Problem 99 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. Let \(f(x)=\sqrt[4]{x+8}-\sqrt[4]{2 x} .\) Find all values of \(x\) for which \(f(x)=0\)

Step-by-Step Solution

Answer

The solution is \(x = 8\), which is valid and not extraneous.

Step 1: Set up the Equation

To solve for the values of \(x\) that satisfy \(f(x) = 0\), set the given function equal to zero: \(\sqrt[4]{x+8} - \sqrt[4]{2x} = 0\).

Step 2: Solve the Equation

Re-arrange the equation to isolate the two radicals: \(\sqrt[4]{x+8} = \sqrt[4]{2x}\). Next, raise both sides to the fourth power to eliminate the radicals: \[ (\sqrt[4]{x+8})^4 = (\sqrt[4]{2x})^4 \] which simplifies to \(x + 8 = 2x\).

Step 3: Simplify the Equation

Subtract \(x\) from both sides of the equation to solve for \(x\): \[ x + 8 - x = 2x - x \]Simplifying, we are left with: \[ 8 = x \].

Step 4: Verify Solution

Substitute \(x = 8\) back into the original function to verify it is not extraneous:\[ f(8) = \sqrt[4]{8+8} - \sqrt[4]{2 \times 8} = \sqrt[4]{16} - \sqrt[4]{16} = 2 - 2 = 0 \].Since substituting \(x = 8\) results in \(f(x) = 0\), the solution \(x = 8\) is valid and not extraneous.

Key Concepts

Radical Equations

Radical equations are equations where the variable is inside a radical, usually a square root, cube root, or any higher-order root. Solving these equations often involves isolating the radical expression and then eliminating the radical by raising both sides of the equation to the power that matches the root.

• Identify the radical part of the equation.
• Isolate the radical on one side of the equation.
• Raise both sides of the equation to a power that will remove the radical.

Once the radical is removed, you can solve the resulting equation like any other algebraic equation. It's essential to check for extraneous solutions after solving, since raising both sides of the equation to a power can introduce solutions that don't actually satisfy the original equation.

Extraneous Solutions

Extraneous solutions are apparent solutions generated when solving radical equations that do not satisfy the original equation. They often appear when both sides of an equation are raised to an even power. These solutions are not true because they arise from the process of solving the equation rather than from the equation itself.
Here’s how extraneous solutions can occur:

• When both sides of an equation are squared or raised to any even power.
• This operation can introduce solutions that work for the transformed equation but not for the original one.
• Always substitute your solutions back into the original equation to determine their validity.

Extraneous solutions are why it's crucial to verify each potential solution by plugging it back into the original equation to see if it results in a true statement.

Function Verification

Function verification is the process of confirming that a proposed solution to an equation or function actually works. After solving, substitute the potential solution back into the original equation to ensure it satisfies the equation fully.
Steps for verifying solutions include:

• Take the found solution and substitute it back into the original function or equation.
• Simplify the expression to check if it equals zero or the expected result.
• If it does, the solution is valid; otherwise, it's extraneous.

Verification is crucial in radical equations due to the possibility of extraneous solutions. By verifying, you ensure that the solutions are genuine and applicable to the original problem.