- #1
velox_xox
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Hi everyone. I'm back and happy to be back. I missed math. Anyways, I am taking a college Algebra class as it has been a while, and I definitely want to make sure I have a good foundation for higher maths. But, I have a conundrum. The teacher today gave this example that I thought I had solved correctly, but I did it differently.
I'm going to post how I solved it and how the teacher solved it.
My Way:
[tex]-9^{-3/2} [/tex]
--
[tex]-9^{-3/2} [/tex]
[tex]=-(\frac{1}{9^3})^{1/2}[/tex]
[tex]=-(\frac{1}{729})^{1/2}[/tex]
[tex]=-(\frac{1}{27})[/tex] ^^
[tex]=±\frac{1}{27}[/tex]
^^ Did I bend the rules here? It's a negative before that... but then a negative of a negative ('cuz of the square root) is a positive, so plus-minus.
TEACHER'S WAY:
[tex]-9^{-3/2} [/tex]
--
[tex]-9^{-3/2} [/tex]
[tex]=\frac{1}{-9^{3/2}}[/tex]
[tex]=\frac{1}{-(9^{1/2})^3}[/tex]
[tex]=\frac{-1}{(3)^3}[/tex]
[tex]=\frac{-1}{27}[/tex]
[tex]=±\frac{1}{27}[/tex]All right, so that's it. The problem solved... my way and the teacher's way. I just want to make sure the answer is correct (never hurts to verify), and that I'm not bending the rules of math again. Thanks in advance for your help.
My Way:
Homework Statement
[tex]-9^{-3/2} [/tex]
Homework Equations
--
The Attempt at a Solution
[tex]-9^{-3/2} [/tex]
[tex]=-(\frac{1}{9^3})^{1/2}[/tex]
[tex]=-(\frac{1}{729})^{1/2}[/tex]
[tex]=-(\frac{1}{27})[/tex] ^^
[tex]=±\frac{1}{27}[/tex]
^^ Did I bend the rules here? It's a negative before that... but then a negative of a negative ('cuz of the square root) is a positive, so plus-minus.
TEACHER'S WAY:
Homework Statement
[tex]-9^{-3/2} [/tex]
Homework Equations
--
The Attempt at a Solution
[tex]-9^{-3/2} [/tex]
[tex]=\frac{1}{-9^{3/2}}[/tex]
[tex]=\frac{1}{-(9^{1/2})^3}[/tex]
[tex]=\frac{-1}{(3)^3}[/tex]
[tex]=\frac{-1}{27}[/tex]
[tex]=±\frac{1}{27}[/tex]All right, so that's it. The problem solved... my way and the teacher's way. I just want to make sure the answer is correct (never hurts to verify), and that I'm not bending the rules of math again. Thanks in advance for your help.