Problem 16 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$(R+9)(R+1)=13$$

Step-by-Step Solution

Answer

The solutions are \(R = -5 + \sqrt{29}\) and \(R = -5 - \sqrt{29}\).

Step 1: Expand the Equation

First, expand the expression \((R+9)(R+1)\) using the distributive property. This gives us the equation:\[ R^2 + R + 9R + 9 = 13 \] Simplifying this, we get:\[ R^2 + 10R + 9 = 13 \]

Step 2: Move Constant Term

Rearrange the equation by moving the constant term on the right side to the left side. This results in:\[ R^2 + 10R + 9 - 13 = 0 \] Simplifying gives:\[ R^2 + 10R - 4 = 0 \]

Step 3: Complete the Square

To complete the square, focus on the quadratic and linear terms. Take the coefficient of \(R\), which is 10, divide by 2, and square it:\[ \left(\frac{10}{2}\right)^2 = 25 \]Add and subtract 25 from the left side to complete the square:\[ R^2 + 10R + 25 - 25 - 4 = 0 \]Simplify to:\[ (R+5)^2 - 29 = 0 \]

Step 4: Solve the Completed Square

Add 29 to both sides of the equation:\[ (R+5)^2 = 29 \]Take the square root of both sides:\[ R+5 = \pm\sqrt{29} \]

Step 5: Solve for R

Isolate \(R\) by subtracting 5 from both expressions to find the possible values of \(R\):\[ R = -5 + \sqrt{29} \] or \[ R = -5 - \sqrt{29} \]

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations by transforming them into a perfect square trinomial. This approach often simplifies the process of finding solutions compared to directly using the quadratic formula. Here's how it works in the context of the exercise.- **Start with the quadratic equation:** In the given problem, after distribution and moving terms around, we arrive at: \[ R^2 + 10R - 4 = 0 \] - **Identify the middle coefficient:** Take the coefficient of the linear term (\(R\)) which is 10. This number plays a crucial role in completing the square.- **Divide and square it:** Divide this coefficient by 2 and then square it. That is, \[ \left(\frac{10}{2}\right)^2 = 25 \] - **Form a perfect square trinomial:** Add and subtract this value (25) to ensure the balance of the equation, leading to \[ R^2 + 10R + 25 - 25 - 4 = 0 \]- **Simplify to a square:** Rearrange into \[ (R+5)^2 - 29 = 0 \]This makes the equation easier to solve by taking its square root.

Factoring Quadratics

Factoring is another method to solve quadratic equations, often through expressing the equation as a product of binomials. This strategy is helpful for checking solutions obtained by other means, such as completing the square.Here's a step-by-step approach to factoring:- **Simple quadratics:** Initially for equations like \[ (R+9)(R+1)=13 \] expanding gives \[ R^2 +10R + 9 =13 \], which directly cannot be factored easily due to the constant term mismatch.- **Move terms to one side:** To attempt factoring, move all terms to one side of the equation: \[ R^2 + 10R - 4 = 0 \]- **Assess for factors:** An equation in the standard form \[ ax^2 + bx + c=0 \] doesn't directly factor into integer roots here, hence requiring completing the square or factoring by guessing and checking or using quadratic formula may sometimes be more applicable.Remember, factoring is most straightforward when the quadratic has whole number roots.

Distributive Property

The distributive property allows simplification of expressions where a single term is multiplied by terms in parentheses. In algebra, it states that: \[ a(b + c) = ab + ac \]In this particular problem, the initial step requires expanding the terms: - **Apply distributive law:** The expression \[ (R+9)(R+1) \] when expanded becomes \[ R \cdot (R+1) + 9 \cdot (R+1) \] Each term inside the parentheses is multiplied individually: \[ R^2 + R + 9R + 9 \]- **Combine like terms:** You then simplify by combining similar terms: \[ R^2 + 10R + 9 \]Understanding and applying the distributive property is critical as it sets the stage for rewriting complex expressions into simpler forms, aiding in solving equations, especially in equation work.

Solving for Variables

Solving for variables is the ultimate goal when working with equations. It involves isolating the variable to determine its value, tailored for quadratic equations in this problem.- **After completing the Square:** Once you achieve \[ (R+5)^2 - 29 = 0 \]- **Isolate squared term:** Move the constant to the opposite side: \[ (R+5)^2 = 29 \]- **Radical expression:** Take the square root of both sides to further simplify: \[ R+5 = \pm \sqrt{29} \] Here, the \(\pm\) symbol indicates two possible values.- **Final isolation:** Subtract 5 from each side to solve for \(R\): \[ R = -5 + \sqrt{29} \] or \[ R = -5 - \sqrt{29} \]This last step provides the solutions needed. Solving for variables is crucial as it translates mathematical expressions into results.