Problem 86 - Elementary Algebra

Question

Solve by completing the square. \(w^{2}=5 w-1\)

Step-by-Step Solution

Answer

The solutions are \( w = \frac{5 + \sqrt{21}}{2} \) and \( w = \frac{5 - \sqrt{21}}{2} \).

Step 1: Move Constant to the Other Side

Start by moving the constant term from the right-hand side to the left-hand side to set up the equation for completing the square. Subtract 5w from both sides: \[ w^2 - 5w = -1 \]

Step 2: Find the Square Term

To complete the square, take half of the coefficient of the linear term (–5), square it, and add that value to both sides of the equation. Half of –5 is –5/2, and \((-5/2)^2 = 25/4\). Add 25/4 to both sides:\[ w^2 - 5w + \frac{25}{4} = -1 + \frac{25}{4} \]

Step 3: Simplify the Equation

Convert -1 to a fraction with a denominator of 4 and add it to 25/4: \[ -1 = -\frac{4}{4} \] So,\[ -\frac{4}{4} + \frac{25}{4} = \frac{21}{4} \] Therefore, the equation becomes:\[ w^2 - 5w + \frac{25}{4} = \frac{21}{4} \]

Step 4: Write the Perfect Square

Rewrite the left-hand side as a perfect square:\[ (w - \frac{5}{2})^2 = \frac{21}{4} \]

Step 5: Solve for w

Take the square root of both sides to solve for w. Remember to consider both the positive and negative roots:\[ w - \frac{5}{2} = \pm \sqrt{\frac{21}{4}} \] \[ w - \frac{5}{2} = \pm \frac{\sqrt{21}}{2} \]Add \(\frac{5}{2}\) to both sides to isolate w:\[ w = \frac{5}{2} \pm \frac{\sqrt{21}}{2} \]

Step 6: Final Solutions

Combine the fractions to write the final solutions:\[ w = \frac{5 + \sqrt{21}}{2} \quad \text{or} \quad w = \frac{5 - \sqrt{21}}{2} \]

Key Concepts

solving quadratic equations

Solving quadratic equations is a critical skill in algebra. One way to solve these equations is by completing the square. This method involves rewriting the equation in a way that forms a perfect square trinomial on one side. In the given exercise, the original equation is:\[ w^{2} = 5w - 1 \]The goal is to transform this equation to make it easy to solve. Completing the square simplifies the process, helping isolate the variable. In this approach, we perform several tasks: moving terms, finding specific values, and simplifying the equation.

algebraic manipulation

Algebraic manipulation is all about rearranging equations to make them easier to solve. In this problem, the first step is to move the constant term to the other side:\[ w^2 - 5w = -1 \]Next, we add and subtract the same value on both sides to form a perfect square trinomial. This doesn't change the equation's equality but transforms its appearance. The manipulation continues until the equation looks like:\[ (w - \frac{5}{2})^2 = \frac{21}{4} \]These steps involve adding precise values to both sides of the equation and simplifying fractions, both fundamental algebraic skills.

perfect square trinomial

Creating a perfect square trinomial is a vital part of completing the square. A perfect square trinomial has the form:\[ (ax + b)^2 \]To transform the equation, we take half of the linear term's coefficient, square it, and add it to both sides. Here, the linear term coefficient is -5. Half of -5 is -5/2, and squaring it gives:\[ \bigg(\frac{-5}{2}\bigg)^2 = \frac{25}{4} \]Adding this to both sides, the equation becomes:\[ w^2 - 5w + \frac{25}{4} = \frac{21}{4} \]We now have a perfect square trinomial on the left side, simplifying our equation.

square root property

The square root property is another essential concept. Once the equation is in the form of a square trinomial, we can apply the property, which states:\[ (w - \frac{5}{2})^2 = \frac{21}{4} \]Taking the square root of both sides gives:\[ w - \frac{5}{2} = \frac{\text{±} \text{sqrt}(21)}{2} \]This step involves calculating both the positive and negative roots. Adding \(\frac{5}{2}\) to both sides helps isolate the variable w. Thus, we get two solutions:\[ w = \frac{5 + \text{sqrt}(21)}{2} \text{ or } w = \frac{5 - \text{sqrt}(21)}{2} \]These steps show how the square root property helps simplify and solve the equation completely.