonymous

4. anonymous

im stuck and need to be done within the hour

5. anonymous

this one factors

6. anonymous

ok

7. anonymous

do you know how it factors?

8. anonymous

ehh not really :(

9. anonymous

me neither i cheat

10. anonymous

:( lol

11. pooja195

|dw:1433297189737:dw|

12. anonymous

factors as $(2x-1)(2x-5)=0$ would you like to know how i know it?

13. pooja195

Or that...

14. anonymous

15. anonymous

i will wait and let @pooja195 explain maybe i will learn something

16. anonymous

i have never figured out a way to factor other than to grind it til i find it

17. pooja195

thanks :) Ok well fist you want to multiply the a and c value what do you get?

18. anonymous

this is algebra 1 incase you where wondering!

19. pooja195

Whats $\huge 5 \times 4$

20. anonymous

20...

21. pooja195

good and our B value is -12 right?

22. anonymous

yes sir

23. pooja195

|dw:1433297395133:dw|

24. anonymous

ok

25. pooja195

Now ask yourself what multiplies into -20 BUT also adds up to -12?

26. pooja195

You will have 2 numbers...

27. anonymous

2 and 10?

28. mathmate

multiplies into 20...

29. anonymous

or 4 and 5

30. pooja195

oh right multiplies into 20 but adds to -12 you are close with 10 and 2

31. pooja195

you need a sign on both numbers...

32. anonymous

-2 and -10?

33. anonymous

or -2 and 10

34. pooja195

yes! :) Now... $\huge~(4x^2-2)+(10x+5)$

35. pooja195

Now find the GCF

36. pooja195

for each...

37. anonymous

After @pooja195 is done, I can summarize it all into one huge message of steps so it won't seem all over the place :)

38. anonymous

-2 for the first and 5 for the second?

39. anonymous

thank you aureyliant!

40. mathmate

$$\huge~(4x^2-2\color{red}{x})+(10x+5)$$

41. anonymous

Are you trying to find the zeros?

42. mathmate

$$\huge~(4x^2-2\color{red}{x})\color{red}{-}(10x\color{red}{-}5)$$

43. pooja195

You have the signs messed up but 2x and 5 is right :) $\huge 2x(2x-1)+5(2x-1)$

44. pooja195

*minus

45. pooja195

xD

46. anonymous

so 2x(2x-1) - 5 (2x-1)?

47. anonymous

i know but im making sure that was correct XD

48. anonymous

(2x - 5)(2x - 1)

49. pooja195

now you just combine the parenthese and the outter numbers (2x-5)(2x-1) and you end up with Sat's grinding :P

50. anonymous

You need to combine the first loner and the second loner and bring them together

51. anonymous

ok is that all to it?

52. pooja195

Then you set both equations to 0

53. anonymous

Trying to factor by splitting the middle term 2.1 Factoring 4x2-12x+5 The first term is, 4x2 its coefficient is 4 . The middle term is, -12x its coefficient is -12 . The last term, "the constant", is +5 Step-1 : Multiply the coefficient of the first term by the constant 4 • 5 = 20 Step-2 : Find two factors of 20 whose sum equals the coefficient of the middle term, which is -12 . -20 + -1 = -21 -10 + -2 = -12 That's it Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and -2 4x2 - 10x - 2x - 5 Step-4 : Add up the first 2 terms, pulling out like factors : 2x • (2x-5) Add up the last 2 terms, pulling out common factors : 1 • (2x-5) Step-5 : Add up the four terms of step 4 : (2x-1) • (2x-5) Which is the desired factorization Equation at the end of step 2 : (2x - 5) • (2x - 1) = 0 Step 3 : Solve (2x-5)•(2x-1) = 0 (This goes towards solving for the zeros but that's only if the Q is asking that, too)

54. pooja195

2x-5=0 2x-1=0

55. anonymous

^^ Yup

56. anonymous

ok

57. pooja195

Solve it .-.

58. anonymous

Now you could either do that or use the quadratic formula

59. anonymous

But thats a story for another day

60. pooja195

^no

61. pooja195

Just set it to 0 its sooooo much more easier

62. anonymous

Solving a Single Variable Equation : 3.2 Solve : 2x-5 = 0 Add 5 to both sides of the equation : 2x = 5 Divide both sides of the equation by 2: x = 5/2 = 2.500 Solving a Single Variable Equation : 3.3 Solve : 2x-1 = 0 Add 1 to both sides of the equation : 2x = 1 Divide both sides of the equation by 2: x = 1/2 = 0.500

63. anonymous

can you guide me through setting through 0?

64. anonymous

It depends on the type of factoring Q Pooja

65. pooja195

2x-5=0 we want to isolate X

66. anonymous

It may not be as easy to split the middle term

67. anonymous

ok

68. pooja195

69. pooja195

that way on the left side the 5 will cancel what does the equation look like now?

70. anonymous

ok so 2x=5?

71. pooja195

Yes now just divde both sides by 2

72. pooja195

what do you get?

73. anonymous

1x=1/2????

74. pooja195

No

75. pooja195

keep the 5

76. anonymous

Now we know the two values for x... also I summarized solving for the zeros above @cacaman

77. pooja195

you get x=5/2

78. anonymous

X=5?

79. anonymous

just kidding

80. pooja195

-_- i gave u the answer :P

81. pooja195

2x-1=0 solve

82. pooja195

I think you answered this one earlier...

83. anonymous

2x=1

84. anonymous

so 1/2

85. pooja195

YES! now what are the 2 answers?

86. pooja195

87. anonymous

5/2 and 1/2

88. pooja195

Good! :)

89. anonymous

lol im nervous!

90. pooja195

why so? :P and why did we choose this do you know?

91. anonymous

no..

92. pooja195

You tell me because i dont know either.

93. anonymous

LOL

94. anonymous

We chose this because it was an easier method in my books?

95. pooja195

Find a better reason -_-

96. pooja195

I guess you can write because it is a easier way to do it :/ and since its a trinomial grouping is involved...

97. Nnesha

there is another method to factor quadratic equation cross multiply instead group method |dw:1433299203705:dw| factor first and last term then cross multiply after that combine them if you get middle term then (2x-5) and (2x-1) are factor of that equation