Roots Of Q² - 125: A Step-by-Step Solution
Hey guys! Let's dive into the fascinating world of quadratic functions and explore how to find their roots. Today, we're tackling the quadratic function f(q) = q² - 125. Finding the roots of a quadratic function is like uncovering the hidden keys that unlock its secrets. These roots, also known as the zeros or x-intercepts, are the values of q that make the function equal to zero. In simpler terms, they're the points where the parabola, the graph of the quadratic function, intersects the x-axis. To find these crucial points, we'll embark on a step-by-step journey, employing algebraic techniques and a dash of mathematical intuition. So, buckle up and get ready to unravel the mysteries of this quadratic equation!
Our main objective here is to solve the equation q² - 125 = 0. This will reveal the values of q that satisfy the equation, and consequently, the roots of our function. There are several methods we can use, but for this particular equation, the easiest and most direct approach is to use the square root property. This property states that if x² = a, then x = ±√a. This is because both the positive and negative square roots of a number, when squared, will give you the original number. This is a super handy tool for solving quadratic equations that are in this specific form, where we have a squared term and a constant term. By applying this property, we can efficiently isolate q and find its possible values. We'll walk through this process meticulously, ensuring every step is crystal clear. Understanding the square root property is crucial not just for this problem, but for a wide range of mathematical challenges you might encounter. So, let's get started and see how this powerful tool helps us find the roots of our quadratic function!
To kick things off, we'll isolate the q² term. This is a classic algebraic maneuver – getting the variable term by itself on one side of the equation. We can achieve this by adding 125 to both sides of the equation q² - 125 = 0. This step is fundamental because it sets the stage for applying the square root property. By adding 125 to both sides, we maintain the balance of the equation, ensuring that the left side simplifies to q² and the right side becomes 125. This gives us the equation q² = 125. Now, we're in a prime position to unleash the power of the square root property and unravel the possible values of q. Remember, the goal here is to get q by itself, and isolating the q² term is a crucial stepping stone in that process. So, we've successfully moved the constant term to the other side, paving the way for the final act – finding the roots! Stay tuned as we take the square root of both sides and unveil the solutions.
Now comes the exciting part – taking the square root of both sides of the equation q² = 125. This is where the magic happens and we start to see the roots emerge. When we apply the square root to both sides, we get √(q²) = ±√125. Remember that crucial '±' sign! This is because, as we mentioned earlier, both the positive and negative square roots of a number, when squared, will give you the original number. For instance, both 5² and (-5)² equal 25. This is a key concept in solving quadratic equations, and it's vital not to overlook the negative root. On the left side, √(q²) simplifies beautifully to q. On the right side, we have ±√125. But wait, we can simplify this even further! The number 125 isn't a perfect square, but it does have a perfect square factor hiding within it. That's where our knowledge of simplifying radicals comes into play. We'll break down 125 into its prime factors, identify the perfect square, and pull it out of the radical. This will give us the roots in their simplest, most elegant form. So, let's dive into simplifying √125 and reveal the final answers!
To simplify √125, we need to find its prime factorization. This means breaking down 125 into its prime factors – the prime numbers that multiply together to give us 125. If you think about it, 125 is divisible by 5. In fact, 125 = 5 * 25. And 25 is also divisible by 5: 25 = 5 * 5. So, the prime factorization of 125 is 5 * 5 * 5, which we can write as 5³. Now, we can rewrite √125 as √(5³) or √(5² * 5). This is where the perfect square comes into play. We have 5² inside the square root, and we know that √(5²) = 5. So, we can pull the 5 out of the square root, leaving us with 5√5. This is the simplified form of √125. Now, we remember the '±' sign we had earlier. So, ±√125 simplifies to ±5√5. This means we have two possible values for q: positive 5√5 and negative 5√5. These are the two roots of our quadratic function f(q) = q² - 125. We've successfully navigated the square root, simplified the radical, and arrived at our final solutions!
Identifying the Correct Roots
Alright, now that we've done the math and simplified our radical, we've arrived at the two roots of our quadratic function: q = 5√5 and q = -5√5. These values represent the points where the parabola defined by f(q) = q² - 125 intersects the x-axis. In other words, these are the values of q that make the function equal to zero. Now, let's take a look at the options presented in the original problem and see which ones match our calculated roots.
