91Ó°ÊÓ

The substitution method is a powerful technique to simplify complex equations. It involves replacing a part of the equation with a new variable. This makes the equation easier to solve.

In this exercise, we have the equation:
\(2 x^{-2} - x^{-1} - 1 = 0\)

We can substitute a simpler variable for the more complex term. Let's set \(y = x^{-1}\). The given equation can now be rewritten using this substitution. Since \(x^{-2} = (x^{-1})^2\), we have \(x^{-2} = y^2\).

Substituting, the equation becomes: \(2y^2 - y - 1 = 0\). This new equation is a quadratic equation, and it is easier to solve compared to the original form.

Using substitution allows us to work with simpler terms and transform the original problem into a more manageable one.

Simplifying an equation involves rewriting it in a simpler, more manageable form, often by using algebraic identities or substitutions. This can make solving the equation much easier.

For our example: \(2 x^{-2} - x^{-1} - 1 = 0\), we can simplify it by substituting \(y\) for \(x^{-1}\), leading to: \(2 y^2 - y - 1 = 0\).

The next step in simplifying is to solve this quadratic equation. We can do that by factoring: \(2y^2 - y - 1 = (2y + 1)(y - 1) = 0\).

By setting each factor equal to zero, we get two solutions: \(y = -1/2\) and \(y = 1\).

Finally, revert the substitution by replacing \(y\) back to \(x^{-1}\), giving us: \(x^{-1} = -1/2\) and \(x^{-1} = 1\).

This means our final solutions are: \(x = -2\) and \(x = 1\). Simplifying in this systematic manner ensures accuracy and clarity.

Reciprocal exponents are exponents that are fractions, typically shown as the reciprocal of an integer. They simplify calculations, especially when dealing with roots or inverse powers.

In our initial equation: \(2 x^{-2} - x^{-1} - 1 = 0\), \(x^{-2}\) and \(x^{-1}\) are examples of reciprocal exponents. Here, \(x^{-1}\) is the same as \(1/x\), and \(x^{-2}\) is \(1/x^2\).

Understanding reciprocal exponents is crucial in simplifying and solving equations involving them. By replacing \(x^{-1}\) with \(y\), complex exponents become easier. In our example, when \(x^{-1} = y\), the original term \(x^{-2}\) turns into \(y^2\).

This substitution transforms the equation into a more familiar quadratic form, emphasizing how reciprocal exponents can simplify intricate algebraic problems.