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anonymous

  • one year ago

Solve the equation using square roots. x^2– 15 = 34 A) z = 1/5 or z = -4 B) z = 1 or z = -4 C) z = 1/5 or z = 4 D) z = 1 or z = 4

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  1. anonymous
    • one year ago
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    @steve816

  2. steve816
    • one year ago
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    What you want to do first is to get the x alone. I'll show you!|dw:1440177092884:dw|

  3. steve816
    • one year ago
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    Now we take the square root on both sides of the equation.|dw:1440177139414:dw|

  4. anonymous
    • one year ago
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    sorry just realized I put up the wrong question, I already solved that one. @steve816

  5. anonymous
    • one year ago
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    5z^2 - 21z + 4 = 0

  6. anonymous
    • one year ago
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    Solve the equation by factoring. 5z^2 - 21z + 4 = 0

  7. Rushwr
    • one year ago
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    wat's the answer u got ?

  8. anonymous
    • one year ago
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    for the first question?

  9. anonymous
    • one year ago
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    \[\pm7\]

  10. Rushwr
    • one year ago
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    the 2nd one

  11. anonymous
    • one year ago
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    I need help with that one

  12. Rushwr
    • one year ago
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    \[5z ^{2}-21z+4=0\]

  13. Rushwr
    • one year ago
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    \[5z ^{2}-21z+4=0\]\[5z ^{2}-20z-1z+4=0\]

  14. Rushwr
    • one year ago
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    did u get tht part?

  15. anonymous
    • one year ago
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    why did u take out the one from 21

  16. Rushwr
    • one year ago
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    it's like this u see the middle number 21z right? 1st we multiply 4 from 5 which is 20 right? Now we know we need 4 numbers right? So using the factors of 20z we are gonna make 21z 20z* 1 is 20z right? but we also know -20z-1z= 21z right? So we substitute those 2 numbers instead of 21z !

  17. Rushwr
    • one year ago
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    I'm not sure if u got this !

  18. anonymous
    • one year ago
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    yeah I kinda do

  19. Rushwr
    • one year ago
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    sure?

  20. Rushwr
    • one year ago
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    \[5z ^{2}-20z-1z+4=0\] \[5z(z-4)-1(z-4)=0\] did u get what I did in this step ?

  21. anonymous
    • one year ago
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    yes, if we multiply the numbers we get \[5z^2 -20z -1z + 4 = 0\]

  22. Rushwr
    • one year ago
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    yeah since we have two (z-4) Now we gonna consider only one of it ! So now:\[5z(z-4)-1z(z-4)=0\]\[(5z-1) (z-4)=0\]

  23. anonymous
    • one year ago
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    I don't understand this.. I have a doctor appointment now, if u can help me later on.. sorry

  24. Rushwr
    • one year ago
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    oki

  25. Nnesha
    • one year ago
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    i like ur pfp. :)

  26. anonymous
    • one year ago
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    I need help with this question.. I have a lot more to do :P

  27. Nnesha
    • one year ago
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    first you need to get `like` terms on one side and variable x on the other side

  28. Nnesha
    • one year ago
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    so move the -15 to the right how would you do that >

  29. Nnesha
    • one year ago
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    ?*

  30. anonymous
    • one year ago
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    x^2 = 34 - 15?

  31. Nnesha
    • one year ago
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    no it's already negative at left side to cancel out you would do `opposite` of `subtraction`

  32. anonymous
    • one year ago
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    to add.. x^2 = 34 + 15?

  33. Nnesha
    • one year ago
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    yes right

  34. Nnesha
    • one year ago
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    and then take square root `both` sides to cancel out square of x.

  35. anonymous
    • one year ago
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    so we don't solve for 34 + 15

  36. Nnesha
    • one year ago
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    we should solve for x (i guess it supposed to be `z`)

  37. Nnesha
    • one year ago
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    you answer choices are wrong....

  38. Nnesha
    • one year ago
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    your*

  39. anonymous
    • one year ago
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    wait Lol the question up there is the wrong question

  40. Nnesha
    • one year ago
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    ye...

  41. anonymous
    • one year ago
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    I put it up there that it was wrong.. sorry just realized we were doing the wrong Q

  42. anonymous
    • one year ago
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    Solve the equation by factoring. 5z^2 - 21z + 4 = 0

  43. Nnesha
    • one year ago
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    there are like 4 method to solve quadratic equations you can use the quadratic formula or you can find factors of that equation

  44. Nnesha
    • one year ago
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    to factor it find two number when you multiply them you should get product of AC and when you add them you should get middle term \[\huge\rm Ax^2+Bx+C=0\] a=leading coefficient b=middle term c=constant term so what is A and C in that equation .

  45. anonymous
    • one year ago
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    21z + 4 ?

  46. Nnesha
    • one year ago
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    hmm nope |dw:1440193043766:dw|

  47. anonymous
    • one year ago
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    Ooh A

  48. anonymous
    • one year ago
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    5z^2 + 4

  49. Nnesha
    • one year ago
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    no just the coefficient so A = 5 b=21 c=4 now multiply a times c = 5 times 4 = 20 find two number when you multiply them you should get 20 but when you add or subtract them you should get middle term which is 21

  50. anonymous
    • one year ago
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    I don't really get it, what am I supposed to do now?

