Problem 80 - Algebra and Trigonometry

Question

Solve: \((x-1)^{2}=5\) (Section \(1.5, \text { Example } 2)\)

Step-by-Step Solution

Answer

The solution of the equation \((x-1)^{2}=5\) are \(x=1+\sqrt{5}\) and \(x=1-\sqrt{5}\).

Step 1: Apply the square root to both sides

Take the square root of the both sides of the equation \((x-1)^{2}=5\), these results in two equations: \(x-1= \sqrt{5}\) and \(x-1= -\sqrt{5}\) because \(\sqrt{a^2}\) can be either \(a\) or \(-a\).

Step 2: Solve for x

Solving the above equations for x gives \(x=1+\sqrt{5}\) for the first equation and \(x=1-\sqrt{5}\) for the second equation.

Step 3: Verify the solution

Substitute \(x=1+\sqrt{5}\) and \(x=1-\sqrt{5}\) into the original equation. For \(x=1+\sqrt{5}\), you get \((1+\sqrt{5}-1)^{2}=(\sqrt{5})^{2}=5\), which matches the original equation. For \(x=1-\sqrt{5}\), you get \((1-\sqrt{5}-1)^{2}=(-\sqrt{5})^{2}=5\), which also matches the original equation. Therefore, the solutions are verified.

Key Concepts

Square Root Property

The square root property is a powerful tool in solving quadratic equations. It states that if you have an equation of the form \((x - h)^2 = k\), then taking the square root of both sides gives two possible equations:

• \(x - h = \sqrt{k}\)
• \(x - h = -\sqrt{k}\)

This duality occurs because both a number and its negative have the same square.
In the original exercise with \((x-1)^2=5\), we apply this property and get \(x-1 = \sqrt{5}\) and \(x-1 = -\sqrt{5}\). This results in two possible values for \(x\) once we solve both equations.
Using the square root property effectively requires recognizing when an equation is in the right format and thinking about the implications of taking a square root.

Solving Equations

Solving equations involves finding values for the variable that make the equation true. Once you've applied the square root property and arrived at the equations \(x-1=\sqrt{5}\) and \(x-1=-\sqrt{5}\), the next step is straightforward. You want to solve for \(x\) by isolating it on one side of the equation.

• Start by adding 1 to both sides to get \(x=1+\sqrt{5}\) and \(x=1-\sqrt{5}\).
• These solutions represent the values of \(x\) that will satisfy the original equation.

These steps test our ability to perform basic algebraic manipulations.
The simplicity of this problem helps build understanding of more complex algebraic concepts over time.

Verification of Solutions

Verification of solutions is a vital step in solving equations. It ensures that the operations performed have led to valid solutions. By substituting the solutions back into the original equation, you verify their correctness:

• For \(x=1+\sqrt{5}\), substitute back to check \([1+\sqrt{5} - 1]^2 = (\sqrt{5})^2 = 5\).
• For \(x=1-\sqrt{5}\), verify similarly: \([1-\sqrt{5} - 1]^2 = (-\sqrt{5})^2 = 5\).

Both instances result in true statements, confirming that the solutions satisfy the original equation.
This process is crucial not just for accuracy, but it also builds confidence in solving equations and reinforces understanding of algebraic principles.