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mathmath333

  • one year ago

solve for \(x\)

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x^2-|5x+8|>0\hspace{.33em}\\~\\ \end{align}}\)

  2. misty1212
    • one year ago
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    HI!!

  3. misty1212
    • one year ago
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    work in cases solve if \(x<-\frac{8}{5}\) and if \(x>-\frac{8}{5}\)

  4. mathmath333
    • one year ago
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    i worked m onfised

  5. misty1212
    • one year ago
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    ??

  6. mathmath333
    • one year ago
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    *confused

  7. misty1212
    • one year ago
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    if \(x>-\frac{8}{5}\) then \[|5x+8|=5x+8\] right ?

  8. misty1212
    • one year ago
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    if it is not clear how i got that say so and i will explain

  9. mathmath333
    • one year ago
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    i dk how u got it

  10. misty1212
    • one year ago
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    the absolute value is a piece wise function

  11. misty1212
    • one year ago
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    let me write it as one hold on

  12. mathmath333
    • one year ago
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    ok i got it

  13. misty1212
    • one year ago
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    ok so if \(5x+8>0\) then \(|5x+8|=5x+8\) and \(5x+8>0\) if \(x>-\frac{8}{5}\) so case 1:\(x>-\frac{8}{5}\) you solve \[x^2-(5x+8)>0\] or \[x^2-5x-8>0\]

  14. misty1212
    • one year ago
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    once you solve that, the actual answer is the intersection of your solution and \(x>-f\rac{8}{5}\) then repeat with \(x<-\frac{8}{5}\) solving \[x^2+5x+8>0\]

  15. misty1212
    • one year ago
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    ach

  16. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x>\dfrac{5+\sqrt{57}}{2},\ x<\dfrac{5-\sqrt{57}}{2} \hspace{.33em}\\~\\ \end{align}}\)

  17. misty1212
    • one year ago
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    intersection of your solution and \(x>-\frac{8}{5}\)

  18. misty1212
    • one year ago
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    looks good to me

  19. mathmath333
    • one year ago
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    lol m still confused

  20. mathmath333
    • one year ago
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    this is solution of \(x^2-5x-8>0\) \(\large \color{black}{\begin{align} x>\dfrac{5+\sqrt{57}}{2},\ x<\dfrac{5-\sqrt{57}}{2} \hspace{.33em}\\~\\ \end{align}}\)

  21. mathmath333
    • one year ago
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    @phi

  22. mathmath333
    • one year ago
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    for \(x^2+5x+8>0\) \(\large \color{black}{\begin{align} -\infty<x<+\infty \hspace{.33em}\\~\\ \end{align}}\)

  23. phi
    • one year ago
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    You began by assuming x >= -8/5 = -1.6 you found either x <= -1.275 (roughly) so -1.6 <= x < -1.275 or x > 6.275 now you have to check the other condition, where x <= -8/5

  24. mathmath333
    • one year ago
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    for \(x<-8.5\) \(-\infty<x<-8/5\)

  25. phi
    • one year ago
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    in other word, when (5x-8) is negative, we can drop the absolute value sign, and negate the value inside. (for example, if we know we have a negative number, say -2 inside the abs value: | -2 | we know this is the same as - (-2) = +2 ) so x^2 - -(5x+8)>0 or x^2 +5x+8> 0 complete the square x^2 + 5x +25/4 +8 > 25/4 (x+5/2)^2 > 25/4 - 8 or \[ (x+5/2)^2 > -7/4 \] which will be true for all x (the left side will be 0 or bigger for any x) but we assumed x <= -8/5 so the solution is x <= -8/5

  26. phi
    • one year ago
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    now combine -1.6 <= x < -1.275 or x > 6.275 and x<= -8/5 = -1.6 we get \[ -\infty < x < \frac{5-\sqrt{57}}{2} \text{ or } x >\frac{5+\sqrt{57}}{2} \]

  27. phi
    • one year ago
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    Does this stuff make sense? If you are missing some idea, you should ask.

  28. mathmath333
    • one year ago
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    no question , thnx

  29. anonymous
    • one year ago
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    http://openstudy.com/study#/updates/5593fc3ae4b0989cc8780f4d

  30. mathmath333
    • one year ago
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    how do u approximate \(\sqrt{57}\) without calculater

  31. phi
    • one year ago
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    I don't. but I know its bigger than 7 (7*7= 49) and less than 8 (8*8=64)

  32. phi
    • one year ago
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    I only used decimals because it was easier to type (and easier to compare to -8/5)

  33. mathmath333
    • one year ago
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    ok

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