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TeemoTheTerific

  • one year ago

Simplify (2^1/2 - 2^-1/2)^2 ill write it neater. Please help

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  1. TeemoTheTerific
    • one year ago
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    |dw:1359872594130:dw|

  2. TeemoTheTerific
    • one year ago
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    i know formula a^2 - 2ab + b^2

  3. TeemoTheTerific
    • one year ago
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    i dont get it :P

  4. TeemoTheTerific
    • one year ago
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    |dw:1359872887253:dw|btw original question is

  5. TeemoTheTerific
    • one year ago
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    thats original quesiton

  6. hartnn
    • one year ago
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    |dw:1359872899922:dw|

  7. TeemoTheTerific
    • one year ago
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    yes your right now hartnn :)

  8. hartnn
    • one year ago
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    |dw:1359872998532:dw|

  9. hartnn
    • one year ago
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    can you simplify that now ?

  10. TeemoTheTerific
    • one year ago
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    answer 5?

  11. TeemoTheTerific
    • one year ago
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    is there a way to do this question using indices laws?

  12. TeemoTheTerific
    • one year ago
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    cause my teacher prefers that :P

  13. hartnn
    • one year ago
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    yes, 5 is correct.

  14. hartnn
    • one year ago
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    i have used law od indices \(2^{1/2}.2^{1/2}= 2^{1/2+1/2}=2^1=2\)

  15. hartnn
    • one year ago
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    |dw:1359873234483:dw|

  16. hartnn
    • one year ago
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    |dw:1359873310197:dw|

  17. hartnn
    • one year ago
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    got that^ ?

  18. TeemoTheTerific
    • one year ago
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    yes thank you so much

  19. hartnn
    • one year ago
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    welcome ^_^

  20. whpalmer4
    • one year ago
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    \[(2^{1/2} - 2^{-1/2})^2 = (2^{1/2}-2^{-1/2})(2^{1/2}-2^{-1/2})=\]\[2^{1/2}*2^{1/2}-2^{-1/2}2^{1/2} - 2^{-1/2}2^{1/2} + 2^{-1/2}2^{-1/2} = \]\[2^1-2^0-2^0+2^{-1} = 2 - 1 - 1 + \frac{1}{2} = \frac{1}{2}\]

  21. whpalmer4
    • one year ago
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    Maybe easier to write \(a = 2^{1/2}\) and \(b = a^{-1/2}\) then \[(a-b)^2 = a^2 - 2ab + b^2 = (2^{1/2})^{1/2} - 2*2^{1/2}*2^{-1/2} + (2^{-1/2})^2 =\]\[2^1 - 2*2^0 + 2^{-1} = 2^{-1} = \frac{1}{2}\] and use some different properties :-)

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