Problem 70 - Elementary Algebra

Question

Solve by completing the square. \(w^{2}+8 w=65\)

Step-by-Step Solution

Answer

The solutions are \( w = 5 \) and \( w = -13 \).

Step 1: Move the constant term to the other side

To complete the square, first move the constant term (65) to the other side of the equation. This gives us: \[ w^2 + 8w = 65 \] becomes: \[ w^2 + 8w - 65 = 0 \]

Step 2: Set the equation equal to zero

Since we want to solve for \( w \), we should set the equation equal to zero: \[ w^2 + 8w - 65 = 0 \]

Step 3: Isolate the variable terms

Rearrange the equation so the variable terms are on one side and the constant term is on the other side: \[ w^2 + 8w = 65 \]

Step 4: Complete the square

To complete the square, add and subtract \( (\frac{b}{2})^2 \). In this case, \( b = 8 \). Therefore, add and subtract \( (\frac{8}{2})^2 = 16 \) to the left side: \[ w^2 + 8w + 16 - 16 = 65 \] which simplifies to: \[ (w + 4)^2 - 16 = 65 \]

Step 5: Solve the resulting equation

Add 16 to both sides to isolate the perfect square trinomial: \[ (w + 4)^2 = 81 \] Now take the square root of both sides: \[ w + 4 = \pm 9 \]

Step 6: Solve for the variable w

Solve the two equations resulting from taking the square root: \[ w + 4 = 9 \] and \[ w + 4 = -9 \] Now solve each equation: \[ w = 5 \] and \[ w = -13 \]

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is squared, as in the general form: \[ ax^2 + bx + c = 0 \] In these equations,

• \(a\) represents the coefficient of the squared term;
• \(b\) is the coefficient of the linear term;
• \(c\) is the constant term.

A quadratic equation can have two solutions, found through various methods, one of which is completing the square. Understanding how to manipulate these elements is key to solving quadratic equations.

Algebraic Techniques

Algebraic techniques involve using algebraic rules and properties to manipulate and simplify equations. When solving quadratic equations by completing the square, several algebraic techniques are used:

• Rearranging terms: Move terms around to isolate variable terms.
• Adding and subtracting terms: Add and subtract the same value to both sides of an equation to maintain equality.
• Factoring: Recognize and create perfect square trinomials.

For example, in the given equation \(w^2 + 8w = 65\), we need to move the constant to the left and perform arithmetic manipulations to transform the equation into a perfect square trinomial. These steps include:

• Adding and subtracting \(16\).
• Rewriting the equation in squared form \((w + 4)^2\).

Using these techniques systematically leads to solving the equation efficiently.

Solving Equations

Solving equations is about finding the values of the variable that make the equation true. Completing the square is one technique for solving quadratic equations. Here are the steps applied to our original equation:

1. **Rearrange the equation**: Move the constant term to the other side.
\[ w^2 + 8w = 65 \]2. **Complete the square**: Add and subtract \(\left( \dfrac{b}{2} \right)^2\) on the left.
\[ w^2 + 8w + 16 - 16 = 65 \]3. **Form a perfect square**: Rewrite the left side as a square of a binomial.\[ (w + 4)^2 - 16 = 65 \]4. **Isolate the perfect square**: Add 16 to both sides to isolate the squared term.\[ (w + 4)^2 = 81 \]5. **Solve for the variable**: Take the square root of both sides.\[ w + 4 = \pm 9 \]Solve the resulting linear equations:

• \(w + 4 = 9\) implies \(w = 5\)
• \(w + 4 = -9\) implies \(w = -13\)

Through this method, we find that the solutions to the equation are \(w = 5\) and \(w = -13\). This is a systematic way to ensure the correct values are determined effectively.