Using exponents what is the simplified form of the expression 12x^5/6x^2
1 answer:
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Answer:
Considerations:
Also,
A positive number divided by a positive number results in a positive quotient, and its opposite is always negative.
The fraction can be described as a positive number divided by a negative number, which always result in a negative quotient.
The fractions have same value, so Liam is correct.
Isolate the root expression:
![\sqrt[3]{x+1}+2=0\implies\sqrt[3]{x+1}=-2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B1%7D%2B2%3D0%5Cimplies%5Csqrt%5B3%5D%7Bx%2B1%7D%3D-2)
Take the third power of both sides:
![\sqrt[3]{x+1}=-2\implies(\sqrt[3]{x+1})^3=(-2)^3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B1%7D%3D-2%5Cimplies%28%5Csqrt%5B3%5D%7Bx%2B1%7D%29%5E3%3D%28-2%29%5E3)
Simplify:
![(\sqrt[3]{x+1})^3=(-2)^3\implies x+1=-8](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bx%2B1%7D%29%5E3%3D%28-2%29%5E3%5Cimplies%20x%2B1%3D-8)
Isolate and solve for
:
Since the cube root function is bijective, we know this won't be an extraneous solution, but it doesn't hurt to verify that this is correct. When, we have
![\sqrt[3]{-9+1}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-9%2B1%7D%3D%5Csqrt%5B3%5D%7B-8%7D%3D%5Csqrt%5B3%5D%7B%28-2%29%5E3%7D%3D-2)
as required.
Take the third power of both sides:
Simplify:
Isolate and solve for
Since the cube root function is bijective, we know this won't be an extraneous solution, but it doesn't hurt to verify that this is correct. When
as required.