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Solve (z + 6)2 = 5 A. {-6+√5,6-√5} B. {-6±√5} C. {√5±-6} D. {6±√5}
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is the 2 supposed to be a squared?
Yes
Okay. Thanks
Rule you can multiple exponents on both sides of an equation so to solve for a: a^2 = b \[a^{2*(1/2)} = b^{1*(1/2)}\] \[a^{1} = b^{1/2}\] \[a = b^{1/2}\]
It might also help to remember BEDMAS order of operations.
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\[\sqrt{(Z+6)^{2}}=\sqrt{5}\] Then solve for Z.
Would it be D?
I think you made a mistake in your algebra
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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