Problem 72 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ 3 \sqrt{x}=\sqrt{3 x+54} $$

Step-by-Step Solution

Answer

The solution is \(x = 9\) and it is not extraneous.

Step 1: Isolate Square Roots

Start by examining the given equation: \[3 \sqrt{x} = \sqrt{3x + 54}\]Ensure that both square roots are isolated on either side of the equation.

Step 2: Square Both Sides

To eliminate the square roots, square both sides of the equation. This will give us:\[(3 \sqrt{x})^2 = (\sqrt{3x + 54})^2\]Simplifying both sides, we get:\[9x = 3x + 54\]

Step 3: Solve the Linear Equation

Subtract \(3x\) from both sides of the equation to isolate the variable term:\[9x - 3x = 54\]This simplifies to:\[6x = 54\]Next, divide both sides by 6 to find \(x\):\[x = \frac{54}{6} = 9\]

Step 4: Check for Extraneous Solutions

Substitute \(x = 9\) back into the original equation to verify it satisfies the equation:\[3 \sqrt{9} = \sqrt{3 \times 9 + 54}\]Calculate both sides:Left side: \(3 \times 3 = 9\)Right side: \(\sqrt{81} = 9\)Since both sides are equal, \(x = 9\) is a valid solution and not extraneous.

Key Concepts

Extraneous Solutions

Extraneous solutions are solutions that arise when solving equations but do not satisfy the initial equation. When we solve radical equations by squaring both sides, we might inadvertently introduce such solutions.

This happens because squaring an equation can sometimes increase the number of solutions. To counteract this, always substitute your solutions back into the original equation to verify their validity. This ensures each solution truly works and is not extraneous.

For example, in the given exercise, we found a solution for the equation after squaring and simplifying. However, by checking back, we confirmed that it was indeed a valid solution. Thus, being mindful of extraneous solutions helps in establishing accurate solutions.

Square Roots

Square roots are fundamental in understanding and solving radical equations. A square root, denoted as \( \sqrt{x} \), is a value that, when multiplied by itself, gives the number \( x \). Solving radical equations often involves isolating terms that contain square roots.

In the provided equation \( 3 \sqrt{x} = \sqrt{3x + 54} \), both terms involve square roots. To simplify, square both sides. This process removes square roots, leaving a polynomial equation that's easier to solve. It's crucial to handle the square roots properly to avoid errors and ensure correctness.

Remember, checking your solution in the context of square roots is vital as sometimes squaring can hide certain features of the original equation.

Linear Equations

Once the square roots are eliminated from the equation, the problem often simplifies to a linear equation. Linear equations involve terms where the variable is raised to the first power. They take the general form of \( ax + b = c \).

Solving a linear equation like \( 6x = 54 \) is straightforward. Isolate the variable by performing inverse operations, such as subtracting, adding, multiplying, or dividing both sides by the same number. Here, dividing both sides by 6 gives \( x = 9 \).

Linear equations are simpler and don't introduce extraneous solutions, unlike squaring equations with radicals. Therefore, once simplified to a linear equation, the solution process is much more direct and reliable.