anonymous
  • anonymous
Solve (z + 6)2 = 5 A. {-6+√5,6-√5} B. {-6±√5} C. {√5±-6} D. {6±√5}
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
is the 2 supposed to be a squared?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Okay. Thanks

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Australopithecus
  • Australopithecus
Rule you can multiple exponents on both sides of an equation so to solve for a: a^2 = b \[a^{2*(1/2)} = b^{1*(1/2)}\] \[a^{1} = b^{1/2}\] \[a = b^{1/2}\]
Australopithecus
  • Australopithecus
It might also help to remember BEDMAS order of operations.
anonymous
  • anonymous
\[\sqrt{(Z+6)^{2}}=\sqrt{5}\] Then solve for Z.
anonymous
  • anonymous
Would it be D?
Australopithecus
  • Australopithecus
I think you made a mistake in your algebra
anonymous
  • anonymous
No, because if you subtract 6 from both sides the square root of 5 is not what is subtracted you subtract the 6.

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