Let's analyze the table and use it to solve the given equations step-by-step.
### Step 1: Understanding the Table
We are given a table with expressions involving [tex]\( x \)[/tex]. Let's recall the values from the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 \\
\hline
x^2+x & -6 & 2 & 0 & 0 & 2 & 6 \\
\hline
x^3+x & -30 & -10 & -2 & 0 & 2 & 10 \\
\hline
\end{array}
\][/tex]
### Step 2: Solving [tex]\( x^2 + x = 2 \)[/tex]
We need to find the [tex]\( x \)[/tex] values for which [tex]\( x^2 + x = 2 \)[/tex].
From the table, we have:
- When [tex]\( x = -3 \)[/tex], [tex]\( x^2 + x = -6 \)[/tex]
- When [tex]\( x = -2 \)[/tex], [tex]\( x^2 + x = 2 \)[/tex]
- When [tex]\( x = -1 \)[/tex], [tex]\( x^2 + x = 0 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( x^2 + x = 0 \)[/tex]
- When [tex]\( x = 1 \)[/tex], [tex]\( x^2 + x = 2 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( x^2 + x = 6 \)[/tex]
By comparing the values, we see that [tex]\( x^2 + x = 2 \)[/tex] occurs when:
- [tex]\( x = -2 \)[/tex]
- [tex]\( x = 1 \)[/tex]
So, the solutions for [tex]\( x^2 + x = 2 \)[/tex] are [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex].
### Step 3: Solving [tex]\( x^3 + x = 2 \)[/tex]
We need to find the [tex]\( x \)[/tex] values for which [tex]\( x^3 + x = 2 \)[/tex].
From the table, we have:
- When [tex]\( x = -3 \)[/tex], [tex]\( x^3 + x = -30 \)[/tex]
- When [tex]\( x = -2 \)[/tex], [tex]\( x^3 + x = -10 \)[/tex]
- When [tex]\( x = -1 \)[/tex], [tex]\( x^3 + x = -2 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( x^3 + x = 0 \)[/tex]
- When [tex]\( x = 1 \)[/tex], [tex]\( x^3 + x = 2 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( x^3 + x = 10 \)[/tex]
By comparing the values, we see that [tex]\( x^3 + x = 2 \)[/tex] occurs when:
- [tex]\( x = 1 \)[/tex]
So, the solution for [tex]\( x^3 + x = 2 \)[/tex] is [tex]\( x = 1 \)[/tex].
### Final Solutions:
- For [tex]\( x^2 + x = 2 \)[/tex], the solutions are [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex].
- For [tex]\( x^3 + x = 2 \)[/tex], the solution is [tex]\( x = 1 \)[/tex].
