Problem 56 - Algebra and Trigonometry

Question

\(47-70\) 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: Isolate the Squared Expression

To solve the equation, start by isolating the squared expression on one side. Given the equation \(3(x-5)^2 = 15\), divide both sides by 3 to simplify:\[(x-5)^2 = \frac{15}{3}\]which simplifies to:\[(x-5)^2 = 5\]

Step 2: Solve the Squared Equation

Take the square root of both sides to solve for \(x-5\). Remember, taking a square root introduces both a positive and negative solution:\[x-5 = \pm \sqrt{5}\]This provides two separate equations to solve:1. \(x-5 = \sqrt{5}\)2. \(x-5 = -\sqrt{5}\)

Step 3: Solve for x

For each equation from step 2, solve for \(x\):**Equation 1:**\[x - 5 = \sqrt{5}\]Add 5 to both sides:\[x = \sqrt{5} + 5\]**Equation 2:**\[x - 5 = -\sqrt{5}\]Add 5 to both sides:\[x = -\sqrt{5} + 5\]

Step 4: Simplify the Solutions

Finally, express the solutions in the simplest form:The solutions are:1. \(x = 5 + \sqrt{5}\)2. \(x = 5 - \sqrt{5}\)

Key Concepts

Isolating the squared term

The first step in solving equations with a squared term is to isolate this part of the equation. This means getting the squared term by itself on one side. It's similar to cleaning up your workspace before starting a project. In our example, the equation is given as:
\(3(x-5)^2 = 15\).
To isolate \((x-5)^2\), divide both sides of the equation by 3.
This operation simplifies the equation to:

• \((x-5)^2 = \frac{15}{3}\)

This simplifies further to:

• \((x-5)^2 = 5\)

By isolating the squared term, we ensure that the next operational steps can be applied more effectively, allowing us to solve for the unknown variable.

Taking square roots

Once the squared term is isolated, the next step involves taking the square roots of both sides of the equation. This operation will help us to 'unlock' the squared variable. With our example,
\((x-5)^2 = 5\), we take the square root.
This results in:

• \(x-5 = \pm \sqrt{5}\)

It's essential to note that taking the square root of a number introduces both a positive and a negative solution.
This is because both a positive and a negative number, when squared, will give you the same positive result.
As you continue solving, you'll have two paths to follow.

Positive and negative solutions

With the square root providing two outcomes, we must solve for both.
You now have two separate but equally important equations:

• \(x-5 = \sqrt{5}\)
• \(x-5 = -\sqrt{5}\)

These equations reflect the dual nature of squares: each square root has two solutions.
To solve these, simply add 5 to both sides, completing the process for each equation:

• For \(x-5 = \sqrt{5}\), adding 5 gives \(x = \sqrt{5} + 5\)
• For \(x-5 = -\sqrt{5}\), adding 5 gives \(x = -\sqrt{5} + 5\)

This thorough approach ensures you capture all possible solutions.

Simplifying equations

Simplification is final but crucial.
Simplifying ensures the solution is as clear and concise as possible.
For our equations, simplification mainly involved expressing the solutions in a clear form.

• The solution \(x = 5 + \sqrt{5}\) comes from solving \(x - 5 = \sqrt{5}\), by adding 5 to both sides.
• Similarly, \(x = 5 - \sqrt{5}\) originates from the process in \(x - 5 = -\sqrt{5}\).

Each step hones in on getting the purest and neatest form of the solution. Simplifying is about ensuring that anyone can easily identify the solutions without unnecessary complexity.
Thus, our final answers are presented clearly and without additional fuss.