Problem 35 - Algebra 2

Question

Find the two real-number solutions of each equation. $$ x^{2}=0.25 $$

Step-by-Step Solution

Answer

The two real number solutions to the equation are \(x=0.5\) and \(x=-0.5\).

Step 1: Analysis of the Problem

The problem involves a simple quadratic equation \(x^{2}=0.25\). Based on the square root principle, we will apply the square root on both sides of the equation to isolate the variable \(x\).

Step 2: Applying the Square Root Principle

By applying the square root to both sides of the equation, remember that the solution can be both positive and negative which results in: \[x=\pm 0.5\]

Key Concepts

Square Root Principle

When you encounter a quadratic equation like \(x^2 = a\), the square root principle is a powerful tool. It helps us solve for \(x\) by taking the square root of both sides of the equation. Here's how it works:
To apply the square root principle, perform the following steps:

• Take the square root of both sides of the equation. In our example, we get \(x = \pm\sqrt{0.25}\).
• Remember that a square root has two possible values: a positive value and a negative value. Hence, \(x\) can be \(0.5\) or \(-0.5\).

Using the square root principle, we can quickly find the solutions of a quadratic equation when it is in the form of \(x^2 = a\). This principle provides a straightforward way to find solutions without having to factor the equation or use the quadratic formula.

Real-Number Solutions

In mathematics, real numbers include all the numbers on the number line. This includes rational numbers (like fractions and integers) as well as irrational numbers. When solving quadratic equations, we often look for real-number solutions which can be positive, negative, or zero.
For the equation \(x^2 = 0.25\), finding real-number solutions involves recognizing that both \(0.5\) and \(-0.5\) are valid answers. These solutions occur when we calculate the square root of \(0.25\) and include both the positive and negative roots.Quadratic equations might sometimes have solutions that are purely imaginary or complex, but in this case, both solutions are real numbers. Therefore, the final solutions are tangible, measurable values that can be represented on the number line.

Variable Isolation

Isolating the variable in an equation is a crucial step in solving algebraic equations. It involves performing operations on the equation in such a way that \(x\) or any other variable stands alone on one side of the equation.
Let's see how it works in the context of our quadratic equation \(x^2 = 0.25\):

• Initially, the variable \(x^2\) is equal to \(0.25\).
• To isolate \(x\), apply the square root principle to both sides, giving \(x = \pm 0.5\).
• This step effectively separates the variable \(x\) from any other numbers or coefficients, making it easier to pinpoint the values of \(x\).

Variable isolation is a key part of algebra that ensures solutions are accurate and correctly reflect the original equation's requirements.