Problem 29 - A Graphical Approach to College Algebra

Question

Solve each equation by hand. Do not use a calculator. $$x^{2 / 3}-x^{1 / 3}-6=0$$

Step-by-Step Solution

Answer

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

Step 1: Identify Substitution

To simplify the equation, use a substitution. Let \(y = x^{1/3}\). This transforms the equation into \(y^2 - y - 6 = 0\).

Step 2: Solve the Quadratic Equation

Solve the transformed quadratic equation \(y^2 - y - 6 = 0\). First, find two numbers that multiply to -6 and add to -1. These numbers are 2 and -3.

Step 3: Factor the Equation

Factor the quadratic equation: \((y - 3)(y + 2) = 0\). This gives the solutions \(y = 3\) and \(y = -2\).

Step 4: Back Substitute for x

Recall that \(y = x^{1/3}\). Therefore, set up the equations: \(x^{1/3} = 3\) leading to \(x = 3^3 = 27\), and \(x^{1/3} = -2\) leading to \(x = (-2)^3 = -8\).

Step 5: Verify Solutions

To verify, substitute back into the original equation. For \(x = 27\): \(27^{2/3} - 27^{1/3} - 6 = 9 - 3 - 6 = 0\). For \(x = -8\): \((-8)^{2/3} - (-8)^{1/3} - 6 = 4 + 2 - 6 = 0\).

Step 6: Conclusion

Both solutions satisfy the original equation. Therefore, the solutions are \(x = 27\) and \(x = -8\).

Key Concepts

Substitution Method

When solving complicated equations, simplifying them using substitution can make the process easier. In our equation, \(x^{2 / 3}-x^{1 / 3}-6=0\), direct solving is not straightforward. That's where substitution comes in handy. We set \(y = x^{1/3}\) to transform the equation into a simpler quadratic form: \(y^2 - y - 6 = 0\).
This substitution changes the expression for \(x\) into a similar expression for the new variable \(y\), allowing you to apply quadratic solving techniques to a seemingly more complex expression. Once you solve for \(y\), you can then switch back to the original variable, \(x\). Remember, substituting not only simplifies the expression but also maintains the values needed to satisfy the original equation.
This method is especially useful when dealing with fractional powers or multiple terms of the same variable, as it can effectively turn a problem that seems unsolvable into a standard form that is easier to handle.

Factoring Quadratics

Factoring is one of the easiest and most commonly used techniques to solve quadratic equations. The goal is to express the quadratic as a product of two binomials, which can then be set to zero to find the roots. In our transformed equation, \(y^2 - y - 6 = 0\), we look for two numbers whose product is \(-6\) and sum is \(-1\). These numbers are \(2\) and \(-3\).
With these numbers, we can factor the quadratic equation as \((y - 3)(y + 2) = 0\). This equation can now be solved by applying the zero-product property, leading to the solutions: \(y = 3\) and \(y = -2\).
Factoring quadratics is an invaluable tool in algebra because it breaks down the equation into smaller, more manageable parts. Furthermore, understanding how to factor quickly and accurately will serve you well across many different areas of math.

Verification of Solutions

Once you have obtained solutions from your factorized equation, it's essential to verify these solutions by substituting them back into the original equation. This step ensures that no mathematical errors were made during transformation, factorization, or back substitution.
For the solutions \(x = 27\) and \(x = -8\) obtained through iterative processes:

• Substitute \(x = 27\) into the original equation: \(27^{2/3} - 27^{1/3} - 6 = 9 - 3 - 6 = 0\). This confirms \(x = 27\) as a valid solution.
• Similarly, substitute \(x = -8\) back: \((-8)^{2/3} - (-8)^{1/3} - 6 = 4 + 2 - 6 = 0\). This verifies \(x = -8\) as well.

Verifying solutions is crucial because subtle errors in math can lead to incorrect results. By confirming that the math checks out, you ensure that the solutions are accurate and fully satisfy the original equation.