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Melissa_Something

  • one year ago

Solve the inequality :(

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  1. Melissa_Something
    • one year ago
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    \[x^4-32x^2-144>0\]

  2. Melissa_Something
    • one year ago
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    I don't even know where to start..

  3. Loser66
    • one year ago
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    let t = x^2 to take it down

  4. Melissa_Something
    • one year ago
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    Is that substituting?

  5. Loser66
    • one year ago
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    yup

  6. Melissa_Something
    • one year ago
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    What do we do with it? :s

  7. Loser66
    • one year ago
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    do it first and I guide you more.

  8. Melissa_Something
    • one year ago
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    \[t^2-3xt-144>0\]

  9. Loser66
    • one year ago
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    t^2 -32t -144 >0

  10. Melissa_Something
    • one year ago
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    Oh yeah typo

  11. Loser66
    • one year ago
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    ok, now solve for quadratic to get t

  12. Melissa_Something
    • one year ago
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    um for x do I change it to t?

  13. Loser66
    • one year ago
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    we will be back to x later. Now, just solve for t

  14. Melissa_Something
    • one year ago
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    for a i just put t? it looks weird lol

  15. Loser66
    • one year ago
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    |dw:1444598561849:dw|

  16. Melissa_Something
    • one year ago
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    -32 = -1,024 .. am I doing this right?

  17. Loser66
    • one year ago
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    complete the square for the left hand sides.

  18. Melissa_Something
    • one year ago
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    Oh so I have to solve this!! Okay so for the den of 2a I do 2^2? Or 2x^2

  19. Loser66
    • one year ago
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    |dw:1444598640614:dw|

  20. Loser66
    • one year ago
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    solve for t, then plug back to t = x^2 to solve for x. I let you finish the stuff.

  21. Melissa_Something
    • one year ago
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    Okay, but where did you get 16

  22. Loser66
    • one year ago
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    From " complete the square" procedure. Rule: 32/2 =16, hence 16 is the number I need to + its square and then - its square.

  23. Loser66
    • one year ago
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    |dw:1444599011779:dw|

  24. Melissa_Something
    • one year ago
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    So I didnt need to plug this into a quadratic?

  25. Loser66
    • one year ago
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    You did. But the way to solve this problem is not just this, you have many many steps to get the answer. I know it is not easy but no choice for you. Take steps. Understand each step first. So far so good now?

  26. Melissa_Something
    • one year ago
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    I just dont know what I was supposed to do first

  27. Loser66
    • one year ago
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    Ok, I solve it for you as sample. You follow each step and see the reason why we have to do it.

  28. Melissa_Something
    • one year ago
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    Okay, Im just not good at math :(

  29. Loser66
    • one year ago
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    \( t^2 -32 t- 144 > 0\\t^2 -32t +16^2 -16^2 -144 >0\\(t-16)^2 > 400 \\(t-16)^2 >20^2\)

  30. Loser66
    • one year ago
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    Hence \(t -16 > 20 ~~or ~~t-16 <-20 \) Case 1: if t -16 >20 , then t -16 -20 > 0 . That is t -36 >0 Recall \( t = x^2\) , hence \(x^2 -36 > 0\) You can't solve directly from here. You MUST derive like \(x^2 -36 > 0\\(x-6)(x + 6 ) >0 \) For this case, both (x-6) and (x+6) must be the same sign. That is if (x-6) >0, (x+6) >0 also. Because (+) *(+) >0 if (x-6) <0, then (x+6) <0 also, since (-)*(-) = positive >0

  31. Loser66
    • one year ago
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    If both are >0 , that is x-6 >0, x > 6 AND x +6 >0, x > -6 , combine them, you just have x > 6

  32. Melissa_Something
    • one year ago
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    Woah :o, but how did you get 400?

  33. Loser66
    • one year ago
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    -16^2 -144 = -400 , move to other sides to get 400|dw:1444599855908:dw|