Problem 81 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Equation

Start by rewriting the equation in terms of a new variable for simplicity. Let \( y = x^{-1} \). This makes \( x^{-2} = y^2 \). Substitute these into the equation to get a quadratic equation: \[ 12y^2 - 17y - 5 = 0 \]

Step 2: Solve the Quadratic Equation

Solve the quadratic equation \( 12y^2 - 17y - 5 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 12 \), \( b = -17 \), and \( c = -5 \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \) as follows: - \( b^2 = (-17)^2 = 289 \)- \( 4ac = 4 \times 12 \times (-5) = -240 \)- The discriminant is \( 289 - (-240) = 289 + 240 = 529 \).

Step 4: Apply the Quadratic Formula

Use the quadratic formula: \[ y = \frac{17 \pm \sqrt{529}}{24} \]Since \( \sqrt{529} = 23 \), the solutions are:\[ y_1 = \frac{17 + 23}{24} = \frac{40}{24} = \frac{5}{3} \] \[ y_2 = \frac{17 - 23}{24} = \frac{-6}{24} = -\frac{1}{4} \].

Step 5: Solve for \( x \)

Rewrite \( y = x^{-1} \) for both solutions of \( y \): - If \( y_1 = \frac{5}{3} \), then \( x_1 = \frac{3}{5} \).- If \( y_2 = -\frac{1}{4} \), then \( x_2 = -4 \).

Step 6: Verify the Solutions

Double-check by substituting \( x_1 = \frac{3}{5} \) and \( x_2 = -4 \) back into the original equation to ensure both satisfy it:- Substitute \( x_1 = \frac{3}{5} \) into the original equation and confirm that it returns a true statement.- Substitute \( x_2 = -4 \) and confirm that it also satisfies the original equation.

Key Concepts

Substitution Method in Algebra

Solving complex algebraic equations can become much simpler with the substitution method. When you encounter an equation involving fractions with variables in the denominator, assigning a new variable might help.
In the original problem, we had the equation with terms like \( x^{-1} \) and \( x^{-2} \). By setting \( y = x^{-1} \), you convert \( x^{-2} \) into \( y^2 \). This smart step transforms the equation into a more familiar quadratic form:

• Substitute \( x^{-1} \) with \( y \).
• Change \( x^{-2} \) to \( y^2 \).

So, the equation \( 12 x^{-2}-17 x^{-1}-5=0 \) becomes \( 12y^2 - 17y - 5 = 0 \). This makes the problem easier to approach using quadratic methods.

Discriminant Calculation

The discriminant in a quadratic equation gives crucial information about the nature of the roots. For any quadratic equation in the form \( ax^2 + bx + c = 0 \):

• The formula to find the discriminant is \( b^2 - 4ac \).

This value tells us whether there are two real solutions, one real solution, or complex solutions to the equation.

• A positive discriminant means two distinct real roots.
• Zero indicates exactly one real root (also known as a repeated root).
• A negative discriminant results in complex roots.

In our exercise, when calculating the discriminant for \( 12y^2 - 17y - 5 = 0 \):

• \( b^2 = 289 \)
• \( 4ac = -240 \)
• Discriminant \( = 289 + 240 = 529 \)

With \( 529 \) being a perfect square, the equation has two real and distinct roots.

Quadratic Formula Usage

The quadratic formula is your go-to tool for solving any quadratic equation. It is expressed as: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula helps find the values of \( y \) that satisfy the equation. By plugging in the values of \( a \), \( b \), and \( c \), you can find the solutions.
In our problem:

• \( a = 12 \)
• \( b = -17 \)
• \( c = -5 \)

Applying these into the formula, you have:

• \( y = \frac{17 \pm 23}{24} \)

The solutions are:

• \( y_1 = \frac{5}{3} \) and \( y_2 = -\frac{1}{4} \)

Finally, revert back using substitution. For each solution of \( y \), swap back to the original variable \( x \) using \( y = x^{-1} \). This gives the final solutions:

• If \( y_1 = \frac{5}{3} \), \( x_1 = \frac{3}{5} \)
• If \( y_2 = -\frac{1}{4} \), \( x_2 = -4 \)

Thus, helping you solve the original equation proficiently.