Problem 89 - Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt[4]{10 y+6}=2 \sqrt[4]{y} $$

Step-by-Step Solution

Answer

The solution is \(y = 1\). No solutions are extraneous.

Step 1: Eliminate Radicals

Raise each side of the equation to the power of 4 to eliminate the fourth roots.\[ (\sqrt[4]{10y+6})^4 = (2\sqrt[4]{y})^4 \]This simplifies to:\[ 10y + 6 = 16y \]

Step 2: Simplify the Equation

Rearrange the terms to isolate a single variable on one side of the equation.\[ 10y + 6 = 16y \]Subtract \(10y\) from both sides:\[ 6 = 6y \]

Step 3: Solve for y

Divide both sides of the equation by 6 to find the value of \(y\).\[ \frac{6}{6} = \frac{6y}{6} \]This gives:\[ y = 1 \]

Step 4: Check for Extraneous Solutions

Substitute \(y = 1\) back into the original equation to verify it satisfies the equation.Original equation:\[ \sqrt[4]{10(1)+6} = 2\sqrt[4]{1} \]Simplify:\[ \sqrt[4]{16} = 2 \times 1 \]\[ 2 = 2 \]Since the left-hand side equals the right-hand side, \(y = 1\) is not extraneous.

Key Concepts

Extraneous Solutions

When solving equations, especially those that involve roots or radicals, sometimes we obtain solutions that do not satisfy the original equation. These are referred to as extraneous solutions. They arise due to the process of squaring both sides or applying other even-powered operations that may introduce "false" solutions.
Here's how extraneous solutions can occur:

• When both sides of an equation are raised to an even power, this can potentially create solutions that weren't possible in the original form of the equation.
• It's crucial to substitute the found solutions back into the original equation to confirm they are accurate and not merely artifacts of the manipulation process.

In the provided problem, after solving the equation and finding a solution, we checked it by substituting it back into the original equation. This is an essential step to ensure that the solution is real and not extraneous.

Radical Equations

Radical equations are those that contain terms with roots, such as square roots or fourth roots. Solving these involves isolating the radical on one side and then eliminating it by raising both sides of the equation to the necessary power.
Steps to solve a basic radical equation include:

• Isolate the radical: Make sure the term with the radical is by itself on one side of the equation.
• Eliminate the radical: Raise both sides of the equation to a power that matches the root. For example, square both sides to eliminate a square root.
• Solve the resulting equation: Once the radical is removed, solve the equation like any other algebraic equation.
• Check your solutions: Substitute your solutions back into the original equation to ensure they are not extraneous.

In the example provided, the equation involved fourth roots, which were eliminated by raising both sides to the fourth power.

Fourth Roots

The fourth root of a number is a value that, when multiplied by itself four times, gives the original number. In algebra, taking the fourth root is often written as \( \sqrt[4]{x} \) or \( x^{1/4} \).
Key points about fourth roots include:

• The fourth root is similar to a square root, but it involves four instances of multiplication to achieve the original number, rather than two.
• Solving equations with fourth roots typically involves raising both sides to the fourth power to eliminate the root, making it easier to solve.
• It is important to note the nature of the numbers under the root. For non-negative numbers, raising both sides to the fourth power is straightforward; however, care needs to be taken with negative numbers as they can introduce complex solutions.

In the exercise, both sides of the equation were powers by raising to the fourth power, which eliminated the fourth roots and allowed us to find the solution efficiently.