Problem 78 - Intermediate Algebra

Question

Solve each equation. $$ x^{-2}+4 x^{-1}-12=0 $$

Step-by-Step Solution

Answer

Solutions are \( x = \frac{1}{2} \) and \( x = -\frac{1}{6} \).

Step 1: Substitute the variable

Since we have terms with negative exponents, let's start by substituting a new variable. Let \( y = x^{-1} \). This means \( y = \frac{1}{x} \) and \( y^2 = (x^{-1})^2 = x^{-2} \). The equation then becomes \( y^2 + 4y - 12 = 0 \).

Step 2: Solve the quadratic equation

Now solve the quadratic equation \( y^2 + 4y - 12 = 0 \) using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -12 \). Calculate the discriminant: \( b^2 - 4ac = 16 + 48 = 64 \). Substitute back into the formula gives: \( y = \frac{-4 \pm 8}{2} \). Thus, \( y = 2 \) or \( y = -6 \).

Step 3: Substitute back to find x

Now, substitute back to find \( x \). Since \( y = x^{-1} \), if \( y = 2 \), then \( x^{-1} = 2 \) or \( x = \frac{1}{2} \). If \( y = -6 \), then \( x^{-1} = -6 \) or \( x = -\frac{1}{6} \).

Step 4: Write the solutions

Therefore, the solutions to the original equation are \( x = \frac{1}{2} \) and \( x = -\frac{1}{6} \).

Key Concepts

Understanding Negative Exponents

Negative exponents might seem a bit tricky, but they're quite simple once you understand how they work. Essentially, a negative exponent means that you take the reciprocal of the base and raise it to the positive of that exponent.
For example, if you have \( x^{-n} \), this means \( \frac{1}{x^n} \). It's a way to express division in exponent form.

Here are a few handy tips to remember about negative exponents:

• \( a^{-n} = \frac{1}{a^n} \)
• If the exponent is negative, think of it as 'flipping' the base and changing the exponent to a positive number.
• Negative exponents only change the position of the base number--from the numerator to the denominator, or vice versa.

These basic rules will help you tackle problems involving negative exponents with ease.

Using the Substitution Method

The substitution method is a powerful tool for simplifying equations, especially when dealing with complex terms like negative exponents. By substituting a tricky expression with a simpler variable, you can transform the problem into a form that's easier to handle.

For instance, in the original exercise, substituting \( y = x^{-1} \) made it possible to simplify the equation from \( x^{-2} + 4x^{-1} - 12 = 0 \) to \( y^2 + 4y - 12 = 0 \). This step is crucial because:

• It changes a complex equation involving negative exponents to a simpler quadratic equation.
• It allows us to apply well-known techniques like the quadratic formula to find solutions.

By using substitution, you can simplify equations into a format that is more familiar and less intimidating to solve.

Solving with the Quadratic Formula

The quadratic formula is a key tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This allows us to find the values of \( y \) that satisfy the equation.

To use the quadratic formula, follow these steps:

• Determine the coefficients \( a \), \( b \), and \( c \) from your equation.
• Calculate the discriminant, \( b^2 - 4ac \). The discriminant determines the nature of the roots:
• If it's positive, there are two distinct real roots.
• If it's zero, there is one real root.
• If it's negative, there are no real roots.
• Plug the values back into the quadratic formula to find \( y \).

In our example, with \( y^2 + 4y - 12 = 0 \), we find \( y = 2 \) and \( y = -6 \) using this method.

Solving Equations Efficiently

When solving equations, it's essential to follow a structured approach. After simplifying an equation using substitution, you'll often find yourself with a standard form that you can solve using familiar techniques.

Here are some general steps to follow when solving equations:

• Simplify the equation by combining like terms and using techniques like substitution, if necessary.
• Determine the method you'll use to solve it: this could be factoring, using the quadratic formula, or graphing.
• Verify your solutions by substituting them back into the original equation to ensure they hold true.

In the provided exercise, once we obtained the solutions \( y = 2 \) and \( y = -6 \) using the quadratic formula, we substituted back to find \( x \). This process of back-substitution is critical to ensure that the solutions satisfy the initial problem statement. By following these steps diligently, you can enhance your problem-solving skills and approach each equation with confidence.