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dddan

  • 3 years ago

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

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  1. tejasree
    • 3 years ago
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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

  2. tejasree
    • 3 years ago
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    \[\sqrt{x} = -\sqrt{x-36}\]

  3. dddan
    • 3 years ago
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    what about the 2?

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

  5. tejasree
    • 3 years ago
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    what 2

  6. tejasree
    • 3 years ago
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    ?

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

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

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

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

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

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

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

  14. tejasree
    • 3 years ago
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    no no!!

  15. tejasree
    • 3 years ago
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    wait a sec!

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

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

  18. tejasree
    • 3 years ago
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    you following?

  19. tejasree
    • 3 years ago
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    64 = X-36 64 + 36 = X - 36 + 36 100 = X

  20. tejasree
    • 3 years ago
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    thank you! :)

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