91Ó°ÊÓ

Negative exponents can initially seem confusing, but they follow a straightforward rule. When you see an expression like \( x^{-n} \), it means that you take the reciprocal of the base raised to the positive of that exponent. In simpler terms, \[x^{-n} = \frac{1}{x^n}\]This means that any number with a negative exponent flips its position in a fraction. It turns the base from numerator to denominator. For example:

• \( 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \)
• \( 5^{-1} = \frac{1}{5} \)

Using this understanding, when the problem statement says \( x^{-1} \) and \( x^{-2} \), you simply rewrite them as \( \frac{1}{x} \) and \( \left(\frac{1}{x}\right)^2 \), respectively. This simplification process is crucial for solving algebraic equations involving negative exponents effectively.

A quadratic equation is a polynomial that's characterized by its highest degree of 2. The general form is \[ax^2 + bx + c = 0\]where \( a \), \( b \), and \( c \) are constants. Solving quadratic equations is a fundamental skill in algebra, as they appear frequently in various mathematical problems. There are several methods to solve quadratic equations:

• Factoring: Rewriting the quadratic expression as a product of two binomials.
• Completing the Square: Rewriting the equation to make it into a perfect square trinomial.
• Quadratic Formula: Using the formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]which solves any quadratic equation directly.

In our exercise, the substitution of \( x^{-1} = y \) transformed the equation into a quadratic form \( y^2 - y - 42 = 0 \), which was then solved by factoring.

Factorization is a technique where an expression is rewritten as the product of simpler expressions. In algebra, factorizing quadratic equations means expressing them as the product of two binomials. Consider the quadratic equation \[y^2 - y - 42 = 0\]To factorize, we look for two numbers that multiply to give \(-42\) (the constant term) and add to give \(-1\) (the coefficient of \(y\)).

• These numbers are \(-7\) and \(6\) because \(-7 \times 6 = -42\) and \(-7 + 6 = -1 \).

Thus, the factorization of the quadratic is:\[(y - 7)(y + 6) = 0\]This process of finding two numbers that fit both criteria is essential for solving quadratic equations by factorization.

To solve equations algebraically means to find the values of the variables that make the equation true. This usually involves a series of logical steps, using mathematical operations. Let's briefly go through these steps using our original problem.

• Identify and Simplify: We first rewrite any complex parts using substitutions. In this case, negative exponents were expressed in a simpler form with a new variable \( y \).
• Solve the New Equation: Once simplified, solve the equation using appropriate methods for the type of equation you have—in our case, factorization of the quadratic equation.
• Back-Substitute: Reintroduce the original variables to find solutions to the original equation. Here, we found \( x \) by substituting back \( y = x^{-1} \).
• Validation: Always check the solutions by substituting them back into the original equation to ensure they work—which they did here for both solutions \( x = \frac{1}{7} \) and \( x = -\frac{1}{6} \).

Solving algebraically ensures that you fully understand the relationship between the variables and the equation at hand.