anonymous
  • anonymous
sqrt x+2+sqrt x=4 lol i keep asking for help on this
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dumbcow
  • dumbcow
is that sqrt(x) + 2 or sqrt(x+2)
dumbcow
  • dumbcow
well if its the former, x = 1 if the latter, x =49/16
anonymous
  • anonymous
x=49/16 can you explain how you got this please?

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anonymous
  • anonymous
\[\sqrt{x+2} + \sqrt{x} = 4\] \[\iff \sqrt{x+2} = 4-\sqrt{x} \] 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.
anonymous
  • anonymous
16-8sqrt x+x can you explain that more
anonymous
  • anonymous
(a-b)^2 = a^2 - 2ab + b^2 let a = 4 , b = sqrt(x)
anonymous
  • anonymous
You're welcome. ¬_¬
anonymous
  • anonymous
is this a formula (a-b)^2 = a^2 - 2ab + b^2
anonymous
  • anonymous
Not a formula so much - well, it is an identity, but it's more just common sense of expanding brackets.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
i meant \[\sqrt{x+2}+\sqrt{x} -4\]
anonymous
  • anonymous
Im still confused .......
anonymous
  • anonymous
Im still confused .......

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