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some.random.cool.kid

  • one year ago

can someone help me with quadratics please? you can make your own examples and guide me to the the answer.

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  1. phi
    • one year ago
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    can you be more specific?

  2. some.random.cool.kid
    • one year ago
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    ok well I'm working on quadratics functions and i need help solving them. I don't have any specific examples I just want to practice them.

  3. phi
    • one year ago
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    can you post a problem ?

  4. some.random.cool.kid
    • one year ago
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    ok

  5. some.random.cool.kid
    • one year ago
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    y = -16x^2 + 96x for instance this one : above

  6. phi
    • one year ago
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    and what is the question?

  7. some.random.cool.kid
    • one year ago
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    kimberly tracks free fall of object she models a graph I don't have the graph but...

  8. some.random.cool.kid
    • one year ago
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    if this doesn't help I can try and find some other one ...

  9. some.random.cool.kid
    • one year ago
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    i have a perfect one 𝑔𝑔 βˆ’ 8𝑔𝑔 βˆ’ 20.

  10. phi
    • one year ago
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    some background: a quadratic is the name for a "second order polynomial", which means the highest exponent is 2 the simplest quadratic is y= x^2 the most general form is \[ a x^2 + b x +c \] where a, b, and c are numbers. If you graph a quadratic, you get a parabola A parabola looks like a U or a \( \cap\)

  11. some.random.cool.kid
    • one year ago
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    yeah negative and positive ...

  12. some.random.cool.kid
    • one year ago
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    those were whats on the graph ...

  13. some.random.cool.kid
    • one year ago
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    what about factoring quad's though?

  14. some.random.cool.kid
    • one year ago
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    lets say I want to use this .. Factor 𝑦𝑦 + 11𝑦𝑦 + 24.

  15. some.random.cool.kid
    • one year ago
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    thats one y each not two

  16. phi
    • one year ago
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    to factor quadratics, and see how to do it, it is better to start backwards say you have (x+a)(x+b) can you multiply that out ?

  17. some.random.cool.kid
    • one year ago
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    /uh well I can multiply x and x by distributing right?

  18. phi
    • one year ago
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    use the distributive property to see it more clearly, temporarily rename (x+b) as B (x+a)*(x+b) becomes (x+a)*B now distribute the B what do you get ?

  19. some.random.cool.kid
    • one year ago
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    uh well i see that you can do x^a'b?

  20. some.random.cool.kid
    • one year ago
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    imlost

  21. phi
    • one year ago
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    B*(x+a) now multiply B times each term inside the parens (this is called "distributing")

  22. some.random.cool.kid
    • one year ago
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    yes but I'm not sure whether its x^b + a^b or ...

  23. some.random.cool.kid
    • one year ago
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    theres no variable for B so idk what I'm multiplying.

  24. phi
    • one year ago
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    the ^ means exponents. it is neither try it with numbers 2*( 3+5) multiply 2 times each number inside the parens

  25. some.random.cool.kid
    • one year ago
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    see now that theres variables it looks more reasonable that what i was understanding ... this is 6 + 10 = 16 after you've distributed.

  26. some.random.cool.kid
    • one year ago
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    /wasnt*

  27. phi
    • one year ago
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    yes, and of course it is the same as 2* (3+5) = 2* 8= 16 so 16 both ways. if you have letters, multiplying is easy... for example x times y is written xy just like 2 times 3 is 2*3 (getting 6 is the next step, "simplifying" ) or 2 times x is written 2x

  28. some.random.cool.kid
    • one year ago
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    well I write that earlier I thought you disagreed with it...

  29. some.random.cool.kid
    • one year ago
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    wrote*

  30. phi
    • one year ago
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    try B*(x+a)

  31. some.random.cool.kid
    • one year ago
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    so this is bx + ab?

  32. some.random.cool.kid
    • one year ago
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    or xb*

  33. phi
    • one year ago
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    yes. But I used B, because B is short for (x+a) and bx= xb (order does not matter when multiplyiing: 2*3= 3*2)

  34. phi
    • one year ago
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    so B*(x+a)= Bx + Ba and if we replace B with (x+b) (x+b)(x+a)= (x+b)x +(x+b)a or (x+b)(x+a)= x(x+b) +a(x+b)

  35. phi
    • one year ago
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    *B is short for (x+b) (not x+a )

  36. some.random.cool.kid
    • one year ago
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    /so much wisdom its making my brain fall out xD. thanks tough ... so what if you have me those letters mixed with integers also what do i do then?

  37. some.random.cool.kid
    • one year ago
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    gave*

  38. some.random.cool.kid
    • one year ago
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    though*

  39. phi
    • one year ago
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    Unfortunately, you have to know this stuff before tackling quadratics. you may not get it in one session, but you will get closer. meanwhile (x+b)(x+a)= x(x+b) +a(x+b) now distribute x in the first x(x+b) and distribute the a in the second can you do that ?

  40. some.random.cool.kid
    • one year ago
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    I can certainly try.

  41. some.random.cool.kid
    • one year ago
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    uh so i think its x^2 + b and ax + b?

  42. some.random.cool.kid
    • one year ago
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    or am i missing something?

  43. phi
    • one year ago
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    if the problem was 2(2+3) you did 2*2 + 3

  44. some.random.cool.kid
    • one year ago
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    yeah? thats 4 + 3? but Im not sure what I missed in the first one?

  45. phi
    • one year ago
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    2(2+3) is 10 2*5=10 or distributing: 2*2 + 2*3 = 4+6 = 10

  46. some.random.cool.kid
    • one year ago
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    oh distribute both...

