To solve the equation [tex]\((x+1)^2 - 2 = \frac{2}{x}\)[/tex] algebraically:
1. Start with the given equation:
[tex]\[
(x+1)^2 - 2 = \frac{2}{x}
\][/tex]
2. Expand the left-hand side:
[tex]\[
(x + 1)^2 = x^2 + 2x + 1
\][/tex]
[tex]\[
(x + 1)^2 - 2 = x^2 + 2x + 1 - 2 = x^2 + 2x - 1
\][/tex]
3. So, the equation simplifies to:
[tex]\[
x^2 + 2x - 1 = \frac{2}{x}
\][/tex]
4. Multiply every term by [tex]\(x\)[/tex] to eliminate the fraction:
[tex]\[
x \cdot (x^2 + 2x - 1) = x \cdot \frac{2}{x}
\][/tex]
[tex]\[
x^3 + 2x^2 - x = 2
\][/tex]
5. Rearrange all terms to one side to set the equation to zero:
[tex]\[
x^3 + 2x^2 - x - 2 = 0
\][/tex]
We need to solve the cubic equation:
6. Factor the cubic equation:
By using polynomial division or the Rational Root Theorem, we determine possible factors. After checking, we get:
[tex]\[
(x+2)(x+1)(x-1) = 0
\][/tex]
7. Set each factor to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x + 2 = 0 \implies x = -2
\][/tex]
[tex]\[
x + 1 = 0 \implies x = -1
\][/tex]
[tex]\[
x - 1 = 0 \implies x = 1
\][/tex]
Therefore, the solutions to the equation [tex]\((x+1)^2 - 2 = \frac{2}{x}\)[/tex] are:
[tex]\[
x = -2, \quad x = -1, \quad x = 1
\][/tex]
Now, selecting from the table:
[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline
\multicolumn{4}{|c|}{ Solution Options } \\
\hline
x=-1 & x=2 & x=0.5 & x=-2 \\
\hline
x=0 & x=1 & x=0.25 & x=-0.5 \\
\hline
\end{tabular}
\][/tex]
The correct solutions from the table are:
[tex]\[
x = -1, \quad x = -2, \quad x = 1
\][/tex]
