Problem 34 - Algebra for College Students

Question

Solve each equation. Don't forget to check each of your potential solutions. \(\sqrt{x^{2}+2 x+1}=x+3\)

Step-by-Step Solution

Answer

The solution is \(x = -2\).

Step 1: Understand the Equation

The given equation is \( \sqrt{x^2 + 2x + 1} = x + 3 \). This involves a square root on the left side and a linear expression on the right side.

Step 2: Square Both Sides

In order to eliminate the square root, square both sides of the equation. The equation becomes: \[ (\sqrt{x^2 + 2x + 1})^2 = (x + 3)^2 \] which simplifies to: \[ x^2 + 2x + 1 = (x + 3)^2 \]

Step 3: Expand the Right Side

Expand the expression \((x + 3)^2\) on the right side of the equation: \[ x^2 + 2x + 1 = x^2 + 6x + 9 \]

Step 4: Simplify the Equation

Subtract \(x^2\) from both sides of the equation to get: \[ 2x + 1 = 6x + 9 \] Now, move all terms involving \(x\) to one side and constants to the other side by subtracting \(2x\) and \(9\) from both sides: \[ 1 - 9 = 6x - 2x \] Thus, \[ -8 = 4x \]

Step 5: Solve for x

To isolate \(x\), divide both sides by 4: \[ x = \frac{-8}{4} \] This simplifies to: \[ x = -2 \]

Step 6: Verify the Solution

Substitute \(x = -2\) back into the original equation to verify it satisfies the equation: Left Side: \(\sqrt{(-2)^2 + 2(-2) + 1} = \sqrt{4 - 4 + 1} = \sqrt{1} = 1\) Right Side: \(-2 + 3 = 1\) Both sides are equal, so the solution \(x = -2\) is correct.

Key Concepts

Solving Equations

Solving equations is a fundamental skill in algebra, which involves finding the value of the variable that makes the equation true. In this exercise, we encountered a square root equation, which includes a square root expression on one side. The key to solving such equations is to isolate the square root and eliminate it by squaring both sides of the equation. This transforms the equation into a more familiar form.
The original equation given was \( \sqrt{x^2 + 2x + 1} = x + 3 \). By squaring both sides:

• The left side becomes \( x^2 + 2x + 1 \) because squaring a square root cancels the root itself.
• The right side becomes \((x + 3)^2\), a quadratic expression which can be expanded.

Once the equation is squared, we expand and simplify any terms to isolate the variable on one side.By rearranging and solving the linear equation that forms, we find \( x = -2 \) as the solution.Revisiting the steps without variables helps to understand better: squaring cancels roots, expansion involves multiplying, and isolating x involves basic arithmetic operations.

Verification of Solutions

Once a solution is found, verifying it is essential to ensure its accuracy. Verification involves substituting the solution back into the original equation. This step checks if both sides of the equation hold true with the proposed value.
For the solution \( x = -2 \) obtained from the equation:

• Substitute \( x = -2 \) into the left-hand side as \( \sqrt{(-2)^2 + 2(-2) + 1} = \sqrt{4 - 4 + 1} = \sqrt{1} = 1 \).
• Substitute \( x = -2 \) into the right-hand side it becomes \( -2 + 3 = 1 \).

Since both the left and right sides simplify to the same value, the solution \( x = -2 \) is confirmed to be correct. Verification helps to identify any potential issues such as extraneous solutions, especially in equations involving square roots where squaring can introduce invalid results.

Algebraic Manipulation

Algebraic manipulation involves altering equations to make the problem easier to solve. It typically includes operations like expansion, factoring, and rearranging terms. In solving square root equations, these techniques become crucial.
In this example:

• We expanded \((x + 3)^2\) into \(x^2 + 6x + 9\) to simplify the equation.
• Then, we used subtraction to move terms \(x^2\) and constants strategically, making it easier to isolate the variable.

By recognizing patterns, like the difference of squares, or by using standard formulas, algebraic manipulation streamlines solving processes.
For similar problems, remember the goal is to transform the equation into a simpler linear form, where solving for \(x\) becomes straightforward. Algebraic techniques help break down complex expressions, aiding us to find accurate solutions efficiently.