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Mimi_x3

  • 4 years ago

Can you please help me with this maths question 1. The cubic polynomial P(x)= x^3 +rs^2 +sx+t , where r,s and t are real numbers has three real zeroes,1,a, and -a (a) find the value of s + t

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  1. myininaya
    • 4 years ago
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    is the polynomial written correctly?

  2. Mimi_x3
    • 4 years ago
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    woops my bad sorry P(x) = x^3 +rx^2 +sx +t

  3. helpme1234
    • 4 years ago
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    Colorful pi can you help me?

  4. myininaya
    • 4 years ago
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    let me try to help him and then i will look at your question

  5. Mimi_x3
    • 4 years ago
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    him ? are yu referring to me ?

  6. myininaya
    • 4 years ago
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    is this algebra?

  7. Mimi_x3
    • 4 years ago
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    noo , its a polynomial question

  8. myininaya
    • 4 years ago
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    so what class is it then?

  9. Mimi_x3
    • 4 years ago
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    class ? what do yu mean ? Its from the topic polynomials

  10. myininaya
    • 4 years ago
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    \[P(x)=(x-1)(x-a)(x+a)=x^3+rs^2+sx+t\]

  11. Mimi_x3
    • 4 years ago
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    is that the answer ?

  12. myininaya
    • 4 years ago
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    so I guess you are doing it for fun? \[P(x)=(x-1)(x^2-a^2)=x^3+rs^2+sx+t\] no i just thought u would like to see the steps

  13. Mimi_x3
    • 4 years ago
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    no , im not doing it for fun , its my maths assignment LOL

  14. myininaya
    • 4 years ago
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    \[P(x)=x(x^2-a^2)-1(x^2-a^2)=x^3+rsx^2+sx+t\]

  15. myininaya
    • 4 years ago
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    so the class is called polynomials lol?

  16. Mimi_x3
    • 4 years ago
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    what does yu mean by class ?

  17. myininaya
    • 4 years ago
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    \[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t\] people go to class to learn and get assignments lol

  18. Mimi_x3
    • 4 years ago
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    no the class is called maths LOL

  19. myininaya
    • 4 years ago
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    \[P(x)=x^3-x^2-a^2x+a^2=x^3+rsx^2+sx+t\]

  20. Mimi_x3
    • 4 years ago
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    2 unit maths to be specific

  21. myininaya
    • 4 years ago
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    ok we are almost done with this problem can you guess what to do next?

  22. myininaya
    • 4 years ago
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    we want both sides to be the same

  23. Mimi_x3
    • 4 years ago
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    collecting like terms ?

  24. myininaya
    • 4 years ago
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    ok the the coefficient of x^2 on left hand side is -1 the coefficient of x^2 on the other sides is rs so what can you say about rs and -1?

  25. myininaya
    • 4 years ago
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    they must be equal

  26. myininaya
    • 4 years ago
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    so we need that for the other terms as well

  27. myininaya
    • 4 years ago
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    we have -1=rs -a^2=s a^2=t

  28. polpak
    • 4 years ago
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    I seriously think I'm being trolled here.. but just to ease my tired brain.. \[\sqrt{25} \ne -5\]

  29. myininaya
    • 4 years ago
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    you are right polpak

  30. Mimi_x3
    • 4 years ago
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    wait , dnt yu factorise it further to (x^2-a^2)(x-1)

  31. myininaya
    • 4 years ago
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    that =5

  32. myininaya
    • 4 years ago
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    factor?

  33. polpak
    • 4 years ago
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    Someone was trying to tell me that it equals both 5 and -5

  34. myininaya
    • 4 years ago
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    ok the objective is to find s+t

  35. polpak
    • 4 years ago
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    But I can only really prove it by going to the definition. Which they don't accept.

  36. Mimi_x3
    • 4 years ago
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    ohh yeah , my bad

  37. polpak
    • 4 years ago
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    Sorry for bothering you mimi, but myin rarely watches group chat (with good reason)

  38. myininaya
    • 4 years ago
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    what do you think s+t is based on what i gave you above? hint we don't need that first equation rs=-1

  39. Mimi_x3
    • 4 years ago
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    yu talking to me or polpak ? LOL

  40. myininaya
    • 4 years ago
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    you

  41. Mimi_x3
    • 4 years ago
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    lols , tbh im lost , i dont get it xDD

  42. myininaya
    • 4 years ago
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    do you understand when I wrote P as P(x)=(x-1)(x-a)(x+a)?

  43. Mimi_x3
    • 4 years ago
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    yeah , i dont understand the part -1 = rs -a^2 = s a^2 = t where did yu get it from ?

  44. myininaya
    • 4 years ago
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    both sides had to be equal we were looking at \[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \]

  45. myininaya
    • 4 years ago
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    look at the coefficients of each term on both sides we want each term to be the same as it is on the other side

  46. myininaya
    • 4 years ago
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    so if we look at the x^3's they are good they both have coefficient 1 what is the coefficient of x^2 on left hand side ? what is the coefficient of x^2 on right hand side?

  47. Mimi_x3
    • 4 years ago
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    ohh i get it rs = -1

  48. myininaya
    • 4 years ago
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    yes now what is the coefficient of x on the left hand side? what is the coefficient of x on the right hand side?

