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dddan

how do sove for the solutions for ; the square root of x+ the square root of x-36 which then =0?

  • one year ago
  • one year ago

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  1. tejasree
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    Subtract the square root of x-36 from both sides! now: the square root of x = -the square root of x-36

    • one year ago
  2. tejasree
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    \[\sqrt{x} = -\sqrt{x-36}\]

    • one year ago
  3. dddan
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    what about the 2?

    • one year ago
  4. tejasree
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    Square both of them\[(\sqrt{x})^{2} = (-\sqrt{x-36})^{2}\]

    • one year ago
  5. tejasree
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    what 2

    • one year ago
  6. tejasree
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    ?

    • one year ago
  7. dddan
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    im so sry the whole thing is equal to 2 not 0 im so sry

    • one year ago
  8. tejasree
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    ok...then it is: \[(\sqrt{x})^{2} = (2-\sqrt{x-36})^{2}\]

    • one year ago
  9. tejasree
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    \[x = 4 - 2\sqrt{x-36} - 2\sqrt{x-36} + x - 36\]

    • one year ago
  10. tejasree
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    \[x = -32 + x + -4\sqrt{x-36}\]

    • one year ago
  11. tejasree
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    \[x - x + 32 = -4\sqrt{x-36}\]

    • one year ago
  12. tejasree
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    \[32 = -4\sqrt{x-36}\]

    • one year ago
  13. tejasree
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    \[(32)^{2} = (-4\sqrt{x-36})^{2}\]

    • one year ago
  14. tejasree
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    no no!!

    • one year ago
  15. tejasree
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    wait a sec!

    • one year ago
  16. tejasree
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    Instead of square rooting 32 and the other, first divide -4 from both sides!

    • one year ago
  17. tejasree
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    \[(-8)^{2} = (\sqrt{x-36})^{2}\]

    • one year ago
  18. tejasree
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    you following?

    • one year ago
  19. tejasree
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    64 = X-36 64 + 36 = X - 36 + 36 100 = X

    • one year ago
  20. tejasree
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    thank you! :)

    • one year ago
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