91Ó°ÊÓ

A cubic root equation is an algebraic expression where the variable is under a cubic root. The general form is \((a x + b)^{1/3} = c\). These types of equations can look intimidating but are quite manageable. When solving cubic root equations, our goal is to eliminate the cubic root to make the equation simpler.
Consider the problem we worked on: \( (2 x + 1)^{1/3} = -1 \). Here, the expression \( (2 x + 1)^{1/3} \) involves a cubic root.
The cubic root reverses the effect of cubing a number. For example, the cubic root of 8 is 2, because 2 cubed is 8. In our problem, the cubic root is already isolated on one side of the equation. This isolation is key before proceeding further.

Isolating the variable is a critical step in solving any equation. This means we get the variable by itself on one side of the equation. The first step in our problem involves cubing both sides to remove the cubic root:
Starting with:\[\big[(2x+1)^{1/3}\big]^3 = (-1)^3\]
This removes the cubic root because cubing a cubic root gives the original term back. So, \( (2x+1)^{1/3} \) cubed is simply \(2x+1\). On the right side, \(-1\) cubed remains \(-1\). Our equation then simplifies to \( 2x + 1 = -1 \).
After the cubic root is removed, we isolate x by moving the constant term (1 in this case) to the other side. Subtract 1 from both sides:
\( 2x + 1 - 1 = -1 - 1\)
This simplifies to:
\( 2x = -2\)

Solving for the variable (x) means finding the value that makes the equation true. In our isolated term \( 2x = -2 \), x is multiplied by 2. To solve for x, we need to get rid of that multiplication by doing the opposite operation: division.
Divide both sides by 2:
\( \frac{2x}{2} = \frac{-2}{2} \)
This simplifies to:
\( x = -1 \)
So, the value of x that satisfies the equation is -1.

Real solutions refer to solutions that are real numbers as opposed to complex or imaginary numbers. In our example, we found the real solution x = -1. To verify, substitute x back into the original equation:
\[ (2(-1) + 1)^{1/3} = -1 \] becomes \(( -2 + 1)^{1/3} = -1)\), which simplifies to \(-1^{1/3}\rightarrow -1\text{ . }\) This confirms our solution.
Real solutions are the values you will mostly deal with at this stage. Being able to quickly identify and solve for these is crucial in algebra. Always remember to double-check your solution by plugging it back into the original equation to ensure it works.