Problem 9 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by using the square root property. $$(x-2)^{2}=25$$

Step-by-Step Solution

Answer

The solutions are x = 7 and x = -3.

Step 1: Recognize the Equation in a Suitable Form

Identify that the equation \((x-2)^2 = 25\) is already in the form \((x-a)^2 = k\), which is suitable for the square root property.

Step 2: Apply the Square Root Property

Use the square root property: \((x-a)^2 = k\) implies \(x-a = \pm \sqrt{k}\). Applying this gives \(x-2 = \pm \sqrt{25}\).

Step 3: Calculate the Square Roots

Compute \(\sqrt{25}\), which is 5. Therefore, \(x-2 = \pm 5\).

Step 4: Solve for x

Write the two possible equations from \(x-2 = \pm 5\):1. \(x-2 = 5\)2. \(x-2 = -5\)Solve each separately. For \(x-2 = 5\), add 2 to both sides to get \(x = 7\). For \(x-2 = -5\), add 2 to both sides to get \(x = -3\).

Step 5: State the Solutions

The solutions to the equation are \(x = 7\) and \(x = -3\).

Key Concepts

Square Root Property

The square root property is a handy tool when working with quadratic equations, especially those that are already shaped like \((x-a)^2 = k\). This property allows us to remove the square by taking the square root of both sides of the equation. It's highly effective in scenarios where our equation can fit this specific pattern.

• The key is to recognize that if you have an equation like \((x-a)^2 = k\), you can rewrite it as \(x-a = \pm \sqrt{k}\). This step is crucial, as it directly leads us to the solutions of the given equation.
• Applying the square root property greatly simplifies the solving process because it bypasses more complex methods such as factoring, making it a very efficient strategy.

In our example, the equation was given as \((x-2)^2 = 25\). Recognizing that it fits the square root property allows us to split it into two possible equations: \(x-2 = 5\) and \(x-2 = -5\). This transforms the problem into one that just requires basic algebraic manipulation to solve.

Solving Equations

Solving equations is one of the fundamental skills learned in algebra. It's about finding the value of the variable that makes the equation true. When you apply the square root property, solving the equation becomes straightforward.
Once the equation is set in the form \((x-a)^2 = k\), using the square root property reduces it to two linear equations: \(x-a = \sqrt{k}\) or \(x-a = -\sqrt{k}\).

• The next step is to solve these linear equations. Since they are simpler in form, they require only basic operations like addition or subtraction to isolate \(x\).
• From \(x-2 = \pm 5\), we determine: for \(x-2 = 5\), add 2 to get \(x = 7\); and for \(x-2 = -5\), add 2 to get \(x = -3\).

Thus, solving through the square root method can break down what initially seems like a complex problem into manageable steps, providing clear solutions.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. It provides a foundation for understanding more complex mathematical concepts. Key elements include solving for unknown values, like we did with the quadratic equation.
In the context of this problem, algebra involves operations such as:

• Recognizing patterns like \((x-a)^2\), which allows for effective use of the square root property.
• Executing simple arithmetic to shift terms between sides of an equation to solve for the unknown variable \(x\).

Algebraic techniques make it possible to simplify and solve equations that underpin various fields such as science and engineering. Understanding how to manipulate equations using fundamental principles, like those illustrated here, is critical in further mathematical learning and problem-solving.