HOW DO I SOLVE THIS???!!
1 answer:
You might be interested in
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.