  51. Nnesha
    • one year ago
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    what two number would you multiply to get 20 ? but when you add them you should get 21

  52. anonymous
    • one year ago
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    5*4 = 20 2*10 = 20

  53. anonymous
    • one year ago
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    when u add them all up u get 21

  54. Nnesha
    • one year ago
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    yes but \[4+5 \cancel{=} 20\]

  55. Nnesha
    • one year ago
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    no just 2 number

  56. Nnesha
    • one year ago
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    you can't add 4 number to get 20 here is an example let's say a and b are numbers so when i multiply them a times b = 20 i should add `same` two numbers a+b (or a-b) to get middle term

  57. Nnesha
    • one year ago
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    what other two numbers you can multiply to get 20 ?

  58. anonymous
    • one year ago
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    2 and 10

  59. Nnesha
    • one year ago
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    and??

  60. anonymous
    • one year ago
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    1 and 20.

  61. Nnesha
    • one year ago
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    perfect 20 times 1 =20 20+1=21

  62. anonymous
    • one year ago
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    Ohh I see

  63. Nnesha
    • one year ago
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    \[\huge\rm 5z^2+1z+20z+4\] now use group method to factor it

  64. Nnesha
    • one year ago
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    |dw:1440193728090:dw| make a group of two terms what is common in first two terms ?

  65. anonymous
    • one year ago
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    so we add those with the z first?

  66. Nnesha
    • one year ago
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    no if you add them you will get original equation 5z^2+20x+4 :P

  67. Nnesha
    • one year ago
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    find common of `5x^2+1x` and common factor of `20x+4`

  68. anonymous
    • one year ago
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    okay for 5x^2 + 1x its 1?

  69. anonymous
    • one year ago
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    and then 2 for 20x + 4

  70. Nnesha
    • one year ago
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    not just one what about the variable ?

  71. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @proud_yemeniah_ and then 2 for 20x + 4 \(\color{blue}{\text{End of Quote}}\) greatest common factor yes two is common but it's not greatest common

  72. anonymous
    • one year ago
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    x is the common factor?

  73. Nnesha
    • one year ago
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    factors of 20 are: `20,5,4,2,1` factors of 4 are : `4,2,1` which one is greatest `common` factor ?

  74. anonymous
    • one year ago
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    4

  75. Nnesha
    • one year ago
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    yes right

  76. Nnesha
    • one year ago
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    |dw:1440194265638:dw| when you take out 1z from `5z^2+1x` what will have left in the parentheses

  77. Nnesha
    • one year ago
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    remember both parentheses should be the same!

  78. Nnesha
    • one year ago
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    `5z^2+1z` not x :P

  79. anonymous
    • one year ago
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    5z^2

  80. Nnesha
    • one year ago
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    what will get get when you take out z from 5z^2 ?

  81. anonymous
    • one year ago
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    5^2 and then if u solve it it will be 10

  82. Nnesha
    • one year ago
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    in other words divide both terms by common factor

  83. Nnesha
    • one year ago
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    hmm no \[\huge\rm \color{Red}{\frac{ 5z^2 }{ z }+\frac{ 1z }{ z }}+(\frac{ 20z }{ 4 }+\frac{ 4 }{ 4 })\] \[\huge\rm \color{Red}{z(??+??)}+4(??+??)\] divide by common and then write your answer in the parentheses

  84. anonymous
    • one year ago
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    (5^2 + 1) + (5z + 1)

  85. Nnesha
    • one year ago
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    don't forget the common factor that would stay outside the parentheses do you mean 5 to the 2 power ?

  86. anonymous
    • one year ago
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    yes

  87. Nnesha
    • one year ago
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    \[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times z }{ z}\]

  88. Nnesha
    • one year ago
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    can you cancel out z?

  89. anonymous
    • one year ago
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    Umm so we remove the 2 and not the z or both?

  90. Nnesha
    • one year ago
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    \[\frac{ 2 \times 2 }{ 2}\] would you cancel both 2s's or just one ?

  91. anonymous
    • one year ago
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    I'm not sure

  92. Nnesha
    • one year ago
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    \[\frac{ \heartsuit \times \cancel{\heartsuit}}{\cancel{ \heartsuit }}\] one heart canceled with one heart.

  93. Nnesha
    • one year ago
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    just one \[\frac{ 2 \times \cancel{2} }{ \cancel{2}}\] you can multiply 2 at the top you will get 4/2 when u divide them you will get 2 if you cancel out both 2 with one you will get `1` as an answer which is wrong

  94. Nnesha
    • one year ago
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    \[\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times\cancel{ z} }{\cancel{ z}}\]

  95. anonymous
    • one year ago
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    oh okay

  96. Nnesha
    • one year ago
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    now both parentheses are common so you can take it out

  97. Nnesha
    • one year ago
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    wait there is a typo

  98. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @proud_yemeniah_ (5^2 + 1) + (5z + 1) \(\color{blue}{\text{End of Quote}}\) so it should \[\huge\rm z(5z+1)+4(5z+1)\] and the hing i gave you both parentheses should be the same

  99. Nnesha
    • one year ago
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    |dw:1440195435458:dw| when you take out (5z+1) from z(5z+1) what will get