Problem 78 - Algebra for College Students

Question

Solve each equation. $$x^{\frac{2}{3}}+x^{\frac{1}{3}}-6=0$$

Step-by-Step Solution

Answer

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

Step 1: Substitute and Simplify

Let us first introduce a substitution to simplify the equation. Define \( u = x^{1/3} \). Hence, \( u^2 = x^{2/3} \). Substituting these into the original equation gives us: \( u^2 + u - 6 = 0 \). Now we have a quadratic equation in terms of \( u \).

Step 2: Solve the Quadratic Equation

We will now solve the quadratic equation \( u^2 + u - 6 = 0 \). This equation can be solved by factoring. The factors of \( -6 \) that add up to \( 1 \) are \( 3 \) and \( -2 \). Thus, the equation can be factored as \( (u + 3)(u - 2) = 0 \).

Step 3: Find the Roots of the Quadratic Equation

Setting each factor equal to zero gives us the potential solutions for \( u \): \( u + 3 = 0 \) or \( u - 2 = 0 \). Solving these gives \( u = -3 \) and \( u = 2 \).

Step 4: Substitute Back to Find \( x \)

Recall that \( u = x^{1/3} \). Substitute back \( u = -3 \) and \( u = 2 \) to find \( x \). For \( u = -3 \), \( x^{1/3} = -3 \), so \( x = (-3)^3 = -27 \). For \( u = 2 \), \( x^{1/3} = 2 \), so \( x = 2^3 = 8 \).

Step 5: Verify the Solutions

Let's verify both solutions by substituting back into the original equation. For \( x = -27 \), the equation becomes: \( (-27)^{2/3} + (-27)^{1/3} - 6 = 9 - 3 - 6 = 0 \). For \( x = 8 \), the equation becomes: \( 8^{2/3} + 8^{1/3} - 6 = 4 + 2 - 6 = 0 \). Both solutions satisfy the original equation.

Key Concepts

Substitution Method

The substitution method is a powerful tool when solving complex algebraic equations. It involves replacing one part of the equation with a simpler variable to make the equation easier to manage. In our case, the original equation was quite intimidating: \[ x^{\frac{2}{3}} + x^{\frac{1}{3}} - 6 = 0 \] By substituting \( u = x^{1/3} \), we transformed the equation from terms of \( x \) to terms of \( u \). This simplification resulted in: \[ u^2 + u - 6 = 0 \] Using substitution can help make otherwise complex equations more tractable by forming a conventional quadratic equation. This technique is particularly handy when equations involve roots, exponents, or multiple variables.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our example, by using substitution, we arrived at the quadratic equation: \[ u^2 + u - 6 = 0 \] This quadratic equation is essential because the solutions provide the roots that are then used to find the original variable. Solving a quadratic equation can be done through:

• Factoring
• The quadratic formula
• Completing the square

Quadratic equations are instrumental in various fields such as physics, engineering, and finance, where they are used to model various phenomena.

Factoring

Factoring is a method where we express a quadratic equation as a product of its linear factors. For our equation \( u^2 + u - 6 = 0 \), we used factoring to simplify it further. The idea is to find two numbers that multiply to the constant term, and add up to the linear coefficient. Factoring Steps: - Identify two numbers whose product is \(-6\) and sum is \(1\). - These numbers are \(3\) and \(-2\). - This allows us to express the quadratic as: \[ (u + 3)(u - 2) = 0 \] Factoring reveals the solutions (or roots) of the quadratic equation. It's a preferred method when the equation can be simplified neatly, as it is typically faster and provides immediate insight into the structure of the equation.

Verification of Solutions

Once you have potential solutions, it is crucial to verify them by substituting back into the original equation. Verifying ensures that no errors occurred during simplification or solving. For our solutions \( x = -27 \) and \( x = 8 \): - Substituting \( x = -27 \) back gives us: \[ (-27)^{2/3} + (-27)^{1/3} - 6 = 9 - 3 - 6 = 0 \] - Substituting \( x = 8 \) results in: \[ 8^{2/3} + 8^{1/3} - 6 = 4 + 2 - 6 = 0 \] Both computations confirm that these values satisfy the initial equation, proving they are indeed solutions. Verifying solutions is a vital step to prevent mistakes and to ensure the results are consistent with the given problem.