Problem 79 - Intermediate Algebra

Question

Solve each equation. $$ 12 x^{-2}-17 x^{-1}-5=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{3}{5} \) and \( x = -4 \).

Step 1: Substitute a Variable

First, make the equation easier to handle by setting a new variable for substitution. Let \( y = x^{-1} \). This means \( y = \frac{1}{x} \) and \( x^{-2} = y^2 \). Substitute these into the equation:\[ 12y^2 - 17y - 5 = 0 \]

Step 2: Recognize as a Quadratic Equation

Notice that the equation \( 12y^2 - 17y - 5 = 0 \) is a quadratic equation in terms of \( y \). To solve it, identify it in the standard quadratic form \( ay^2 + by + c = 0 \), where \( a = 12 \), \( b = -17 \), and \( c = -5 \).

Step 3: Apply the Quadratic Formula

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( y \). Plug in the values: \[ y = \frac{-(-17) \pm \sqrt{(-17)^2 - 4\cdot12\cdot(-5)}}{2\cdot12} \]Calculate the discriminant:\[ (-17)^2 - 4\cdot12\cdot(-5) = 289 + 240 = 529 \]Continue with:\[ y = \frac{17 \pm \sqrt{529}}{24} \]\[ y = \frac{17 \pm 23}{24} \]

Step 4: Solve for y

Calculate the two possible solutions for \( y \):1. \( y = \frac{17 + 23}{24} = \frac{40}{24} = \frac{5}{3} \)2. \( y = \frac{17 - 23}{24} = \frac{-6}{24} = -\frac{1}{4} \)

Step 5: Substitute Back to Solve for x

Now revert \( y = \frac{1}{x} \) back to \( x \). For each value of \( y \):1. If \( y = \frac{5}{3} \), then \( x = \frac{1}{y} = \frac{3}{5} \).2. If \( y = -\frac{1}{4} \), then \( x = \frac{1}{y} = -4 \).

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental principle in algebra used to solve quadratic equations, which are in the form of \( ax^2 + bx + c = 0 \). This formula allows us to find the roots of the equation, where the parabola crosses the x-axis. It is especially useful when factoring is not possible or straightforward.
To apply the quadratic formula, use:

• \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Plug in the values of \( a \), \( b \), and \( c \) from the quadratic equation. Here, our standard quadratic equation resulting from the substitution is \( 12y^2 - 17y - 5 = 0 \), where \( a = 12 \), \( b = -17 \), and \( c = -5 \).
The formula derives the potential solutions for \( y \) depending on these constants. As a result, you'll calculate two possible solutions using the plus and minus signs within the formula. Each solution corresponds to a crossing point or a zero of the quadratic function. After addressing these solutions, further simplifications can be made to understand the relationship between \( y \) and \( x \), eventually solving the original equation.

Variable Substitution

Variable substitution is a strategic method used in algebra to simplify complex equations and make them more manageable during problem-solving. In the exercise, we are dealing with a reciprocal function of \( x \) which is not straightforward to handle; thus, substitution alleviates this complexity.
Here, we set \( y = x^{-1} \), implying \( y = \frac{1}{x} \). Consequently, \( x^{-2} \) becomes \( y^2 \) due to the property of exponents: \( (x^{-1})^2 = x^{-2} \). This transforms our original equation into a simpler quadratic form, \( 12y^2 - 17y - 5 = 0 \).
By substituting \( x^{-1} \) with \( y \), solving the equation becomes significantly more straightforward. This technique highlights how variable substitution can be instrumental in addressing equations that appear daunting at first glance. It reduces the potential for errors and allows us to leverage other algebraic tools, such as the quadratic formula, more efficiently.

Discriminant Calculation

The discriminant is a vital component in the quadratic formula that helps determine the number and type of solutions a quadratic equation can have. Found under the square root in the quadratic formula, it's given by \( b^2 - 4ac \).
For our quadratic equation \( 12y^2 - 17y - 5 = 0 \), the discriminant calculates as:

• \( (-17)^2 - 4 \cdot 12 \cdot (-5) \)

Breaking this down, we have:

• \( 289 + 240 = 529 \)

Since the discriminant is positive and a perfect square, two distinct real solutions exist for \( y \). If the discriminant were zero, there would be one real solution, implying the vertex of the parabola exactly touches the x-axis. Conversely, a negative discriminant indicates no real solutions, only complex ones. Understanding the role of the discriminant helps predict the nature of solutions and foresights what values to expect for \( x \) or, in some cases, if a solution exists.