Problem 71 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ 2 \sqrt{x}=\sqrt{5 x-16} $$

Step-by-Step Solution

Answer

The proposed solution is \(x = 16\), and it is not extraneous.

Step 1: Square Both Sides

Start by squaring both sides of the equation to eliminate the square roots:\[ (2\sqrt{x})^2 = (\sqrt{5x-16})^2 \] This simplifies to:\[ 4x = 5x - 16 \]

Step 2: Rearrange the Equation

Subtract \(5x\) from both sides to get all the terms involving \(x\) on one side:\[ 4x - 5x = -16 \]Simplify:\[ -x = -16 \]

Step 3: Solve for x

To solve for \(x\), multiply both sides by -1:\[ x = 16 \]

Step 4: Check for Extraneous Solutions

Plug \(x = 16\) back into the original equation to ensure it's a valid solution:\[ 2\sqrt{16} = \sqrt{5(16) - 16} \]Calculate both sides:\[ 2 \times 4 = \sqrt{80 - 16} \]\[ 8 = \sqrt{64} \]\[ 8 = 8 \]Since both sides are equal, \(x = 16\) is not an extraneous solution.

Key Concepts

Understanding Square Roots

Square roots are about finding a number that, when multiplied by itself, gives you the original number. For instance, the square root of 16 is 4, because \(4 \times 4 = 16\). The notation for square roots is the radical sign \(\sqrt{}\). This concept is important when solving equations involving square roots, as it allows us to "undo" or remove the root by squaring both sides of the equation. In our problem, we have \(2\sqrt{x} = \sqrt{5x-16}\). To eliminate these roots, we squared both sides: \((2\sqrt{x})^2 = (\sqrt{5x-16})^2\). This gave us \(4x = 5x - 16\). By understanding square roots, we effectively removed the radical signs so that we could solve the equation like a regular algebraic expression. When dealing with square roots, always remember:

• Squaring both sides helps eliminate the root.
• Ensure both sides of the square root equation are positive, as square roots of negative numbers are not real.
• Double-check your solutions, as squaring can introduce extraneous solutions, which leads us to our next concept.

Recognizing Extraneous Solutions

Extraneous solutions are false positives. They seem like they solve the equation, but they don't meet all the conditions of the original problem. These solutions often arise when both sides of an equation are manipulated, especially when squaring.In our example, after solving \(4x = 5x - 16 \) and arriving at \(x = 16\), it's crucial to check if this solution holds true in the original equation. By plugging \(x = 16\) back into the original problem, and finding that both sides are indeed equal, we confirm it's not an extraneous solution. Here’s a quick guide on handling extraneous solutions:

• After finding a solution, substitute it back into the original equation.
• Verify that both sides of the equation are equal.
• If they aren't, then discard that solution as extraneous.
• Always consider the domain of the original equation when checking, as roots and square terms might have restricted values.

By consistently validating solutions, you can accurately identify and discard any that don't belong.

Mastering Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for unknowns. It’s about applying basic algebra rules like addition, subtraction, multiplication, and division to isolate the variable you're solving for.In our equation after squaring, we had \(4x = 5x - 16\). We rearranged it by subtracting \(5x\) from both sides, giving \(-x = -16\). Then, by multiplying both sides by \(-1\), we found \(x = 16\).Key tips for algebraic manipulation:

• Simplify each equation step-by-step by dealing one operation at a time.
• Put all terms involving the variable you are solving for on one side of the equation.
• Keep equations balanced by performing the same operation on both sides.
• Watch signs when multiplying or dividing by negative numbers.

Improving algebraic skills can transform complex problems into solvable puzzles, enhancing both understanding and problem-solving efficiency.