Problem 18 - Algebra 2 and Trigonometry

Question

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ \sqrt{20-2 x}=\sqrt{5 x-8} $$

Step-by-Step Solution

Answer

The solution set is \(\{4\}.\)

Step 1: Isolate the Square Root Expressions

Start with the given equation: \( \sqrt{20-2x} = \sqrt{5x-8} \). Notice both sides are square root expressions, so let's square both sides to eliminate the square roots.

Step 2: Square Both Sides

Square both sides of the equation to get rid of the square roots: \[(\sqrt{20-2x})^2 = (\sqrt{5x-8})^2 \]This simplifies to: \[20-2x = 5x-8.\]

Step 3: Rearrange the Equation

Rearrange the equation \(20 - 2x = 5x - 8\) to gather all terms with \(x\) on one side and constant terms on the other. Add \(2x\) to both sides and add \(8\) to both sides:\[20 + 8 = 5x + 2x.\] This simplifies to:\[28 = 7x.\]

Step 4: Solve for x

Divide both sides by \(7\) to find the value of \(x\):\[x = \frac{28}{7}.\] This simplifies to:\[x = 4.\]

Step 5: Check the Solution

Substitute \(x = 4\) back into the original equation to verify:\[\sqrt{20 - 2(4)} = \sqrt{5(4) - 8}\]\[\sqrt{20 - 8} = \sqrt{20 - 8}\]\[\sqrt{12} = \sqrt{12}\]Since both sides are equal, \(x = 4\) is a valid solution.

Step 6: Write the Solution Set

The solution set consists of the valid value of \(x\) that satisfies the original equation. Therefore, the solution set is \(\{4\}.\)

Key Concepts

Solving Square Root Equations

Square root equations are equations where the variable is inside a square root symbol. In the original problem, both sides of the equation have square roots:

• \(\sqrt{20 - 2x}\)
• \(\sqrt{5x - 8}\)

To solve these, you first need to isolate the square root expressions. Once that is done, you can square both sides. This is a common technique to eliminate square roots, allowing you to transform the equation into a polynomial equation. Just be careful: squaring can introduce extraneous solutions, so you'll need to verify your answer.

Understanding the Solution Set

The solution set in mathematics is a set of all values that satisfy the given equation. For a quadratic or square root equation, it involves the values of \(x\) that make the equation true when plugged back into the original expression.Once you solve for \(x\), like finding \(x = 4\) in the original problem, you need to verify that these values satisfy the initial equation completely. Only verified solutions are included in the solution set.In our example, after solving, we confirm that the solution set is \(\{4\}\), meaning \(x = 4\) is the only value that satisfies the equation.

Checking Solutions

Checking solutions becomes crucial, especially when dealing with square root and quadratic equations. After finding a potential solution by algebraic manipulation, you should substitute it back into the original equation to verify it.For the problem at hand, substituting \(x = 4\) back into the equation results in both sides equal, \(\sqrt{12} = \sqrt{12}\). This confirms that \(x = 4\) is indeed a valid solution.This verification step ensures that no extraneous solutions sneak in, which can often be the case in quadratic and radical equations due to operations like squaring.

Mastering Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate variables and solve for unknowns. In the problem given, after squaring both sides, the equation turned into:\[20 - 2x = 5x - 8\]From there, you gather like terms on either side by adding \(2x\) to both sides and adding \(8\) to both sides:\[28 = 7x\]This step-by-step rearrangement consistently breaks the problem down into simpler parts. Finally, divide both sides by \(7\) to solve for \(x\):\[x = \frac{28}{7} = 4\]Mastery in these manipulations comes with practice, enabling you to simplify complex problems effectively.