Problem 28 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the equation by extracting square roots. List both the exact solutions and the decimal solutions rounded to the nearest hundredth. $$(x-5)^{2}=25$$

Step-by-Step Solution

Answer

The exact solutions are \( x = 10 \) and \( x = 0 \). In decimal form, these are still 10.00 and 0.00 respectively.

Step 1: Isolate the Square Term

In the given equation, \( (x-5)^{2}=25 \), the square term is already isolated on one side of the equation.

Step 2: Take Square Roots on Both Sides

Taking square root on both sides gives \( x-5 = ± \sqrt{25} \). So, \( x-5 = ± 5 \).

Step 3: Solve for 'x'

Now, solve for 'x'. First, consider the positive root, which gives \( x = 5 + 5 = 10 \). Then consider the negative root, which gives \( x = 5 - 5 = 0 \).

Key Concepts

Extracting Square Roots

Extracting square roots is a method used to solve simple quadratic equations, especially when they are in perfect square form like \( (x-a)^2 = b \). This makes it easier to find the value of the variable by taking the square root of both sides of the equation.
Here’s how it works:

• Ensure the quadratic expression is isolated on one side of the equation. For example, \( (x-5)^2 = 25 \) shows the perfect square already isolated.
• Take the square root of both sides. Remember that taking a square root can result in both positive and negative roots. Therefore, \( \sqrt{25} \) yields \( ±5 \), so the equation transforms to \( x - 5 = ±5 \).
• Solve the resulting simpler equations to find the possible values for the variable.

This method of "undoing" the square enables a direct path to the possible solutions, making it a straightforward method for specific equations.

Exact Solutions

When solving a quadratic equation like \( (x-5)^2 = 25 \) by extracting square roots, you get exact solutions. This means that the solutions are in their simplest, precise forms without approximations.
Continuing with our example:

• After taking the square roots, \( x - 5 = ±5 \), there are two separate equations: \( x-5 = 5 \) and \( x-5 = -5 \).
• Solve both these equations to find the exact solutions:
  \( x = 5 + 5 = 10 \)
  \( x = 5 - 5 = 0 \).

The term "exact solutions" refers to these integer values, which are accurate and precise based on the operations performed. Several equations may naturally yield such neat, whole number results, making them very understandable and relatable.

Decimal Solutions

Decimal solutions are useful when you need to express solutions in approximate or non-integer form, especially when the exact solutions involve irrational numbers. However, in the case of \( (x-5)^2 = 25 \), the equation yields exact integer solutions (0 and 10), so it's not necessary to convert these into decimals.
Still, understanding the concept is important for other scenarios. Imagine you needed to provide an answer in a decimal setting:

• Exact results can be converted to decimals, especially when dealing with non-perfect squares, you may end up with solutions like \( x = ±\sqrt{2} \).
• In such cases, using a calculator, you might round the results, like \( \sqrt{2} \approx 1.41 \), to a specified degree of accuracy, such as to the nearest hundredth.

Being familiar with converting and understanding decimal solutions helps in many practical situations, especially within science and engineering, where precision is often crucial.

Quadratic Equations

Quadratic equations take the form \( ax^2 + bx + c = 0 \) and are a fundamental part of algebra. They can describe parabolic relationships and often appear in various real-life applications, such as projectile motion or optimization problems.
These equations can be solved using various methods, including:

• Factoring: When the equation can be expressed as a product of binomials.
• Completing the Square: A way to transform the equation into a perfect square form.
• Quadratic Formula: A universal method \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) being useful for any quadratic equation.
• Extracting Square Roots: As demonstrated in our original exercise, useful when the quadratic can be succinctly expressed in a squared form.

Understanding these different approaches allows flexibility in tackling quadratic equations, optimizing the approach for particular equations, ensuring the most efficient and convenient solutions.