Problem 126 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt[3]{2 x}=\sqrt{x} $$

Step-by-Step Solution

Answer

Solutions are \( x = 0 \) and \( x = 4 \). Both are valid; no extraneous solutions.

Step 1: Understand the Problem

We are given the equation \( \sqrt[3]{2x} = \sqrt{x} \) and need to find all values of \( x \) that satisfy this equation. Then, we'll check each solution to determine if any are extraneous (don't actually satisfy the original equation).

Step 2: Cube Both Sides

To eliminate the cube root on the left side, cube both sides of the equation:\[ (\sqrt[3]{2x})^3 = (\sqrt{x})^3 \].This simplifies to:\[ 2x = x^{3/2} \].

Step 3: Solve for x

Rearrange the equation to solve for \(x\):\[ x^{3/2} - 2x = 0 \].Factor out \(x\):\[ x(x^{1/2} - 2) = 0 \].So, \(x = 0\) or \(x^{1/2} - 2 = 0\).

Step 4: Solve for Remaining x Values

Solve the second factor: \( x^{1/2} - 2 = 0 \).Add 2 to both sides:\[ x^{1/2} = 2 \].Square both sides to solve for \(x\):\[ x = 4 \].

Step 5: Check for Extraneous Solutions

Substitute \( x = 0 \) and \( x = 4 \) back into the original equation \( \sqrt[3]{2x} = \sqrt{x} \):1. For \( x = 0 \): \( \sqrt[3]{2 \times 0} = \sqrt{0} \Rightarrow 0 = 0 \). This is true.2. For \( x = 4 \): \( \sqrt[3]{2 \times 4} = \sqrt{4} \Rightarrow \sqrt[3]{8} = 2 \). Recall that \( \sqrt[3]{8} = 2 \), so this is also true.Both solutions satisfy the original equation. Thus, there are no extraneous solutions.

Key Concepts

Cube Roots

The cube root of a number is a value that, when multiplied by itself three times, gives the original number.
For example, the cube root of 27 is 3 because 3 × 3 × 3 = 27. In mathematics, cube roots are usually represented by the radical sign with a small 3, like this: \( \sqrt[3]{a} \).
When solving an equation involving cube roots, an effective method is to eliminate the cube root by cubing both sides of the equation.
Doing this removes the radical sign and allows you to work with polynomial equations, which are often easier to handle.

• Cubing both sides of \( \sqrt[3]{2x} = \sqrt{x} \) resulted in \( 2x = x^{3/2} \).
• This is a crucial step in finding potential solutions.

Understanding cube roots is key to solving equations that involve them, as it lays the groundwork for further manipulations and isolating the variable.

Solving Equations

Solving equations is all about finding the unknown value or values that make the equation true.
Equations are like balanced scales; whatever you do to one side, you must do to the other to maintain balance.
In the exercise, after cubing both sides, we ended up with the equation \( 2x = x^{3/2} \).
To solve it, the next step involves rearranging the equation to bring all terms to one side:

• Rewriting it as \( x^{3/2} - 2x = 0 \) allows us to see it in its polynomial form.
• From this point, our goal is to find values of \( x \) that make the equation true.

It's essential to correctly rearrange and simplify equations as this sets the stage for applying further techniques like factoring.

Factoring Equations

Factoring is a technique used to simplify polynomials and find their roots, which are the values that make the equation zero.
In our exercise, the equation \( x^{3/2} - 2x = 0 \) was factored as \( x(x^{1/2} - 2) = 0 \).
Here's how factoring helps in finding solutions:

• By setting each factor equal to zero, we derived the potential solutions: \( x = 0 \) and \( x^{1/2} - 2 = 0 \).
• For the second factor, solving \( x^{1/2} = 2 \) involved squaring both sides to find \( x = 4 \).

Thus, factorization simplified our task to evaluate the original values that satisfy the equation.
Factoring not only reduces complexity but is also invaluable for checking potential solutions for any equation.