Problem 58 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{x+12}+\sqrt{x}=6 $$

Step-by-Step Solution

Answer

The solution is \( x = 4 \).

Step 1: Isolate one square root

First, we'll isolate one of the square root terms. Subtract \( \sqrt{x} \) from both sides of the equation, so that it becomes: \[ \sqrt{x+12} = 6 - \sqrt{x} \]

Step 2: Square both sides of the equation

To eliminate the square root on the left side, square both sides of the equation: \[ (sqrt{x+12})^2 = (6 - \sqrt{x})^2 \] This simplifies to: \[ x+12 = 36 - 12\sqrt{x} + x \]

Step 3: Simplify the equation

Simplify the resulting equation: \[ x + 12 = 36 + x - 12\sqrt{x} \] By subtracting \(x\) from both sides, the equation becomes: \[ 12 = 36 - 12\sqrt{x} \] Subtract 36 from both sides to further simplify: \[ -24 = -12\sqrt{x} \]

Step 4: Solve for \( \sqrt{x} \)

Divide both sides by -12 to solve for \( \sqrt{x} \): \[ \sqrt{x} = 2 \]

Step 5: Find \( x \) by squaring

Square the result from Step 4 to find \( x \): \[ x = 2^2 = 4 \]

Step 6: Check for extraneous solutions

Substitute \( x = 4 \) back into the original equation to verify that it's a valid solution: \[ \sqrt{4+12} + \sqrt{4} = \sqrt{16} + 2 = 4 + 2 = 6 \] Since the original equation is satisfied, \( x = 4 \) is not extraneous.

Key Concepts

Extraneous Solutions

Extraneous solutions are potential solutions that arise during the process of solving an equation but do not actually satisfy the original equation when substituted back into it. They often occur as a result of algebraic manipulations, such as squaring both sides of an equation, which can introduce solutions that aren't valid.

• For example, when solving radical equations, squaring the terms can produce extra solutions, which need to be checked against the original equation.
• To avoid completely relying on the steps of manipulation, verification of proposed solutions is crucial. This helps ensure that the found solutions are truly valid for the initial problem.

Always take extra steps at the end to substitute these solutions back into the original equation, confirming they hold true.

Isolating Square Roots

Isolating square roots is a critical step when working with radical equations. To solve the equation, you need to separate one of the square root terms to simplify your work.

• Begin by arranging your equation such that one of the square roots stands alone. This often involves moving other terms to the other side of the equation.
• This step simplifies the process of squaring, which follows isolation, allowing easier elimination of the radical term.
• In the example equation, \(\sqrt{x+12} = 6 - \sqrt{x}\) is achieved by subtracting \(\sqrt{x}\) from both sides.

Mastering this step sets up the equation for successful solving later on. It simplifies further manipulation and ensures a path to an accurate solution.

Squaring Equation

After isolating one of the square roots, squaring the equation is the next crucial move in handling radical equations. This powerful algebraic tool helps in removing radical signs by raising both sides to the power of 2.

• This operation transforms the equation, often bringing it to a simpler polynomial form easier to handle.
• In the given exercise, squaring led from \(\sqrt{x+12} = 6 - \sqrt{x}\) to \(x+12 = 36 - 12\sqrt{x} + x\).
• Remember, squaring can also add extraneous roots, which makes the verification phase important in final steps.

Use squaring to clear out root terms effectively, bearing in mind it might complicate equations temporarily and always check outcomes against original conditions.

Algebraic Verification

Algebraic verification is the final and essential step to ensure that the solutions proposed indeed satisfy the original equation. This comes after isolating square roots and squaring equations.

• Substitute the solution back into the original equation to verify its correctness.
• For instance, substituting \(x = 4\) back into \(\sqrt{x+12} + \sqrt{x} = 6\) should satisfy the equation: \(\sqrt{16} + 2 = 4 + 2 = 6\).
• Only validated values should be considered actual solutions, while incorrect ones are dismissed as extraneous.

Verification not only confirms correctness but also provides confidence in the solution process, ensuring that no false answers are carried forward.