  47. phi
    • one year ago
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    if we had x(x+b) we multiply x times *each* term inside the parens

  48. some.random.cool.kid
    • one year ago
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    so X^2 and xb?

  49. phi
    • one year ago
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    yes

  50. some.random.cool.kid
    • one year ago
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    ok

  51. phi
    • one year ago
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    (x+b)(x+a)= x(x+b) +a(x+b) (x+b)(x+a)= x^2 +bx + ax + ab

  52. some.random.cool.kid
    • one year ago
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    so what about both numbers and letters can you perhaps explain those?

  53. phi
    • one year ago
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    if we "factor out" an x from the 2 middle terms (the opposite of distribute) we get (x+b)(x+a)= x^2 +(b+a)x + ab

  54. phi
    • one year ago
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    now we can do factoring of a quadratic say we have x^2 +7x +12 and we are trying to find a and b so (x+a)(x+b)= x^2 +7x +12

  55. phi
    • one year ago
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    using the letters x^2 +(b+a)x + ab notice that (a+b) matches the 7 and a*b matches the 12 in x^2 +7x +12

  56. phi
    • one year ago
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    the first thing I do is list all pairs of numbers that when multiplied give us 12 1,12 2,6 3,4

  57. phi
    • one year ago
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    then I add each pair 1+12= 13 2+6= 8 3+4= 7 if I find a pair that add up to the middle number (7 in this case) then we hit the jackpot the factors are (x+3)(x+4) and if we multiply that out we will get x^2 +7x+12

  58. some.random.cool.kid
    • one year ago
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    alright... that was a bit of a long shot does it normally take that much steps?

  59. phi
    • one year ago
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    it is like anything you learn. You probably forgot how hard it was to learn to read. But in school, if you work at this every day, (and not try to learn it all in 1 day as you seem to be trying to do now) , it can be done.

  60. phi
    • one year ago
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    try factoring x^2+4x+4

  61. some.random.cool.kid
    • one year ago
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    well uh ... I assume its 4^2x? for the first part

  62. some.random.cool.kid
    • one year ago
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    then add the 4?

  63. phi
    • one year ago
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    factoring x^2+4x+4 means find a and b so that (x+a)(x+b)= x^2+4x+4

  64. phi
    • one year ago
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    the first step is look at the last number: +4 find all pairs of numbers that multiply to give you 4 1,4 2,2 add each pair 1+4=5 2+2= 4 choose the pair that add up to the middle number (the 4 from 4x) 2,2 are the numbers for a and b thus (x+2)(x+2)= x^2 + 4x + 4

  65. phi
    • one year ago
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    try factoring x^2 +4x + 3

  66. some.random.cool.kid
    • one year ago
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    back

  67. some.random.cool.kid
    • one year ago
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    ok

  68. some.random.cool.kid
    • one year ago
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    wait idk how to factor this my teacher has asked me this before and i failed

  69. phi
    • one year ago
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    x^2 +4x + 3 what is the first step? (look what I posted up above)

  70. some.random.cool.kid
    • one year ago
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    except she did it with a minus

  71. some.random.cool.kid
    • one year ago
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    i really don't know please walk me through ...

  72. some.random.cool.kid
    • one year ago
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    all i know is that I factor it ..

  73. phi
    • one year ago
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    the first step is look at the last number:

  74. some.random.cool.kid
    • one year ago
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    3?

  75. phi
    • one year ago
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    find all pairs of numbers that multiply to give you 3

  76. some.random.cool.kid
    • one year ago
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    3 and 1

  77. phi
    • one year ago
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    now add them. if they add to the "middle number" 4, you found the correct pair

  78. some.random.cool.kid
    • one year ago
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    k

  79. phi
    • one year ago
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    3+1= 4 (which matches the 4 in x^2 + 4x +3) so 3 , 1 are the numbers we want the factors are (x+3)(x+1)

  80. some.random.cool.kid
    • one year ago
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    uh ok

  81. phi
    • one year ago
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    try y^2 + 11y + 24 which you posted at the start

  82. some.random.cool.kid
    • one year ago
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    do I start with 24

  83. phi
    • one year ago
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    yes

  84. some.random.cool.kid
    • one year ago
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    in that case 12 and 2 can work..

  85. some.random.cool.kid
    • one year ago
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    or 24 ad 1

  86. phi
    • one year ago
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    I always start with 1, 2, 3, etc up to the square root of 21 (less than 5)

  87. phi
    • one year ago
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    1,24 2,12 3,8 4,6

  88. some.random.cool.kid
    • one year ago
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    ok

  89. some.random.cool.kid
    • one year ago
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    so i pick 12 and 2

  90. phi
    • one year ago
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    do any of the pairs add up to the middle number in y^2 +11y + 24

  91. some.random.cool.kid
    • one year ago
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    no

  92. some.random.cool.kid
    • one year ago
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    so is it done already

  93. some.random.cool.kid
    • one year ago
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    /thats been simplified?

  94. phi
    • one year ago
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    not yet. recheck each pair

  95. some.random.cool.kid
    • one year ago
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    ok im lost Im not sure I see that that pairs except 12 and 2 equally the 24

  96. phi
    • one year ago
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    add each pair 1,24 2,12 3,8 4,6

  97. some.random.cool.kid
    • one year ago
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    ill be back in two hours

  98. phi
    • one year ago
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    1+24= 25 2+12= 14 3+8= 11 4+6= 10

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