anonymous one year ago Solve (z + 6)2 = 5 A. {-6+√5,6-√5} B. {-6±√5} C. {√5±-6} D. {6±√5}

1. anonymous

is the 2 supposed to be a squared?

2. anonymous

Yes

3. anonymous

Okay. Thanks

4. Australopithecus

Rule you can multiple exponents on both sides of an equation so to solve for a: a^2 = b $a^{2*(1/2)} = b^{1*(1/2)}$ $a^{1} = b^{1/2}$ $a = b^{1/2}$

5. Australopithecus

It might also help to remember BEDMAS order of operations.

6. anonymous

$\sqrt{(Z+6)^{2}}=\sqrt{5}$ Then solve for Z.

7. anonymous

Would it be D?

8. Australopithecus

I think you made a mistake in your algebra

9. anonymous

No, because if you subtract 6 from both sides the square root of 5 is not what is subtracted you subtract the 6.