Problem 16 - Elementary Algebra

Question

Solve the following quadratic equations. \((y-5)^{2}=36\)

Step-by-Step Solution

Answer

The solutions are \( y = 11 \) and \( y = -1 \).

Step 1: Identify the equation components

Recognize that the given equation \( (y-5)^2 = 36 \) is a quadratic equation in the form \( (y-a)^2 = b^2 \). In this case, \( a = 5 \) and \( b = 6 \) because \( 36 = 6^2 \).

Step 2: Apply the square root property

To solve the equation, apply the square root property. This property states that if \( x^2 = k \, then \ x = \pm \sqrt{k} \). Thus, \( (y-5)^2 = 36 \) results in \( y-5 = \pm 6 \).

Step 3: Solve for the positive case

Consider the positive case first: \( y - 5 = 6 \. Add 5 to both sides to solve for \ y \): \ y = 6 + 5 = 11 \.

Step 4: Solve for the negative case

Now consider the negative case: \( y - 5 = -6 \. Add 5 to both sides to solve for \ y \): \ y = -6 + 5 = -1 \.

Step 5: Write the final solutions

Summarize the solutions found from both cases: \( y = 11 \) and \( y = -1 \). hence, the solutions to the equation are \( y = 11 \) and \( y = -1 \).

Key Concepts

Square Root Property

The square root property is a crucial tool for solving quadratic equations. When you have an equation of the form \texttt{x^2 = k}, where \texttt{k} is a positive number, the square root property tells us that \texttt{x} can be both the positive and negative square root of \texttt{k}. This is expressed mathematically as \( x = \pm \sqrt{k} \). For example, if \( x^2 = 36 \), then \( x = 6 \) or \( x = -6 \). This property becomes very handy in solving quadratic equations that are already set up in a squared format.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2, which can be written in the standard form \( ax^2 + bx + c = 0 \). These types of equations often require special techniques to solve. One common technique is factoring, but another useful approach is the square root property, especially when the equation is already in a squared binomial form. In this exercise, we start with the equation \( (y-5)^2 = 36 \), recognizing it as a form of quadratic equation ready to be solved using the square root property.

Algebraic Solutions

Solving quadratic equations often involves multiple algebraic steps. Here’s a breakdown of the steps used in the original solution:

• Step 1: Identify the components of the equation. Recognize the given form \((y-5)^2 = 36 \) and identify the constants \texttt{a} and \texttt{b}.
• Step 2: Apply the square root property. We convert \((y-5)^2 = 36\) to \( y-5 = \pm 6 \).
• Step 3: Solve for each case. We handle the positive case by solving \( y - 5 = 6 \) which simplifies to \( y = 11 \). For the negative case, we solve \( y - 5 = -6 \) which simplifies to \( y = -1 \).
• Step 4: Compile the final solutions. Summarize the results from both cases yielding the solutions \( y = 11 \) and \( y = -1 \).

This step-by-step algebraic process ensures that all potential solutions to the quadratic equation are found.