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Chapter 8: Problem 34

Solve each equation. \(x^{2 / 3}-7 x^{1 / 3}+12=0\)

Short Answer

Expert verified
The solutions are \(x = 27\) and \(x = 64\).

Step by step solution

01

Substitution

Let \( y = x^{1/3} \). This implies that \( y^2 = (x^{1/3})^2 = x^{2/3} \). Substitute these into the equation to get a quadratic equation in terms of \( y \): \( y^2 - 7y + 12 = 0 \).
02

Solve the Quadratic Equation

Factor the quadratic equation \( y^2 - 7y + 12 = 0 \). Notice that it factors to \((y - 3)(y - 4) = 0\).
03

Find the Solutions for y

Using the factored form \((y - 3)(y - 4) = 0\), set each factor equal to zero: \(y - 3 = 0\) and \(y - 4 = 0\). Solving these gives \( y = 3 \) and \( y = 4 \).
04

Back Substitution

Recall that \( y = x^{1/3} \). Substitute \( y = 3 \) back to get \( x^{1/3} = 3 \). Cubing both sides gives \( x = 3^3 = 27 \). Now substitute \( y = 4 \) into \( y = x^{1/3} \) to get \( x^{1/3} = 4 \). Cubing both sides here gives \( x = 4^3 = 64 \).
05

Verification

Verify both solutions by substituting them back into the original equation \( x^{2/3} - 7x^{1/3} + 12 = 0\). For \( x = 27 \): \( 27^{2/3} - 7 \times 27^{1/3} + 12 = 0 \). For \( x = 64 \): \( 64^{2/3} - 7 \times 64^{1/3} + 12 = 0 \). Both satisfy the equation, confirming that both are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful tool for simplifying complex equations by introducing a new variable. When faced with an equation involving a term like \(x^{2/3}\), it can be helpful to switch perspectives by defining a new variable. For instance, we can let \(y = x^{1/3}\), transforming our original equation into a simpler form that鈥檚 easier to handle.
  • By substituting \(y = x^{1/3}\), we automatically also set \(y^2 = x^{2/3}\).
  • This change of variable helps reduce the complexity, turning the equation into a typical quadratic form: \(y^2 - 7y + 12 = 0\).
This method effectively changes the original format of the equation, making it possible to apply straightforward algebraic techniques. The goal is to make the problem easier and more manageable, particularly when typical operations would otherwise be cumbersome.
Factoring Quadratics
Factoring quadratics is a fundamental algebraic skill used to solve quadratic equations like \(y^2 - 7y + 12 = 0\). The quadratic equation needs to be arranged into a form where it can be expressed as the product of two binomials.
  • For the quadratic \(y^2 - 7y + 12\), look for two numbers that multiply to 12 (the constant term) and add to -7 (the coefficient of y).
  • These numbers are -3 and -4, which means the quadratic factors as \((y - 3)(y - 4) = 0\).
Factoring breaks down the quadratic into more manageable linear factors, allowing for simple equations to solve. Good factoring skills transform seemingly complex problems into simple ones, by identifying the right components that fit together.
Back Substitution
Back substitution is the process of returning to the original variable after solving a simpler equation. Once we鈥檝e solved the quadratic equation from the substitution phase, we need to find the corresponding values for the initial variable, \(x\).
  • Solving \((y - 3)(y - 4) = 0\) gives us \(y = 3\) or \(y = 4\).
  • Remembering that \(y = x^{1/3}\), we substitute back: for \(y = 3\), we have \(x^{1/3} = 3\), which gives \(x = 27\).
  • Similarly, for \(y = 4\), we set \(x^{1/3} = 4\), resulting in \(x = 64\).
Back substitution ensures that we have correctly returned to the context of the original problem, confirming our solutions against the initial conditions. This step integrates the new findings within the framework of the original equation, providing a thorough resolution process.

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Most popular questions from this chapter

Solve each equation. Approximate the solutions to the nearest hundredth when appropriate. $$2 x^{2}+0.1 x=0.04$$

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