Problem 62 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(x^{2 / 3}-2 x^{1 / 3}-3=0\)

Step-by-Step Solution

Answer

The solutions are \(x = 27\) and \(x = -1\).

Step 1: Substitute Variable

Let's start by using a substitution to simplify the equation. Letting \[ u = x^{1/3} \] means that \[ u^2 = x^{2/3} \]. Replace the variables in the original equation: \[ x^{2/3} - 2x^{1/3} - 3 = 0 \] becomes \[ u^2 - 2u - 3 = 0 \].

Step 2: Solve the Quadratic Equation

Now we solve the quadratic equation \[ u^2 - 2u - 3 = 0 \]. This can be factored into: \[ (u - 3)(u + 1) = 0 \].

Step 3: Solve for u

Setting each factor equal to zero: \[ u - 3 = 0 \] or \[ u + 1 = 0 \]. Therefore, \[ u = 3 \] or \[ u = -1 \].

Step 4: Substitute Back x

Recall that \[ u = x^{1/3} \]. Thus, \[ x^{1/3} = 3 \] or \[ x^{1/3} = -1 \]. To find x, cube both sides: \[ x = 3^3 = 27 \] and \[ x = (-1)^3 = -1 \].

Step 5: Check Solutions

Substitute \(x = 27\) and \(x = -1\) back into the original equation to ensure they satisfy it: For \(x = 27\): \[ (27)^{2/3} - 2(27)^{1/3} - 3 = 9 - 6 - 3 = 0 \] For \(x = -1\): \[ (-1)^{2/3} - 2(-1)^{1/3} - 3 = 1 + 2 - 3 = 0 \] Since both values satisfy the original equation, they are correct solutions.

Key Concepts

Substitution Method

To solve complex equations, we often use the substitution method to simplify things. This method introduces a new variable to replace a complicated expression. For instance, let's say we have the equation: \[ x^{2/3} - 2x^{1/3} - 3 = 0 \]
The exponents make it tricky to solve directly. Instead, we define a new variable: \[ u = x^{1/3} \]
Consequently, \[ u^2 = x^{2/3} \]
Substituting these into our original equation gives us: \[ u^2 - 2u - 3 = 0 \]
This conversion transforms the equation into a standard quadratic form, making it easier to solve. In our transformed equation, solving for the new variable (u) will eventually lead us back to solutions for x.

Factoring Quadratic Equations

Once we've simplified an equation, the next step is often to factor it. For example, after substitution, our equation became: \[ u^2 - 2u - 3 = 0 \]
Factoring this, we get: \[ (u - 3)(u + 1) = 0 \]
FACTORING involves finding two binomials that multiply to form the quadratic. Here, \((u - 3)\) and \((u + 1)\) multiply to give \(u^2 - 2u - 3\).
Now, we set each factor to zero: \[ u - 3 = 0 \] or \[ u + 1 = 0 \]
This gives us the solutions: \[ u = 3 \] and \[ u = -1 \]
Factoring helps break down the equation, making it simpler to find the values for our new variable (u), and ultimately, for x.

Checking Solutions

To verify the correctness of our solutions, we need to check them by plugging them back into the original equation. From our steps above, we had found: \[ u = 3 \] and \[ u = -1 \]
Now substituting back, recall: \[ u = x^{1/3} \]
Calculating for x: \[ x^{1/3} = 3 \] means \[ x = 3^3 = 27 \]
and \[ x^{1/3} = -1 \] means \[ x = (-1)^3 = -1 \]
Let's check these values in the original equation: For \(x = 27\): \[ 27^{2/3} - 2 \times 27^{1/3} - 3 = 0 \] \[ 9 - 6 - 3 = 0 \] For \(x = -1\): \[ (-1)^{2/3} - 2 \times (-1)^{1/3} - 3 = 0 \] \[ 1 + 2 - 3 = 0 \]
Both values satisfy the original equation, which means our solutions are correct! Always remember to check your solutions to ensure they truly satisfy the initial problem.