  49. Mimi_x3
    • 4 years ago
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    x on the left hand side: 1 i think x on the right hand side 1 i think

  50. myininaya
    • 4 years ago
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    no what is the number in front of x on boths?

  51. myininaya
    • 4 years ago
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    both sides*

  52. Mimi_x3
    • 4 years ago
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    there's no other number

  53. myininaya
    • 4 years ago
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    -a^2? s?

  54. Mimi_x3
    • 4 years ago
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    lols , its a letter not a number

  55. myininaya
    • 4 years ago
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    they are numbers

  56. Mimi_x3
    • 4 years ago
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    they are ? arent they are called pronumerals ?

  57. Mimi_x3
    • 4 years ago
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    are they*

  58. myininaya
    • 4 years ago
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    never heard of this term

  59. Mimi_x3
    • 4 years ago
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    ohh ok

  60. myininaya
    • 4 years ago
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    when you are doing math and you don't see a word and you see a letter it usually represents a number unless it is the word a or i

  61. myininaya
    • 4 years ago
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    becauce a and i can be words

  62. Mimi_x3
    • 4 years ago
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    ohh ok , i never knew that they are numbers xD

  63. myininaya
    • 4 years ago
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    but usually the book will make it clear to you what they are

  64. myininaya
    • 4 years ago
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    yep

  65. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  66. myininaya
    • 4 years ago
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    a^3 is nowhere to be found we aren;t solving equation for x

  67. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  68. Mimi_x3
    • 4 years ago
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    ohh ok

  69. myininaya
    • 4 years ago
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    ok so look again at the polynomial what is the coefficent of x on both sides?

  70. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  71. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  72. Mimi_x3
    • 4 years ago
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    x= s

  73. myininaya
    • 4 years ago
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    ok no \[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \] ^ ^ | | -a^2 = s

  74. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  75. Mimi_x3
    • 4 years ago
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    ohh lols , sorry im so slow of understanding it

  76. myininaya
    • 4 years ago
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    ok do you see the terms that don't an x in em? there terms are called constants and they must be equal of that we have the same polynomial on both sides

  77. myininaya
    • 4 years ago
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    god my english is getting bad

  78. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  79. Mimi_x3
    • 4 years ago
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    lols , probably getting annoyed of me not getting it xD

  80. myininaya
    • 4 years ago
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    no its just late here i been up for along time today

  81. myininaya
    • 4 years ago
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    a long*

  82. polpak
    • 4 years ago
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    Ditto!

  83. myininaya
    • 4 years ago
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    so what are the constants on both sides?

  84. myininaya
    • 4 years ago
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    the terms that don't have a x in them

  85. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  86. Mimi_x3
    • 4 years ago
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    t

  87. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  88. myininaya
    • 4 years ago
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    t and what is it on the other side?

  89. myininaya
    • 4 years ago
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    remember you are looking for non-x terms

  90. Mimi_x3
    • 4 years ago
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    ok , so x= a^3

  91. Mimi_x3
    • 4 years ago
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    a

  92. myininaya
    • 4 years ago
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    a or a^2?

  93. Mimi_x3
    • 4 years ago
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    a

  94. myininaya
    • 4 years ago
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    \[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \] ^ ^ | | a^2 = t

  95. Mimi_x3
    • 4 years ago
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    a^2 my bad

  96. myininaya
    • 4 years ago
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    so what is s again? and what is t again?

  97. Mimi_x3
    • 4 years ago
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    s = 1 t = a^2

  98. myininaya
    • 4 years ago
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    s is not 1

  99. Mimi_x3
    • 4 years ago
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    a^3

  100. myininaya
    • 4 years ago
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    not a^3 scroll up we have it above

  101. Mimi_x3
    • 4 years ago
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    -a^2

  102. Mimi_x3
    • 4 years ago
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    lolss , im so~ slow

  103. myininaya
    • 4 years ago
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    YES! :)

  104. Mimi_x3
    • 4 years ago
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    lols , finally i get it

  105. myininaya
    • 4 years ago
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    ok now s+t is -a^2+a^2 right?

  106. Mimi_x3
    • 4 years ago
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    yeah

  107. myininaya
    • 4 years ago
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    but what is -a^2+a^2?

  108. Mimi_x3
    • 4 years ago
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    so substitute it in s and t

  109. Mimi_x3
    • 4 years ago
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    idk

  110. myininaya
    • 4 years ago
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    -5+5 is

  111. myininaya
    • 4 years ago
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    -6+6 is -24+24 is -999+999 is

  112. Mimi_x3
    • 4 years ago
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    where did 5 come from ?

  113. myininaya
    • 4 years ago
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    i'm just trying to help you think what -a^2+a^2 is

  114. Mimi_x3
    • 4 years ago
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    0 ?

  115. myininaya
    • 4 years ago
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    yes! :)

  116. myininaya
    • 4 years ago
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    and now you are done

  117. Mimi_x3
    • 4 years ago
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    really ? omg ! yay i got it right xD

  118. Mimi_x3
    • 4 years ago
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    thank you very mcuh for helping (:

  119. myininaya
    • 4 years ago
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    np

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