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anonymous

  • 5 years ago

sqrt x+2+sqrt x=4 lol i keep asking for help on this

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  1. dumbcow
    • 5 years ago
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    is that sqrt(x) + 2 or sqrt(x+2)

  2. dumbcow
    • 5 years ago
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    well if its the former, x = 1 if the latter, x =49/16

  3. anonymous
    • 5 years ago
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    x=49/16 can you explain how you got this please?

  4. anonymous
    • 5 years ago
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    \[\sqrt{x+2} + \sqrt{x} = 4\] \[\iff \sqrt{x+2} = 4-\sqrt{x} \] Squaring both sides yields: \[x+2 = 16 - 8\sqrt{x} + x \implies \sqrt x = \frac{7}{4} \] And squaring gives the result. Note, squaring can produce spurious roots, so we must check this back in the original equation, and see that it holds.

  5. anonymous
    • 5 years ago
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    16-8sqrt x+x can you explain that more

  6. anonymous
    • 5 years ago
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    (a-b)^2 = a^2 - 2ab + b^2 let a = 4 , b = sqrt(x)

  7. anonymous
    • 5 years ago
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    You're welcome. ¬_¬

  8. anonymous
    • 5 years ago
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    is this a formula (a-b)^2 = a^2 - 2ab + b^2

  9. anonymous
    • 5 years ago
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    Not a formula so much - well, it is an identity, but it's more just common sense of expanding brackets.

  10. anonymous
    • 5 years ago
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    um one more thing\[\sqrt{x+2} +\sqrt{x-4}\rightarrow \sqrt{x+2}-4-\sqrt{x}\] how did the sqrt of x become negative when you move it

  11. anonymous
    • 5 years ago
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    Errr, what? That didn't happen, at all. Assuming your question was right, the 4 was never under the root... and was on the other side.

  12. anonymous
    • 5 years ago
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    i meant \[\sqrt{x+2}+\sqrt{x} -4\]

  13. anonymous
    • 5 years ago
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    Im still confused .......

  14. anonymous
    • 5 years ago
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    Im still confused .......

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