Solve this equation algebraically, then select the solutions in the table.
[tex](x+1)^2-2=\frac{2}{x}[/tex]
Solution Options
[tex]\begin{array}{l}
x=-1 \\
x=0
\end{array}[/tex]
[tex]\begin{array}{l}
x=2 \\
x=1
\end{array}[/tex]
[tex]x=0.5[/tex]
[tex]x=-2[/tex]
[tex]x=0.25[/tex]
[tex]x=-0.5[/tex]
Answer (1)
Expand and rewrite the equation: ( x + 1 ) 2 − 2 = x 2 ⇒ x 2 + 2 x − 1 = x 2 .
Multiply by x to eliminate the fraction: x 3 + 2 x 2 − x = 2 .
Rearrange into a cubic equation: x 3 + 2 x 2 − x − 2 = 0 .
Factor the cubic equation: ( x − 1 ) ( x + 1 ) ( x + 2 ) = 0 , yielding solutions x = 1 , − 1 , − 2 . The final answer is x = 1 , x = − 1 , x = − 2 .
Explanation
Problem Analysis We are given the equation ( x + 1 ) 2 − 2 = x 2 and asked to solve it algebraically and select the solutions from the given options.
Expanding and Rewriting First, let's expand the left side of the equation: ( x + 1 ) 2 − 2 = x 2 + 2 x + 1 − 2 = x 2 + 2 x − 1 So the equation becomes: x 2 + 2 x − 1 = x 2
Eliminating the Fraction Next, we multiply both sides of the equation by x to eliminate the fraction: x ( x 2 + 2 x − 1 ) = x ( x 2 ) x 3 + 2 x 2 − x = 2
Forming a Cubic Equation Now, we rewrite the equation to obtain a cubic equation: x 3 + 2 x 2 − x − 2 = 0
Factoring the Cubic Equation We can factor the cubic equation by grouping: x 3 + 2 x 2 − x − 2 = x 2 ( x + 2 ) − 1 ( x + 2 ) = ( x 2 − 1 ) ( x + 2 ) = ( x − 1 ) ( x + 1 ) ( x + 2 ) = 0
Solving for x Now, we solve for x by setting each factor to zero: x − 1 = 0 ⇒ x = 1 x + 1 = 0 ⇒ x = − 1 x + 2 = 0 ⇒ x = − 2
Checking the Solutions We need to check if these solutions are valid by plugging them back into the original equation. Also, we must remember that x cannot be 0 since it appears in the denominator of the original equation.
For x = 1 :
( 1 + 1 ) 2 − 2 = 1 2 2 2 − 2 = 2 4 − 2 = 2 2 = 2 (Valid)
For x = − 1 :
( − 1 + 1 ) 2 − 2 = − 1 2 0 2 − 2 = − 2 − 2 = − 2 (Valid)
For x = − 2 :
( − 2 + 1 ) 2 − 2 = − 2 2 ( − 1 ) 2 − 2 = − 1 1 − 2 = − 1 − 1 = − 1 (Valid)
Thus, the solutions are x = 1 , x = − 1 , and x = − 2 .
Final Answer The solutions to the equation ( x + 1 ) 2 − 2 = x 2 are x = 1 , x = − 1 , and x = − 2 .
Examples
Cubic equations can model volumes and rates of change in various real-world scenarios, such as determining the dimensions of a container or analyzing population growth. For instance, if you're designing a rectangular container with specific volume requirements and constraints on its dimensions, you might end up with a cubic equation to solve for the unknown side lengths. Understanding how to solve these equations allows you to optimize designs and predict outcomes in engineering and other practical applications.
