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Solve this equation algebraically, then select the solutions in the table.

[tex]\[ (x+1)^2 - 2 = \frac{2}{x} \][/tex]

\begin{tabular}{|c|c|c|c|}
\hline
\multicolumn{4}{|c|}{Solution Options} \\
\hline
[tex]$x = -1$[/tex] & [tex]$x = 2$[/tex] & [tex]$x = 0.5$[/tex] & [tex]$x = -2$[/tex] \\
\hline
[tex]$x = 0$[/tex] & [tex]$x = 1$[/tex] & [tex]$x = 0.25$[/tex] & [tex]$x = -0.5$[/tex] \\
\hline
\end{tabular}


Sagot :

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]
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