Problem 40 - Precalculus: Functions and Graphs

Question

Solve the equation. \(x^{-2}-2 x^{-1}-35=0\)

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{7}\) and \(x = -\frac{1}{5}\).

Step 1: Rewrite the Equation

Begin by substituting a new variable to simplify the equation. Let \( y = x^{-1} \), which implies that \( x^{-2} = y^2 \). Then the equation \( x^{-2} - 2x^{-1} - 35 = 0 \) becomes:\[ y^2 - 2y - 35 = 0 \] This is a quadratic equation in terms of \( y \).

Step 2: Factor the Quadratic Equation

To solve the quadratic equation \( y^2 - 2y - 35 = 0 \), look for two numbers that multiply to \(-35\) and add to \(-2\). The numbers \(-7\) and \(5\) fit these conditions. Therefore, factor the quadratic as:\[ (y - 7)(y + 5) = 0 \]

Step 3: Solve for the New Variable, y

Set each factor to zero to solve for \( y \):\[ y - 7 = 0 \quad \Rightarrow \quad y = 7 \] \[ y + 5 = 0 \quad \Rightarrow \quad y = -5 \] These are the solutions for \( y \).

Step 4: Substitute Back to Find x

Since \( y = x^{-1} \), solve for \( x \) by substituting back each value of \( y \):- If \( y = 7 \), then \( x^{-1} = 7 \) implies \( x = rac{1}{7} \).- If \( y = -5 \), then \( x^{-1} = -5 \) implies \( x = -rac{1}{5} \).

Step 5: Write Down the Solutions for x

The solutions to the original equation \( x^{-2} - 2x^{-1} - 35 = 0 \) are \( x = rac{1}{7} \) and \( x = -rac{1}{5} \).

Key Concepts

Factoring Quadratics

Factoring is a very important skill when solving quadratic equations like \[ y^2 - 2y - 35 = 0 \]. This is an equation where the highest power of the variable (in this case, \( y \)) is squared.When factoring quadratics, you want to find two numbers that multiply to the constant term, in this instance (-35), and at the same time add up to the coefficient of the linear term (here it's -2).

• Look for two numbers. They should multiply to the end number, (-35).
• Check if they add up to the middle number, (-2).
• -7 and 5 are our lucky numbers here because \((-7) \times 5 = -35\) and \((-7) + 5 = -2\).
• Therefore, break down the quadratic as \[ (y-7)(y+5) = 0 \]. This breaks the problem into easier parts. You can solve each part independently.

Variable Substitution

Variable substitution is a highly useful trick for simplifying complex equations. In problems with inverses or fractions like \( x^{-2} - 2x^{-1} - 35 = 0 \), we substitute the variable to make it look more like a regular quadratic equation. - **Why Substitution?** The original equation involves \( x^{-2} \) and \( x^{-1} \), which can be awkward to deal with if you're not comfortable with negative exponents. Using substitution makes the math much more straightforward.
- **How do we do it?** Let's assign \( y = x^{-1} \). This means \( x^{-2} = y^2 \). Substitute to get a new quadratic: \[ y^2 - 2y - 35 = 0 \]. Now the equation is in a standard form, making it easier to handle.

Inverse Exponent

Dealing with inverse exponents is not as tricky as it might seem. The negative exponent means you are really dealing with the reciprocal of that variable. - In the case of \( x^{-1} \), think of it as \( \frac{1}{x} \). Similarly, \( x^{-2} \) can be rewritten as \( \frac{1}{x^2} \).
- Once we have solved for \( y \) using the quadratic equation, we find possible values \( y = 7 \) and \( y = -5 \). Remember that this relates to \( y = x^{-1} \). So to get \( x \), we take the reciprocal of \( y \).- If \( y = 7 \), then \( x = \frac{1}{7} \).- If \( y = -5 \), then \( x = -\frac{1}{5} \).By managing inverse exponents through this method, you can transform challenging equations into simpler, solvable forms.