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Destinyyyy

  • one year ago

Need help with a few problems

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  1. Destinyyyy
    • one year ago
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    Solve the polynomial equation by factoring and then using the zero-product principle. 36y^3-5=y-180y^2

  2. Destinyyyy
    • one year ago
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    @Nnesha

  3. freckles
    • one year ago
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    First step would be to use some algebra to get every non-zero term on one side.

  4. Destinyyyy
    • one year ago
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    36y^3 +18y^2-y-5=0

  5. freckles
    • one year ago
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    is 180 suppose to be 18 or was the 18 suppose to be 180?

  6. Destinyyyy
    • one year ago
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    180 my bad

  7. freckles
    • one year ago
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    \[36y^3+180y^2-y-5=0\] we could try factoring by grouping

  8. freckles
    • one year ago
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    looking like at 36 and 180 what common factor(s) do you notice?

  9. Destinyyyy
    • one year ago
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    y^2

  10. freckles
    • one year ago
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    ok well yes y^3 and y^2 have common factor y^2 but what about 36 and 180?

  11. Destinyyyy
    • one year ago
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    Um 9

  12. freckles
    • one year ago
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    36 and 180 are even so they also have to at least have a common factor of 2 you should eventually see the common factor between 36 and 180 is 36 since 36=36(1) and 180=36(5)

  13. freckles
    • one year ago
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    so the greatest common factor between 36y^3 and 180y^2 is 36y^2 since 36y^3=36y^2*y and 180y^2=36y^2*5

  14. freckles
    • one year ago
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    so you should be able to write 36y^3+180y^2 as: \[36y^3+180y^2=36y^2(y+5)\]

  15. freckles
    • one year ago
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    -y-5 can also be factored

  16. freckles
    • one year ago
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    do you think you can factor out a -1?

  17. Destinyyyy
    • one year ago
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    Okay so... 36y^2(y+5) -1(y+5)=0

  18. freckles
    • one year ago
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    right

  19. freckles
    • one year ago
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    ok Do you know how to factor something like: \[36y^2\cdot Y-1 \cdot Y\] notice both terms have a common factor of ...

  20. freckles
    • one year ago
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    notice the common factor in both terms is Y

  21. freckles
    • one year ago
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    You can factor out Y

  22. Destinyyyy
    • one year ago
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    y= -5, -1/6, 1/6 Now I have to check them

  23. freckles
    • one year ago
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    \[36y^2 \cdot Y-1 \cdot Y \\ Y(36y^2-1) \\ \text{ try the same thing with } \\ 36y^2(y-5)-1(y-5)\]

  24. Destinyyyy
    • one year ago
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    Oops no checking

  25. freckles
    • one year ago
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    oh you have gotten to the answer part

  26. Destinyyyy
    • one year ago
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    Yeah.. Can you help with a few more???

  27. freckles
    • one year ago
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    I can try

  28. Destinyyyy
    • one year ago
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    Solve the radical equation then check. Square root x+24 - square root x-8 = 4

  29. freckles
    • one year ago
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    \[\sqrt{x+24}-\sqrt{x-8}=4 \text{ is this right? }\]

  30. Destinyyyy
    • one year ago
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    Yes

  31. freckles
    • one year ago
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    Ok the way I solve these is: Isolate a square root then square (since you have two square roots you will have to do that twice)

  32. freckles
    • one year ago
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    So can get sqrt(x+24) by itself

  33. freckles
    • one year ago
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    and then square the equation once you have it by itself

  34. Destinyyyy
    • one year ago
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    Square root x+24 = 4- square root x-8

  35. freckles
    • one year ago
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    well one correction

  36. Destinyyyy
    • one year ago
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    Then i put them in parenthesis and square them by 2

  37. freckles
    • one year ago
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    that minus should be a plus

  38. Destinyyyy
    • one year ago
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    Ohh yeah I just saw that

  39. freckles
    • one year ago
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    \[\sqrt{x+24}-\sqrt{x-8}=4 \\ \text{ Add } \sqrt{x-8 } \text{ on both sides } \\ \sqrt{x+24}=4+\sqrt{x-8} \\ \text{ then square both sides } \\ (\sqrt{x+24})^2=(4+\sqrt{x-8})^2\]

