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Destinyyyy

  • one year ago

Solve the equation by making an appropriate substitution.

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  1. Destinyyyy
    • one year ago
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    x^-2 +3x^-1 -10=0

  2. freckles
    • one year ago
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    let u=x^(-1) then u^2=x^(-2)

  3. Destinyyyy
    • one year ago
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    okay

  4. Destinyyyy
    • one year ago
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    Do i factor now?

  5. freckles
    • one year ago
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    yep yep

  6. freckles
    • one year ago
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    and you should be factoring u^2+3u-10 right now

  7. Destinyyyy
    • one year ago
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    (u-2)(u+5)

  8. freckles
    • one year ago
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    neat so you can solve the equation (u-2)(u+5)=0 for u

  9. Destinyyyy
    • one year ago
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    u=2, -5

  10. freckles
    • one year ago
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    now remember we replace x^(-1) with u so we have to undo that replacement by replacing u with x^(-1)

  11. freckles
    • one year ago
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    \[x^{-1}=2 \text{ or } -5 \\ x^{-1}=2 \text{ or } x^{-1}=-5 \]

  12. freckles
    • one year ago
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    recall x^(-1) is the same thing as saying 1/x

  13. freckles
    • one year ago
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    \[\frac{1}{x}=2 \text{ or } \frac{1}{x}=-5 \]

  14. freckles
    • one year ago
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    can you solve both of those equations for x

  15. Destinyyyy
    • one year ago
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    ?? x^-2=2 you have two -1

  16. freckles
    • one year ago
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    \[\frac{1}{x}=a \\ \text{ multiply both sides by } x \\ 1=ax \\ \text{ divide both sides by } a \\ \frac{1}{a}=x \\ \text{ so assuming } x \neq 0 \text{ and } a \neq 0 \\ \text{ then } \frac{1}{x}=a \implies x=\frac{1}{a}\]

  17. Destinyyyy
    • one year ago
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    Um okay I think I understand that but dont see how that solves this.. :/

  18. Destinyyyy
    • one year ago
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    Dont I just square root both sides?

  19. freckles
    • one year ago
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    you have 1/x=2 so x=1/2

  20. freckles
    • one year ago
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    you have 1/x=-5 so x=?

  21. Destinyyyy
    • one year ago
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    -1/5

  22. freckles
    • one year ago
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    right

  23. Destinyyyy
    • one year ago
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    So thats the answer?

  24. freckles
    • one year ago
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    x=1/2 or x=-1/5 yes

  25. Destinyyyy
    • one year ago
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    Do I check it? Or no?

  26. freckles
    • one year ago
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    another way to solve this is multiply both sides of your equation by x^2 \[x^{-2}+3x^{-1}-10=0 \\ x^2(x^{-2}+3x^{-1}-10)=x^{2}(0) \\ x^{2+(-2)}+3x^{2+(-1)}-10x^{2}=0 \\ x^{0}+3x^{1}-10x^2=0 \\ -10x^2+3x+1=0 \text{ since } x^{0}=1 \text{ and I just reorder terms } \\ -10x^2+5x-2x+1=0 \\ -5x(2x-1)-1(2x-1)=0 \\ (2x-1)(-5x-1)=0 \\ 2x-1=0 \text{ when } x=\frac{1}{2} \\ -5x-1=0 \text{ when } x=\frac{-1}{5}\] there is no harm in checking but the answers are fine we didn't raise both sides to an even power and are answers fit in with domain of the original equation (for this question we definitely did not want x=0 as a solution which we didn't have to worry about here)

  27. freckles
    • one year ago
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    oh I see what you did about you replace u with x^(-2) even though u equals x^(-1)

  28. Destinyyyy
    • one year ago
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    Um okay

  29. freckles
    • one year ago
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    above not about :p

  30. freckles
    • one year ago
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    anyways if you have more questions post a new question I'm gonna take a short break

  31. freckles
    • one year ago
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    unless you have questions on this one then I can stay and answer them real quick

  32. freckles
    • one year ago
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    ok I will be back later! peace for now! :)

  33. Destinyyyy
    • one year ago
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    Sorry didnt see you send back.

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