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anonymous

  • one year ago

WILL MEDAL AND FAN! PLEASE, THIS IS URGENT! Solve √(k-7) = √(k) -1 for k.

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  1. Nnesha
    • one year ago
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    take square both sides to cancel out square root

  2. anonymous
    • one year ago
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    i tried that and i got this \[k-7= k-1\]

  3. anonymous
    • one year ago
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    or is it k+1? i think its K-1

  4. anonymous
    • one year ago
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    can anyone help me please?

  5. freckles
    • one year ago
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    \[\sqrt{k-7}=\sqrt{k}-1 ?\]

  6. anonymous
    • one year ago
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    I'm trying to have it done by the time my best friend gets on.

  7. anonymous
    • one year ago
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    yep!

  8. anonymous
    • one year ago
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    the choices are: 20 16 4 8

  9. Nnesha
    • one year ago
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    oh nvm I didn't notice the parentheses

  10. freckles
    • one year ago
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    square both sides

  11. freckles
    • one year ago
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    you will need this on the right hand side \[(a-b)^2=a^2+b^2-2ab\]

  12. anonymous
    • one year ago
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    \[\sqrt{k-7} = \sqrt{k} -1\]

  13. freckles
    • one year ago
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    have you expanded the right hand side yet?

  14. anonymous
    • one year ago
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    i dont know what that means

  15. anonymous
    • one year ago
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    i got to this point\[k-7 = k-1\]

  16. freckles
    • one year ago
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    \[\sqrt{k-7}=\sqrt{k}-1 \\ \text{ Square both sides } (\sqrt{k-7})^2=(\sqrt{k}-1)^2 \\ k-7=(\sqrt{k}-1)^2\] recall \[(a-b)^2=a^2+b^2-2ab\] you can use this to expand the square part on the right hand side

  17. anonymous
    • one year ago
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    ooooohhhhhhh soooooo\[k+1-2k\]

  18. anonymous
    • one year ago
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    right? so \[k+1-2k=k-7\]

  19. freckles
    • one year ago
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    close

  20. freckles
    • one year ago
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    you had a sqrt( ) on k

  21. freckles
    • one year ago
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    \[(a-b)^2=a^2+b^2-2ab \\ (\sqrt{k}-1)^2=(\sqrt{k})^2 +(1)^2-2(\sqrt{k})(1) \\ (\sqrt{k}-1)^2=k+1-2 \sqrt{k}\]

  22. anonymous
    • one year ago
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    oh right. \[k-7 = k+1 - 2\sqrt{k}\]

  23. freckles
    • one year ago
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    so you really have \[k-7=k+1-2 \sqrt{k} \text{ or subtracting } k \text{ on both sides } -7=1-2 \sqrt{k}\]

  24. freckles
    • one year ago
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    isolate the sqrt(k) part

  25. anonymous
    • one year ago
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    k\[-2\sqrt{k}=-8\]

  26. anonymous
    • one year ago
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    right?

  27. freckles
    • one year ago
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    yeah

  28. freckles
    • one year ago
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    divide both sides by -2

  29. anonymous
    • one year ago
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    \[\sqrt{k}=4\]

  30. anonymous
    • one year ago
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    k=16!

  31. freckles
    • one year ago
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    yep and you can check that by pluggin it in

  32. anonymous
    • one year ago
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    THANK YOUUU soooooooooo muchhh

  33. anonymous
    • one year ago
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    Im a fan a medals

  34. freckles
    • one year ago
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    \[\sqrt{k-7}=\sqrt{k}-1 \\ k=16 \\ \sqrt{16-7}=\sqrt{9}=3 \\ \text{ and the other side } \sqrt{16}-1=4-1=3 \] 3=3 k=16 is definitely a solution since it gives a true equation when pluggin it back in

  35. anonymous
    • one year ago
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    thank youuu

  36. freckles
    • one year ago
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    np

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