  40. freckles
    • one year ago
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    now the left hand side is easy it is just x+24

  41. freckles
    • one year ago
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    right hand side you have to do a bit of multiplying there (distributive property twice) and I know the street language is to use foil

  42. freckles
    • one year ago
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    \[(4+\sqrt{x-8})^2 =(4+\sqrt{x-8})(4+\sqrt{x-8})\]

  43. Destinyyyy
    • one year ago
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    Yeah after this I get a bit mixed up

  44. freckles
    • one year ago
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    \[(a+b)(a+b) \\ a(a+b)+b(a+b) \\ a^2+ab+ba+b^2 \\ a^2+2ab+b^2 \\ \text{ Comparing } (a+b)(a+b) \text{ \to } (4+\sqrt{x-8})(4+\sqrt{x-8}) \\ \text{ we see that } a=4 \text{ and } b=\sqrt{x-8} \\ \text{ so you could plug in into the expanded form of } a^2+2ab+b^2 \]

  45. freckles
    • one year ago
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    replace a with 4 replace b with sqrt(x-8) \[(4)^2+2(4)(\sqrt{x-8})+(\sqrt{x-8})^2 \\ 16+8\sqrt{x-8}+(x-8)\] combine like term terms that is 16-8=?

  46. freckles
    • one year ago
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    do you understand what I did to multiply the right hand side? please let me know if you are still confused about it

  47. Destinyyyy
    • one year ago
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    Not really. I have- 16+4 square root x-8 +4 square root x-8..... So you added the 4s to get 8.. But im confused by the end part

  48. freckles
    • one year ago
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    oh you are confused about combining like terms

  49. freckles
    • one year ago
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    you know if you have 4y+4y this is 8y right?

  50. freckles
    • one year ago
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    4y+4y=4y(1+1)=4y(2)=4(2)y=8y

  51. freckles
    • one year ago
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    or you can just look at it as 4 apples plus 4 apples then you have 8 apples

  52. Destinyyyy
    • one year ago
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    Yes.. Im confused on why square root x-8 isnt x^2+64 .. Why did it stay the same?

  53. Destinyyyy
    • one year ago
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    I know how to do distributive property.

  54. freckles
    • one year ago
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    you mean this: \[(\sqrt{x-8})^2=x-8\]?

  55. Destinyyyy
    • one year ago
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    Nevermind. Whats the next step?

  56. freckles
    • one year ago
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    i guess we can come back to it \[\sqrt{x+24}-\sqrt{x-8}=4 \\ \sqrt{x+24}=4+\sqrt{x-8} \\ (\sqrt{x+24})^2=(4+\sqrt{x- 8})^2 \\ x+24=(4)^2+2(4)\sqrt{x-8}+(\sqrt{x-8})^2 \\ x+24=16+8\sqrt{x-8}+x-8 \\ x+24=8+8 \sqrt{x-8}+x\] well remember how I said we would have to isolate the square root term and then square both sides (we would have to do this twice)

  57. freckles
    • one year ago
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    that is what we have to do here isolate the 8sqrt(x-8) term

  58. freckles
    • one year ago
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    so we need to subtract x and subtract 8 on both sides to do this

  59. Destinyyyy
    • one year ago
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    Okay

  60. Destinyyyy
    • one year ago
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    x+24-8-x= 8 square root x-8

  61. freckles
    • one year ago
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    ok and we do have like terms on on the left

  62. freckles
    • one year ago
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    so we can do x-x and 24-8

  63. Destinyyyy
    • one year ago
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    16= 8 square root x-8

  64. freckles
    • one year ago
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    \[16=8 \sqrt{x-8}\] if you want to you could divide both sides by 8 first then square or you can square then divide both sides by 8^2 honestly you would working with smaller numbers if you go ahead and divide both sides by 8

  65. freckles
    • one year ago
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    \[\frac{16}{8}=\sqrt{x-8} \\ 2=\sqrt{x-8}\] square both sides

  66. Destinyyyy
    • one year ago
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    Okay

  67. Destinyyyy
    • one year ago
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    4= x-8

  68. freckles
    • one year ago
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    right !

