Problem 42 - Precalculus: Functions and Graphs

Question

Solve the equation. \(2 y^{1 / 3}-3 y^{1 / 6}+1=0\)

Step-by-Step Solution

Answer

The solutions are \( y = \frac{1}{64} \) and \( y = 1 \).

Step 1: Simplifying the Exponents

Let's begin by substituting a variable for the fractional exponent. Let \( x = y^{1/6} \). Therefore, \( x^2 = y^{1/3} \). Substitute these into the original equation:\[ 2x^2 - 3x + 1 = 0 \]

Step 2: Factoring the Quadratic

Now, factor the quadratic equation. We need two numbers that multiply to \( 2 \times 1 = 2 \) and add to \( -3 \): \[ 2x^2 - 3x + 1 = (2x - 1)(x - 1) = 0 \]

Step 3: Solving for x

Set each factor from the factored quadratic equation to zero and solve for \( x \):- For \( 2x - 1 = 0 \), we get \( x = \frac{1}{2} \).- For \( x - 1 = 0 \), we get \( x = 1 \).

Step 4: Solving for y

Recall that \( x = y^{1/6} \). Solve for \( y \) for each value of \( x \):- If \( x = \frac{1}{2} \), then \( (y^{1/6}) = \frac{1}{2} \), so \( y = \left(\frac{1}{2}\right)^6 = \frac{1}{64} \).- If \( x = 1 \), then \( (y^{1/6}) = 1 \), so \( y = 1^6 = 1 \).

Step 5: Final Solution

The solutions for the equation \( 2 y^{1 / 3}-3 y^{1 / 6}+1=0 \) are \( y = \frac{1}{64} \) and \( y = 1 \).

Key Concepts

Solving Equations

When tackling equations, it's essential to understand the process of isolating variables to find solutions. Solving equations often involves several steps to simplify and rearrange the terms. This exercise began with the equation \(2 y^{1 / 3}-3 y^{1 / 6}+1=0\). By introducing a substitution, specifically letting \(x = y^{1/6}\), the equation was transformed into a more recognizable form.

• The substitution helped convert fractional exponents into simpler quadratic terms.
• Changing variables can make an equation easier to manage and solve.

After substitution, the problem became a quadratic equation, which is easier to solve using standard algebraic methods. The journey of solving an equation is about systematically reducing its complexity to find solutions.

Fractional Exponents

Fractional exponents can often seem confusing at first, but they have clear rules and make certain calculations easier. A fractional exponent like \(y^{1/6}\) can be understood as the sixth root of \(y\).

• Fractional exponents denote roots, where the denominator indicates the root's degree.
• Here, \(y^{1/3}\) corresponds to the cube root of \(y\), and \(y^{1/6}\) to the sixth root.

Substituting fractional exponents can simplify expressions and equations, converting them into ones with simpler whole number exponents. This makes it easier to apply algebraic techniques, such as factoring and solving quadratic equations. Being comfortable with fractional exponents opens the door to understanding and simplifying many algebraic expressions.

Quadratic Equations

Quadratic equations are a common type of polynomial equation characterized by the form \(ax^2 + bx + c = 0\). The goal is often to factor the equation to find the values of \(x\) where the equation holds true.In this exercise, the transformed equation \(2x^2 - 3x + 1 = 0\) was factored into \((2x - 1)(x - 1) = 0\).

• Factoring involves finding pairs of numbers that, when multiplied, give the constant term and when added, give the linear coefficient.
• The factored form allows us to use the zero product property, which states if \(ab=0\), then \(a=0\) or \(b=0\).

This property simplifies the process of finding the solutions to the equation, which are often called roots or zeros. Understanding quadratics is crucial as they frequently appear in advanced algebra and calculus. Knowing how to solve them using various techniques, such as factoring, the quadratic formula, or completing the square, broadens your mathematical toolkit.