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The substitution method is a powerful technique used to simplify equations, especially when dealing with variables that are difficult to solve directly. This method often involves replacing a complex or awkward expression with a new variable.

In the context of solving quadratic equations, substitution helps convert the original equation into a more familiar form. For instance, consider the equation involving negative powers:

• We started with \( x^{-2} + 3 x^{-1} + 2 = 0 \).
• By setting \( y = x^{-1} \), the expression \( x^{-2} \) becomes \( y^2 \).
• This substitution transforms the original equation into the quadratic \( y^2 + 3y + 2 = 0 \).

This transformation allows one to leverage more straightforward methods, such as factoring, to solve the now simpler equation. Once solved, it's crucial to substitute back the original variable to find the actual solution to the problem.

Factoring is a method used to break down quadratic equations into a product of simpler expressions that can be set to zero for easy solving. The goal here is to express the quadratic equation in the form \((y + a)(y + b) = 0\).

In the given equation, \( y^2 + 3y + 2 = 0 \), we aim to find two numbers that:

• Multiply together to give 2 (the constant term).
• Add up to 3 (the coefficient of \(y\)).

These numbers are 1 and 2. Hence, the quadratic factors as \((y + 1)(y + 2) = 0\).

By converting the quadratic into its factored form, we can immediately identify potential solutions for \(y\). This is because if the product of two terms is zero, at least one of those terms must be zero. Therefore, we solve the separate equations \(y + 1 = 0\) and \(y + 2 = 0\) to find possible values for \(y\).

The process of solving equations involves finding the values of the variables that satisfy the equation's conditions. Once the equation is in a solvable form, such as factored or simplified, extracting these values becomes more straightforward.

In our exercise, after factoring the quadratic equation to \((y + 1)(y + 2) = 0\), we solved it by setting each factor to zero:

• \(y + 1 = 0\) results in \(y = -1\).
• \(y + 2 = 0\) gives \(y = -2\).

These solutions for \(y\) must be translated back to solutions for the original variable \(x\).

Given \(y = x^{-1}\), translating solutions involves finding the reciprocal of \(y\):

• \(y = -1\) converts to \(x = -1^{-1} = -1\).
• \(y = -2\) translates to \(x = -\frac{1}{2}\).

To confirm, substituting these values back into the original equation is vital to ensure they satisfy the equation.