  69. Destinyyyy
    • one year ago
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    x=12

  70. freckles
    • one year ago
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    right and then you check solution

  71. Destinyyyy
    • one year ago
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    Its true

  72. freckles
    • one year ago
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    hey @Destinyyyy so do we need to talk about this part: \[(4+\sqrt{x-8})^2 \]

  73. Destinyyyy
    • one year ago
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    Nope I figured it out

  74. Destinyyyy
    • one year ago
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    Next problem... I couldn't find any examples to help me I wasn't sure if Im suppose to minus the x to the other side or what.. x- square root 4x-8 =5

  75. freckles
    • one year ago
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    ok sometimes it is easier to look at it in a not so ugly way well in my opinion it is easier to look at it like this: \[(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b) \\ (a+b)^2=a^2+ab+ba+b^2 \\ (a+b)^2=a^2+2ab+b^2 \ \text{ and after doing the multiplication, plug \in the terms }\] ok but anyways on to next problem

  76. freckles
    • one year ago
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    \[x-\sqrt{4x-8}=5\] again you want to isolate the square root term

  77. freckles
    • one year ago
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    and then square both sides

  78. freckles
    • one year ago
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    you could subtract x on both sides to do this but then you would have a - in front of the square root which is fine you can can square both sides without multiplying both sides by -1

  79. freckles
    • one year ago
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    \[-\sqrt{4x-8}=5-x\] like you can go ahead and square both sides here

  80. Destinyyyy
    • one year ago
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    Ohh okay.. I was wondering about the negative

  81. freckles
    • one year ago
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    or \[\sqrt{4x-8}=x-5 \text{ and then \square both sides } \] pick your favorite

  82. Destinyyyy
    • one year ago
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    -4x-8= 25-10x+1

  83. freckles
    • one year ago
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    \[- \sqrt{4x-8}=5-x \\ \text{ the \square both sides } \\ (-\sqrt{4x-8})^2=(5-x)^2 \\ (-1 \sqrt{4x-8})^2=(5-x)^2 \\ (-1)^2 (\sqrt{4x-8})^2=(5-x)^2 \\ (1) (4x-8)=(5-x)^2 \\ 4x-8=(5-x)^2 \] it looks like you might have expanded the right hand side a bit off

  84. freckles
    • one year ago
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    \[(5-x)^2=(5-x)(5-x)=5(5-x)-x(5-x)=25-5x-5x+x^2 \\=25-10x+x^2 \]

  85. freckles
    • one year ago
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    your equation should look like this instead: \[4x-8=25-10x+x^2 \]

  86. freckles
    • one year ago
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    now if you put all non-zero terms on one side you will see you have a quadratic equation to solve (you might already see you have a quadratic before even doing that)

  87. Destinyyyy
    • one year ago
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    x^2 -14x+33=0

  88. freckles
    • one year ago
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    right and this problem made it easy on us because x^2-14x+33 is factorable

  89. Destinyyyy
    • one year ago
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    x=11,3 now to check

  90. freckles
    • one year ago
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    which of those are actually a solution?

  91. Destinyyyy
    • one year ago
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    3

  92. freckles
    • one year ago
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    hmm...

  93. freckles
    • one year ago
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    are you sure?

  94. freckles
    • one year ago
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    \[x-\sqrt{4x-8}=5 \\ \text{ put in 11 }\\ 11-\sqrt{4(11)-8}=11-\sqrt{44-8}=11-\sqrt{36}=11-6=5 \text{ check mark } \\ \text{ now put \in 3} \\ 3 -\sqrt{4(3)-8}=3-\sqrt{12-8}=3-\sqrt{4}=3-2=1 \text{ no check mark }\]

  95. freckles
    • one year ago
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    the only one that gives you a true equation is x=11

  96. freckles
    • one year ago
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    since you end up with 5=5

  97. freckles
    • one year ago
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    x=3 is an extraneous solution (meaning not actually a solution but it is a solution to an equation we "reduced" our equation to) since 1=5 isn't true

  98. Destinyyyy
    • one year ago
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    Yeah. I didnt bring the negative over

  99. freckles
    • one year ago
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    any questions on those 3 we did together?

  100. Destinyyyy
    • one year ago
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    Nope

  101. freckles
    • one year ago
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    ok enjoy your fun math adventures

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