Problem 75 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$x^{2 / 3}=2 x^{1 / 3}$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = 8\).

Step 1: Introduce substitution

Let’s introduce a substitution to simplify the equation. Let set \(y = x^{1/3}\). Hence, \(y^2 = (x^{1/3})^2 = x^{2/3}\). Now replace \(x^{2/3}\) and \(x^{1/3}\) in the original equation with \(y\), giving us: \(y^2 = 2y\).

Step 2: Rewrite the equation

Next, we need to rewrite the equation in standard quadratic form: \(y^2 - 2y = 0\).

Step 3: Factor the quadratic equation

Factor the quadratic equation: \(y(y - 2) = 0\). This gives us two possible solutions: \(y = 0\) or \(y = 2\).

Step 4: Reverse the substitution

Reverse the substitution \(y = x^{1/3}\) to find \(x\). If \(y = 0\), then \(x^{1/3} = 0\), so \(x = 0\). If \(y = 2\), then \(x^{1/3} = 2\), so \(x = 2^3 = 8\).

Step 5: Write the final solution

The solutions to the equation \(x^{2/3} = 2x^{1/3}\) are \(x = 0\) and \(x = 8\).

Key Concepts

Substitution

Substitution is a technique used to simplify complex equations by replacing variables or expressions with simpler ones. In our exercise, we replaced the term \(x^{1/3}\) with \(y\). This turned a complicated radical equation into a simpler quadratic equation. By substituting, the original equation \(x^{2/3} = 2 x^{1/3}\) becomes \(y^2 = 2y\). Substitution helps to transform the equation into a form that is easier to solve. This method can be particularly useful in dealing with radical expressions and polynomials.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They can be solved using factoring, completing the square, or the quadratic formula. In our exercise, after substitution, we end up with the quadratic equation \(y^2 - 2y = 0\). This is already in its standard form and ready to be solved. Quadratic equations are fundamental in algebra and appear frequently, so mastering the various methods to solve them is crucial.

Factoring

Factoring is the process of breaking down an equation into simpler components called factors. Factoring is often the first method tried because it can be straightforward and quick. In our exercise, we factor the quadratic equation \(y^2 - 2y = 0\) into \(y(y - 2) = 0\). Here, the equation is broken down into two factors: \(y\) and \(y - 2\). Setting each factor equal to zero gives us the solutions \(y = 0\) or \(y = 2\). Factoring is a powerful tool because it often simplifies the solving process considerably.