Problem 60 - Algebra and Trigonometry

Question

The given equation involves a power of the variable. Find all real solutions of the equation. \(3(x-5)^{2}=15\)

Step-by-Step Solution

Answer

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

Step 1: Simplify the Equation

Start by simplifying the equation. Divide both sides of the equation by 3 to get \((x - 5)^2 = 5\).

Step 2: Eliminate the Exponent

To solve for \(x\), take the square root of both sides of the equation. This gives you \(x - 5 = \pm \sqrt{5}\).

Step 3: Solve for x

Add 5 to both sides of the equation to isolate \(x\), giving the solutions \(x = 5 + \sqrt{5}\) and \(x = 5 - \sqrt{5}\).

Step 4: Verify the Solutions

Substitute the solutions back into the original equation to ensure they satisfy the equation. Both \(3(5 + \sqrt{5} - 5)^2 = 15\) and \(3(5 - \sqrt{5} - 5)^2 = 15\) are true.

Key Concepts

Powers of Variables

In the context of the equation provided, powers of variables come into play when we see expressions like \((x - 5)^2\). Here, the variable \(x\) is part of a term that is squared. This means that the term is multiplied by itself. The power of the variable, in this case, is 2, indicating that we are dealing with a quadratic equation.
Quadratic equations often appear in the form \(ax^2 + bx + c = 0\), where the highest power of \(x\) is 2. This is what gives these equations their name, with "quad" referring to the square. In solving quadratic equations, isolating the term with the squared variable is a crucial first step. In our exercise, we started by simplifying \(3(x-5)^2=15\) to get \((x-5)^2 = 5\), making it easier to handle by focusing on the power of the variable directly.
Understanding powers allows us to apply appropriate mathematical operations, like taking square roots, to solve equations. Recognizing and managing powers is essential for simplifying and solving equations efficiently.

Real Solutions

Real solutions refer to the values of the variable that satisfy an equation in the real number system. Real numbers include all numbers that can be found on the number line, including both positive and negative numbers, zero, and all the fractions and decimal numbers in between.
When we seek real solutions for an equation like \((x-5)^2=5\), we are looking for all x-values that, when substituted back into the equation, make the equality true. In this example, simplifying and solving the equation gives us \(x = 5 + \sqrt{5}\) and \(x = 5 - \sqrt{5}\). Both of these solutions are real numbers, as square roots of positive numbers result in real numbers.
Confirming these solutions involves substituting them back into the original equation to verify that the left side equals the right side. This validation ensures that no mistakes were made during the simplification process, and it confirms that the solutions are indeed real and correct.

Square Roots

Square roots are involved when we eliminate the power of a variable to solve equations. The square root of a number \(n\), denoted as \(\sqrt{n}\), is a value that, when multiplied by itself, gives \(n\). For example, the square root of 9 is 3, because \(3 \times 3 = 9\).
In our exercise, to eliminate the exponent on \((x-5)^2\), we take the square root of both sides of the equation \((x-5)^2=5\). This operation gives us \(x-5 = \pm \sqrt{5}\). The "±" symbol indicates that both the positive and negative square roots are considered because both are needed to explore all real solutions.
It is important to remember that taking square roots is a pivotal step in solving quadratic equations. It helps simplify expressions and is key to finding possible real solutions. Always check each potential solution by substituting it back into the original equation to ensure accuracy.