Problem 27 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ 2 x(x+2)=(x-1)(x+2) $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -2\).

Step 1: Expand Both Sides

Expand both sides of the equation: Left Side: \(2x(x + 2) = 2x^2 + 4x\) Right Side: \((x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2\)

Step 2: Set the Equation to Zero

Set the expanded equation to zero by moving all terms to one side:\(2x^2 + 4x = x^2 + x - 2\)Subtract \(x^2 + x - 2\) from both sides:\(2x^2 + 4x - x^2 - x + 2 = 0\)

Step 3: Simplify the Equation

Combine like terms:\(2x^2 - x^2 + 4x - x + 2 = 0\)This simplifies to:\(x^2 + 3x + 2 = 0\)

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \(x^2 + 3x + 2 = 0\):Find two numbers that multiply to 2 and add up to 3. These numbers are 1 and 2.Thus, \((x + 1)(x + 2) = 0\)

Step 5: Solve for x

Set each factor equal to zero and solve for \(x\):\(x + 1 = 0\) gives \(x = -1\)\(x + 2 = 0\) gives \(x = -2\)

Step 6: Check Solutions

Verify each solution by substituting back into the original equation:For \(x = -1\):Left Side: \(2(-1)((-1) + 2) = 2(1) = 2\)Right Side: \((-1 - 1)((-1) + 2) = (-2)(1) = -2\)Since Left Side ≠ Right Side, \(x = -1\) is not a valid solution.For \(x = -2\):Left Side: \(2(-2)((-2) + 2) = 2(0) = 0\)Right Side: \((-2 - 1)((-2) + 2) = (-3)(0) = 0\)Since Left Side = Right Side, \(x = -2\) is a valid solution.

Key Concepts

Factoring Quadratics

Factoring quadratics is a key method for solving quadratic equations. Quadratic equations typically have the form \(ax^2 + bx + c = 0\). The goal is to express the quadratic as a product of two binomials. By doing so, we can find the equation's solutions, or roots, by setting each binomial to zero. In our specific equation, \(x^2 + 3x + 2 = 0\), we need to find two numbers that multiply to the constant term, which is 2, and add to the linear coefficient, which is 3.

• The numbers are 1 and 2, since \(1 \times 2 = 2\) and \(1 + 2 = 3\).
• This allows us to write the equation as \((x + 1)(x + 2) = 0\).

By setting each factor equal to zero, \(x + 1 = 0\) and \(x + 2 = 0\), we solve for \(x\) to find the potential solutions. This step is crucial in determining the values of \(x\) that satisfy the equation.

Equation Expansion

Equation expansion is the process of removing parentheses by distributing factors across terms inside the parentheses. This helps simplify the equation and prepare it for further solving steps. Starting with the original equation, \(2x(x+2) = (x-1)(x+2)\), we'll expand both sides:

• The left side becomes \(2x(x+2) = 2x^2 + 4x\), by multiplying \(2x\) with both \(x\) and \(2\).
• The right side needs more steps: \((x-1)(x+2)\) is expanded by using the distributive property.
• First, multiply \(x\) by \(x+2\), giving \(x^2 + 2x\).
• Then, multiply \(-1\) by \(x+2\), giving \(-x - 2\).
• Combine these results to get \(x^2 + x - 2\).

With both sides expanded, the next step is to set all terms to one side of the equation (usually left side), leading us to simplify and factor to solve.

Solution Verification

Solution verification involves checking whether the obtained solutions satisfy the original equation. This step ensures that any mistakes in calculations are caught and corrected. After solving the factored equation \((x + 1)(x + 2) = 0\), we found potential solutions \(x = -1\) and \(x = -2\). To verify:

• For \(x = -1\): Substitute back into the original equation.
  Left side: \(2(-1)((-1) + 2) = 2 \times 1 = 2\)
  Right side: \((-1 - 1)((-1) + 2) = (-2) \times 1 = -2\).
  The sides are not equal, so \(x = -1\) is not valid.
• For \(x = -2\): Substitute to verify.
  Left side: \(2(-2)((-2) + 2) = 2 \times 0 = 0\)
  Right side: \((-2 - 1)((-2) + 2) = (-3) \times 0 = 0\).
  The sides are equal, confirming \(x = -2\) as a valid solution.

Verification confirms solution accuracy, helping ensure mathematical reliability in results.