We were given the following options:
- q = 5√5
- q = -5√5
- q = 3√5
- q = -3√5
- q = 25√5
By comparing our calculated roots (q = 5√5 and q = -5√5) with the options provided, we can clearly see that the first two options, q = 5√5 and q = -5√5, are the correct ones. The other options, q = 3√5, q = -3√5, and q = 25√5, do not satisfy the equation q² - 125 = 0. If you were to plug these values back into the original equation, you would find that the equation does not hold true. This is a great way to check your work and ensure that you've found the correct roots. We can confidently select q = 5√5 and q = -5√5 as the two roots of the quadratic function f(q) = q² - 125.
Why These Are the Roots: A Deeper Understanding
Let's solidify our understanding by taking a moment to explore why these specific values, q = 5√5 and q = -5√5, are the roots of the function f(q) = q² - 125. Remember, roots are the values of q that make the function equal to zero. So, if we plug these values back into the equation, we should get zero. Let's test it out!
First, let's substitute q = 5√5 into the function:
f(5√5) = (5√5)² - 125
To square 5√5, we square both the 5 and the √5: (5²)(√5)² = 25 * 5 = 125. So, we have:
f(5√5) = 125 - 125 = 0
As expected, the function equals zero when q = 5√5. Now, let's do the same for q = -5√5:
f(-5√5) = (-5√5)² - 125
Squaring -5√5, we get: (-5)²(√5)² = 25 * 5 = 125. So, we have:
f(-5√5) = 125 - 125 = 0
Again, the function equals zero! This confirms that both q = 5√5 and q = -5√5 are indeed the roots of the quadratic function f(q) = q² - 125. This exercise demonstrates the fundamental definition of a root and reinforces the solution we found algebraically. It's always a good practice to plug your solutions back into the original equation to verify their correctness. This not only builds confidence in your answer but also deepens your understanding of the concepts involved.
Visualizing the Roots: Connecting Algebra to Geometry
To truly grasp the concept of roots, it's incredibly helpful to visualize them graphically. Quadratic functions, as we mentioned earlier, have a characteristic U-shaped curve called a parabola. The roots of the quadratic function are the points where this parabola intersects the x-axis. Imagine the graph of f(q) = q² - 125. This parabola opens upwards (since the coefficient of the q² term is positive) and its vertex (the lowest point on the curve) is below the x-axis. Because of this, the parabola intersects the x-axis at two distinct points. These two points of intersection are precisely the roots we calculated: q = 5√5 and q = -5√5.
If you were to plot the graph of f(q) = q² - 125, you would see these points clearly marked where the parabola crosses the x-axis. This visual representation connects the algebraic solution we found to the geometric interpretation of the roots. It also provides a more intuitive understanding of what roots represent. A quadratic function can have two roots, one root (if the vertex touches the x-axis), or no real roots (if the parabola doesn't intersect the x-axis). The number of roots depends on the discriminant of the quadratic equation, which is a topic for another time! But for now, understanding that roots are x-intercepts on the graph of the parabola is a powerful tool for visualizing and interpreting quadratic functions.
Key Takeaways and Beyond
So, guys, we've successfully navigated the process of finding the roots of the quadratic function f(q) = q² - 125. We started by understanding the definition of roots, then employed the square root property to solve the equation q² - 125 = 0. We simplified the radical expression √125 to arrive at our final solutions: q = 5√5 and q = -5√5. We verified these solutions by plugging them back into the original equation and confirmed that they indeed make the function equal to zero. Finally, we connected our algebraic solution to the geometric representation of roots as the x-intercepts of the parabola.
This journey has reinforced several important concepts:
- Roots of a quadratic function: The values of the variable that make the function equal to zero.
- Square root property: A powerful tool for solving equations of the form x² = a.
- Simplifying radicals: Breaking down radicals to their simplest form.
- Connecting algebra and geometry: Visualizing roots as x-intercepts on the graph of the parabola.
Understanding these concepts is crucial for tackling more complex quadratic equations and related problems. This is just the beginning of our exploration into the fascinating world of quadratic functions. There's so much more to discover, including the quadratic formula, completing the square, and applications of quadratic functions in various fields. But for now, you've mastered the fundamentals of finding roots using the square root property. Keep practicing, keep exploring, and keep having fun with math!