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mathmath333

  • one year ago

solve for \(x\)

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} (x-1)\sqrt{x^2-x-2}\geq 0\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    i got \(\large \color{black}{\begin{align} x\geq 1 \cup \left(x\geq 2 \cup x\leq -1\right)\hspace{.33em}\\~\\ \end{align}}\)

  3. freckles
    • one year ago
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    \[x^2-x-2=(x-2)(x+1) \\ (x-1) \sqrt{(x-1)(x-2)} \ge 0\] we should consider the domain of the square root function there on that square root function's domain the square root function will be positive so then you just need to consider when (x-1) is positive and find the intersection of these sets

  4. mathmath333
    • one year ago
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    how to solve this \(\large \color{black}{\begin{align} x\geq 1 \cup \left(x\geq 2 \cup x\leq -1\right)\hspace{.33em}\\~\\ \end{align}}\)

  5. freckles
    • one year ago
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    is that one symbol is suppose to be an intersect sign ?

  6. mathmath333
    • one year ago
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    i havent studied intersection of sets

  7. mathmath333
    • one year ago
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    i just tried and posted

  8. freckles
    • one year ago
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    we should also go back to include when the function is 0 but I think you mean \[(x \ge 1) \cap (x \ge 2 \cup x \le -1)\] draw a number line in keep in mind that x can be 1,-1,2 since that makes the function 0 but anyways it might helped to also graph that... |dw:1436976303765:dw|

  9. freckles
    • one year ago
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    notice the elements in common between x>=1 and also (x ge 2 union x<=-1)

  10. freckles
    • one year ago
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    |dw:1436976374511:dw|

  11. freckles
    • one year ago
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    |dw:1436976397456:dw| but we also want to include these little endpoints because at these endpoints we have the thingy is 0

  12. freckles
    • one year ago
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    the reason x<-1 doesn't work is because x-1 is negative there the reason 1<x<2 doesn't work is because that isn't in the domain of the square root function

  13. freckles
    • one year ago
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    anyways intersect just means you are trying to find the common elements

  14. mathmath333
    • one year ago
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    u mean the answer is \(x\geq 2\)

  15. mathmath333
    • one year ago
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    but wolfram also gives \(x=-1\) which is not in ur intersecton

  16. freckles
    • one year ago
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    you need to include the zeros @mathmath333

  17. freckles
    • one year ago
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    x>=2 or x=1 or x=-1

  18. freckles
    • one year ago
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    this is what I meant by the statement "but we also want to include these little endpoints because at these endpoints we have the thingy is 0 "|dw:1436977880095:dw| and the drawing that went along with it

  19. freckles
    • one year ago
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    if the thing was just \[(x-1) \sqrt{(x+1)(x-2)} >0 \] the answer would have been to just look at the interesections

  20. freckles
    • one year ago
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    the intersection of when x>1 and the domain of the square root function

  21. freckles
    • one year ago
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    but we also had a equal sign in there

  22. freckles
    • one year ago
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    \[(x-1) \sqrt{(x+1)(x-2)} >0 \text{ gives the solution } x>2 \\ (x-1)\sqrt{(x+1)(x-2)}=0 \text{ when } x=\pm 1 \text{ or } x=2 \\ \]

  23. freckles
    • one year ago
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    therefore the solution to \[(x-1) \sqrt{(x+1)(x-2)} \ge 0 \text{ is } x>2 \text{ or } x= \pm 1 \text{ or } x=2 \\ \text{ clean it up a little } x \ge 2 \text{ or } x=\pm 1 \]

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