Problem 37 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ x^{2 / 3}-8 x^{1 / 3}+15=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 27 \) and \( x = 125 \).

Step 1: Identify the Substitution

The given equation is \( x^{2/3} - 8x^{1/3} + 15 = 0 \). Notice that the exponents are multiples of \(1/3\). Let \( u = x^{1/3} \), which implies \( u^2 = x^{2/3} \). Substitute \( u \) into the equation to make it a quadratic equation: \( u^2 - 8u + 15 = 0 \).

Step 2: Solve the Quadratic Equation

Now, solve the quadratic equation \( u^2 - 8u + 15 = 0 \). This can be factored as \( (u - 3)(u - 5) = 0 \). Set each factor equal to zero to find the solutions for \( u \): \( u - 3 = 0 \) and \( u - 5 = 0 \). Therefore, \( u = 3 \) or \( u = 5 \).

Step 3: Back-Substitute to Find \( x \)

Recall that \( u = x^{1/3} \). Therefore, we substitute back to find \( x \):- For \( u = 3 \): \( x^{1/3} = 3 \) which gives \( x = 3^3 = 27 \).- For \( u = 5 \): \( x^{1/3} = 5 \) which gives \( x = 5^3 = 125 \).

Step 4: Verify the Solutions

Substitute each solution back into the original equation to verify:- For \( x = 27 \): \( 27^{2/3} - 8 \times 27^{1/3} + 15 = 9 - 24 + 15 = 0 \).- For \( x = 125 \): \( 125^{2/3} - 8 \times 125^{1/3} + 15 = 25 - 40 + 15 = 0 \).Both values satisfy the original equation, confirming that the solutions are correct.

Key Concepts

Exponentiation in Quadratic Equations

Exponentiation is a mathematical process that involves raising a number to a certain power. In the context of quadratic equations, exponentiation allows us to express variables as powers, which can simplify or transform the equation into a more workable form. In the given exercise:

• The equation is initially of the form \(x^{2/3} - 8x^{1/3} + 15 = 0\).
• Notice how exponents are fractions, which indicates the roots of numbers.

Understanding exponentiation helps us break down these complex expressions:

• The notation \(x^{2/3}\) implies the cube root of \(x\) squared \((\sqrt[3]{x})^2\).
• Similarly, \(x^{1/3}\) is simply the cube root of \(x\).

This exponentiation restructuring is crucial for the algebraic steps, making further manipulation and solving of the equation possible.

Solving Equations Through Factoring

Once we substitute and simplify the equation using exponentiation, we encounter a new equation that must be solved. Solving equations, often involving a step known as factoring, is pivotal in finding solutions. For our exercise, it involves solving the quadratic equation.

• We rewrote the equation as \(u^2 - 8u + 15 = 0\) where \(u = x^{1/3}\).
• This equation is a classic quadratic, allowing us to use factoring techniques.

Factoring is the process of expressing an equation as the product of its linear components:

• Here, the equation factors as \((u - 3)(u - 5) = 0\).
• The solutions are found by setting each factor to zero: \(u - 3 = 0\) or \(u - 5 = 0\).
• This gives solutions \(u = 3\) and \(u = 5\).

Factoring simplifies the process by breaking down the equation into smaller, easily manageable parts, revealing straightforward solutions.

Using Algebraic Substitution to Simplify

Algebraic substitution is a method used to simplify complex equations by replacing one part with a simpler variable. This is particularly useful when dealing with equations that have fractional exponents or are otherwise non-linear. In our problem, substitution plays a key role:

• We start with \(x^{2/3} - 8x^{1/3} + 15 = 0\), which might be intimidating initially.
• By letting \(u = x^{1/3}\), the equation transforms to \(u^2 - 8u + 15 = 0\), a much simpler quadratic form.

Substitution streamlines the solving process by transforming complex parts into more familiar and tractable expressions:

• It allows us to apply our knowledge of solving quadratic equations directly.
• Once solved, we reverse substitute to find the original variable: \(x = 3^3\) and \(x = 5^3\).

This method is powerful for handling equations that involve higher degrees or non-standard formats, making challenging problems approachable.