Solving $x^2 - 0.75x + 0.125 = 0$ In [-2, 2]

Hey everyone! Today, let's dive into the world of quadratic equations and explore how to solve one specific example within a given interval. We'll be focusing on the equation $x^2 - 0.75x + 0.125 = 0$ and determining its solutions within the interval of [-2, 2]. This is a classic math problem that combines algebraic manipulation with a touch of interval analysis. So, grab your calculators (or your trusty mental math skills!) and let's get started!

Understanding Quadratic Equations

Before we jump into solving our particular equation, let's take a moment to refresh our understanding of quadratic equations in general. A quadratic equation is a polynomial equation of the second degree, meaning the highest power of the variable (in our case, x) is 2. The standard form of a quadratic equation is given by:

$ax^2 + bx + c = 0$

where a, b, and c are constants, and a is not equal to 0 (otherwise, it would be a linear equation). These constants play a crucial role in determining the shape and position of the parabola that represents the quadratic equation when graphed.

Methods for Solving Quadratic Equations

There are several methods we can use to find the solutions (also called roots or zeros) of a quadratic equation. The most common ones are:

1. Factoring: This method involves breaking down the quadratic expression into a product of two linear expressions. If we can factor the equation, we can easily find the solutions by setting each linear factor equal to zero and solving for x.

2. Completing the Square: This technique involves manipulating the equation algebraically to create a perfect square trinomial on one side. It's a powerful method that can be used even when factoring isn't straightforward.

3. Quadratic Formula: This is a general formula that provides the solutions for any quadratic equation, regardless of whether it can be factored easily. It's derived by completing the square on the standard form of the equation and is a reliable tool for solving any quadratic equation. The formula is:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The Discriminant

An important part of the quadratic formula is the discriminant, which is the expression under the square root: $b^2 - 4ac$. The discriminant tells us about the nature of the solutions:

• If $b^2 - 4ac > 0$, the equation has two distinct real solutions.
• If $b^2 - 4ac = 0$, the equation has one real solution (a repeated root).
• If $b^2 - 4ac < 0$, the equation has two complex solutions.

Understanding the discriminant can help us anticipate the type of solutions we'll find before we even apply the quadratic formula.

Solving $x^2 - 0.75x + 0.125 = 0$

Now that we've reviewed the basics, let's tackle our specific equation: $x^2 - 0.75x + 0.125 = 0$. Our goal is to find the values of x that satisfy this equation, but we also need to consider the interval [-2, 2]. This means we're only interested in solutions that fall within this range.

Identifying Coefficients

First, let's identify the coefficients a, b, and c in our equation. Comparing it to the standard form $ax^2 + bx + c = 0$, we have:

• a = 1
• b = -0.75
• c = 0.125

These coefficients will be essential when we apply the quadratic formula.

Using the Quadratic Formula

Since factoring might not be immediately obvious, and completing the square can be a bit cumbersome, let's use the quadratic formula. Plugging our coefficients into the formula, we get:

$x = \frac{-(-0.75) \pm \sqrt{(-0.75)^2 - 4(1)(0.125)}}{2(1)}$

Let's simplify this step-by-step:

1. Simplify the negative signs:

   $x = \frac{0.75 \pm \sqrt{(-0.75)^2 - 4(1)(0.125)}}{2}$

2. Calculate the square and the product:

   $x = \frac{0.75 \pm \sqrt{0.5625 - 0.5}}{2}$

3. Subtract inside the square root:

   $x = \frac{0.75 \pm \sqrt{0.0625}}{2}$

4. Take the square root:

   $x = \frac{0.75 \pm 0.25}{2}$

Now we have two possible solutions:

• $x_1 = \frac{0.75 + 0.25}{2} = \frac{1}{2} = 0.5$
• $x_2 = \frac{0.75 - 0.25}{2} = \frac{0.5}{2} = 0.25$

Checking the Solutions

We've found two solutions: $x_1 = 0.5$ and $x_2 = 0.25$. But remember, we need to consider the interval [-2, 2]. Both of these solutions fall within this interval, so they are valid solutions to the equation within the given domain.

Interval Consideration [-2, 2]

The interval [-2, 2] is crucial because it restricts the possible values of x that we consider as solutions. This interval represents all real numbers between -2 and 2, inclusive (the square brackets indicate that -2 and 2 are included). In the context of our problem, it means we are only interested in the roots of the quadratic equation that lie within this range.

Why Intervals Matter

In many real-world applications, intervals are essential because they represent physical constraints or limitations. For example, if we were modeling the height of a projectile, negative values of time might not make sense. Similarly, if we were calculating the dimensions of a rectangle, negative lengths wouldn't be physically possible.

In our case, the interval [-2, 2] might represent a specific domain of interest for a particular problem. Without this constraint, the quadratic equation might have solutions that are mathematically valid but not relevant to the situation we're modeling.

Visualizing the Solutions

Think of the graph of the quadratic equation, which is a parabola. The solutions we found, $x_1 = 0.5$ and $x_2 = 0.25$, are the points where the parabola intersects the x-axis. The interval [-2, 2] acts like a window, showing us only the portion of the parabola within that range. Any intersection points outside this window are not considered solutions for our problem.

Conclusion

So, there you have it! We've successfully solved the quadratic equation $x^2 - 0.75x + 0.125 = 0$ within the interval [-2, 2]. We found two solutions, $x_1 = 0.5$ and $x_2 = 0.25$, both of which lie within the specified interval. This exercise demonstrates the power of the quadratic formula and the importance of considering intervals when solving mathematical problems. Remember, understanding the context of a problem and any constraints it imposes is just as crucial as the algebraic manipulations themselves. Keep practicing, and you'll become a quadratic equation-solving pro in no time!