Problem 84 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(3 y^{-2}-y^{-1}-2=0\) \(\left(\text {Hint: Use } x^{-n}=\frac{1}{x^{n}}\right)\)

Step-by-Step Solution

Answer

The only valid solution is \( y = 1 \).

Step 1: Convert Negative Exponents

Rewrite the equation using the hint provided: \[ \frac{3}{y^2} - \frac{1}{y} - 2 = 0 \] This substitution helps in simplifying the fractions.

Step 2: Eliminate Fractions by Finding a Common Denominator

The common denominator for the fractions \(\frac{3}{y^2}\) and \(\frac{1}{y}\) is \(y^2\). Multiply each term by \(y^2\) to eliminate the fractions: \[ 3 - y - 2y^2 = 0 \] This step eliminates the fractions, making it easier to solve.

Step 3: Rearrange into Quadratic Form

Arrange the terms into a standard quadratic form: \[ -2y^2 - y + 3 = 0 \] This rearrangement sets it up for using the quadratic formula.

Step 4: Use the Quadratic Formula

Apply the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve the equation:Where \(a = -2\), \(b = -1\), and \(c = 3\).Substitute these values into the formula: \[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(3)}}{2(-2)} \] Simplify to find: \[ y = \frac{1 \pm \sqrt{1 + 24}}{-4} \] \[ y = \frac{1 \pm \sqrt{25}}{-4} \] \[ y = \frac{1 \pm 5}{-4} \]

Step 5: Calculate the Solutions

Continue solving by evaluating the two possible solutions: \[ y = \frac{1 + 5}{-4} = -\frac{6}{4} = -\frac{3}{2} \]\[ y = \frac{1 - 5}{-4} = \frac{-4}{-4} = 1 \]So, the two solutions are \( y = -\frac{3}{2} \) and \( y = 1 \).

Step 6: Verify Solutions for Extraneous Roots

Substitute each solution back into the original equation to check if they are valid. For \( y = -\frac{3}{2} \): \[ 3 \left(-\frac{3}{2}\right)^{-2} - \left(-\frac{3}{2}\right)^{-1} - 2 = 0 \] \( y = -\frac{3}{2} \) results in undefined due to negative base in powers above in real number context.For \( y = 1 \):\[ 3(1)^{-2} - (1)^{-1} - 2 = 0 \] \[ 3 - 1 - 2 = 0 \] is true. Only \( y = 1 \) is valid as real number solution.

Key Concepts

Understanding Negative Exponents

Negative exponents can often appear daunting but they can be simplified with a few simple steps. When you encounter a negative exponent such as in the expression \( x^{-n} \), you can rewrite it as \( \frac{1}{x^{n}} \). This transformation allows you to switch from a 'division' perspective to a 'fraction' perspective.
Why is this useful? Knowing how to handle negative exponents helps in rewriting equations to a form that is easier to work with.

• If you have \( y^{-2} \), this becomes \( \frac{1}{y^2} \).
• Similarly, \( y^{-1} \) becomes \( \frac{1}{y} \).

In our exercise, using these transformations helped convert the original equation into a form that was more manageable for solving: \[ \frac{3}{y^2} - \frac{1}{y} - 2 = 0 \]This sets the stage for further algebraic manipulation and solution.

Recognizing and Handling Extraneous Solutions

Extraneous solutions can sneak into your work when solving equations, especially during steps like squaring both sides or dealing with complex fractions. They are solutions derived mathematically correct but do not satisfy the original equation. This can occur due to restrictions in the domain such as division by zero or undefined expressions in real numbers.
Why watch out for them? It's crucial because they can mislead you into believing you've found more solutions than actually exist.
To avoid falling for extraneous solutions, verify each potential solution by substituting it back into the original equation.

• If it satisfies the equation, it is a real solution.
• If not, it's extraneous and should be discarded.

In our case study, when substituting \( y = -\frac{3}{2} \) back into \( 3 y^{-2} - y^{-1} - 2 = 0 \), the calculations lead to undefined outcomes because of negative bases in powers, indicating this solution is extraneous.

The Quadratic Formula: Solving Quadratic Equations

The quadratic formula is a powerful tool for finding solutions to quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula directly computes the roots, which are the solutions where the quadratic expression equals zero.
When and how to use it: The quadratic formula is especially helpful when factoring is difficult or when the quadratic doesn't neatly rearrange into recognizable binomials.

• Identify \( a, b, \) and \( c \) from your equation.
• Substitute these into the formula to find possible solutions.

In our problem, \( a = -2 \), \( b = -1 \), and \( c = 3 \). Plugging these into the formula gives us:\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(3)}}{2(-2)} \]Simplifying yields two solutions: \( y = 1 \) and \( y = -\frac{3}{2} \). Only \( y = 1 \) is valid upon verification, underscoring the importance of checking for extraneous solutions.