Problem 76 - Intermediate Algebra

Question

Solve each equation. $$ x^{\frac{2}{3}}+x^{\frac{1}{3}}-6=0\left[\text { Hint: Let } y=x^{\frac{1}{3}} .\right] $$

Step-by-Step Solution

Answer

The solutions are \( x = -27 \) and \( x = 8 \).

Step 1: Introduce the substitution

Let's substitute \( y = x^{\frac{1}{3}} \). This means that \( y^2 = x^{\frac{2}{3}} \). Now, rewrite the original equation in terms of \( y \).

Step 2: Rewrite the equation

The original equation \( x^{\frac{2}{3}} + x^{\frac{1}{3}} - 6 = 0 \) becomes \( y^2 + y - 6 = 0 \) after substitution.

Step 3: Factor the quadratic equation

Factor the quadratic equation \( y^2 + y - 6 = 0 \). It can be factored as \( (y + 3)(y - 2) = 0 \).

Step 4: Solve for y

Set each factor equal to zero and solve for \( y \):1. \( y + 3 = 0 \) gives \( y = -3 \)2. \( y - 2 = 0 \) gives \( y = 2 \).

Step 5: Solve for x

Since \( y = x^{\frac{1}{3}} \), we find \( x \) by cubing each solution for \( y \):1. If \( y = -3 \), then \( x = (-3)^3 = -27 \).2. If \( y = 2 \), then \( x = 2^3 = 8 \).

Key Concepts

Quadratic Equations

Quadratic equations are vital in algebra, representing equations where the highest degree is 2. A typical quadratic equation takes the form \( ax^2+bx+c=0 \). In this exercise, we used a substitution method, turning a more complex equation into a simpler quadratic form. This transformation made it feasible to apply familiar quadratic solution techniques.

Quadratic equations can be solved using various methods, including:

• Factoring
• Completing the square
• Quadratic formula
• Graphing

In this instance, factoring became crucial once the substitution was applied. By transforming \( x^{\frac{2}{3}}+x^{\frac{1}{3}}-6=0 \) into \( y^2+y-6=0 \), the equation could be factored easily. Recognizing quadratic form is the initial step that paved the way for subsequent techniques like factoring to be used effectively.

Factoring

Factoring involves breaking down expressions into simpler "factors" that can be multiplied to obtain the original expression. It's a core technique in solving quadratic equations. In our exercise, the expression \( y^2 + y - 6 = 0 \) was factored into \( (y + 3)(y - 2) = 0 \). By setting each factor equal to zero, we found possible values for \( y \).

When factoring, one of the main methods is to look for two numbers that multiply to the constant term (in this case, -6) and add to the coefficient of the linear term (1). Here, 3 and -2 fulfilled those conditions:

• 3 \( \times \) -2 = -6
• 3 + (-2) = 1

Breaking down quadratic expressions into factors drastically simplifies solving equations, as it narrows down solutions to testing possible scenarios by setting factors to zero.

Radical Equations

Radical equations include variables within a radical, such as a square root or cube root. Solving these equations often involves eliminating the radical by raising both sides of the equation to a power that matches the radical's degree. In this exercise, the expression \( x^{\frac{1}{3}} \) signifies a cube root of \( x \).

A useful strategy for radical expressions is substitution, which transforms complex terms into simple forms using temporary variables (like \( y = x^{\frac{1}{3}} \)). This allowed us to handle the equation as a standard quadratic, moving from \( x^{\frac{2}{3}} + x^{\frac{1}{3}} - 6 = 0 \) to the simpler \( y^2 + y - 6 = 0 \).

After finding \( y \), reversing the substitution involved cubing the values, thereby extracting solutions for \( x \). This process showcases how tackling radical equations often necessitates both substitution and wise manipulation of powers to simplify and solve effectively.