Solving Quadratic Equations X^2 - 10x + 25 = 0 And H^2 + 6h = 16
In the realm of mathematics, quadratic equations hold a significant place, serving as fundamental tools in various fields ranging from physics and engineering to economics and computer science. These equations, characterized by the presence of a squared variable, often present intriguing challenges and offer a gateway to understanding more complex mathematical concepts. In this comprehensive guide, we delve into the intricacies of solving two specific quadratic equations: x^2 - 10x + 25 = 0 and h^2 + 6h = 16. We will explore different methods, unravel the underlying principles, and provide step-by-step solutions to empower you with the skills to conquer similar problems.
Understanding Quadratic Equations
Before we embark on the journey of solving these particular equations, let's establish a firm understanding of what quadratic equations are and their general form. A quadratic equation is a polynomial equation of the second degree, meaning the highest power of the variable is 2. The standard form of a quadratic equation is expressed as:
ax^2 + bx + c = 0
where 'a', 'b', and 'c' are constants, and 'x' represents the variable we aim to solve for. The coefficients 'a', 'b', and 'c' play a crucial role in determining the nature and solutions of the quadratic equation. The coefficient 'a' cannot be zero, as this would reduce the equation to a linear form.
Quadratic equations can have up to two solutions, also known as roots or zeros. These solutions represent the values of 'x' that satisfy the equation, making the expression equal to zero. Finding these solutions is the primary goal when solving a quadratic equation. There are several methods available to achieve this, each with its own strengths and applicability depending on the specific equation. We will explore some of these methods as we tackle our two example equations.
Solving x^2 - 10x + 25 = 0
Let's begin with the first equation: x^2 - 10x + 25 = 0. This equation presents a unique characteristic that allows us to solve it using a simple and elegant method: factoring. Factoring involves expressing the quadratic expression as a product of two linear expressions. In this case, we can recognize that the given equation is a perfect square trinomial.
A perfect square trinomial is a trinomial that can be factored into the square of a binomial. The general form of a perfect square trinomial is:
(x + a)^2 = x^2 + 2ax + a^2
(x - a)^2 = x^2 - 2ax + a^2
Comparing our equation, x^2 - 10x + 25 = 0, with the second form, we can identify that it fits the pattern. Specifically, we can see that:
- 2a = 10, which implies a = 5
- a^2 = 25
Therefore, we can rewrite the equation as:
(x - 5)^2 = 0
Now, to solve for 'x', we simply take the square root of both sides:
√(x - 5)^2 = √0
x - 5 = 0
Finally, we isolate 'x' by adding 5 to both sides:
x = 5
Thus, the equation x^2 - 10x + 25 = 0 has a single solution, x = 5. This solution is a repeated root, meaning it occurs twice. This is characteristic of perfect square trinomials.
Solving h^2 + 6h = 16
Now, let's move on to the second equation: h^2 + 6h = 16. This equation, unlike the previous one, is not immediately recognizable as a perfect square trinomial. Therefore, we need to employ a different strategy to solve it. The most common and versatile method for solving quadratic equations is the quadratic formula. However, before we apply the quadratic formula, it's often beneficial to try factoring or completing the square, as these methods can sometimes be more efficient.
In this case, let's first try rearranging the equation to the standard form:
h^2 + 6h - 16 = 0
Now, we look for two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2. Therefore, we can factor the equation as:
(h + 8)(h - 2) = 0
To find the solutions, we set each factor equal to zero:
h + 8 = 0 or h - 2 = 0
Solving for 'h' in each case, we get:
h = -8 or h = 2
Thus, the equation h^2 + 6h = 16 has two distinct solutions: h = -8 and h = 2.
Alternative Method: Completing the Square
While we successfully solved the second equation by factoring, let's explore another method known as completing the square. This method is particularly useful when factoring is not straightforward or when we want to rewrite the quadratic equation in a specific form.
To complete the square, we start by moving the constant term to the right side of the equation:
h^2 + 6h = 16
Next, we take half of the coefficient of the 'h' term (which is 6), square it (which is (6/2)^2 = 9), and add it to both sides of the equation:
h^2 + 6h + 9 = 16 + 9
Now, the left side of the equation is a perfect square trinomial:
(h + 3)^2 = 25
Taking the square root of both sides, we get:
√(h + 3)^2 = √25
h + 3 = ±5
This gives us two equations:
h + 3 = 5 or h + 3 = -5
Solving for 'h' in each case, we get:
h = 2 or h = -8
As we can see, completing the square yields the same solutions as factoring, confirming the accuracy of our results.
The Quadratic Formula: A Universal Solution
While factoring and completing the square are valuable techniques, they may not always be applicable or the most efficient methods. The quadratic formula, on the other hand, provides a universal solution for any quadratic equation in the standard form ax^2 + bx + c = 0. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
To apply the quadratic formula to our second equation, h^2 + 6h - 16 = 0, we identify the coefficients:
- a = 1
- b = 6
- c = -16
Substituting these values into the quadratic formula, we get:
h = (-6 ± √(6^2 - 4 * 1 * -16)) / (2 * 1)
h = (-6 ± √(36 + 64)) / 2
h = (-6 ± √100) / 2
h = (-6 ± 10) / 2
This gives us two solutions:
h = (-6 + 10) / 2 = 2
h = (-6 - 10) / 2 = -8
Again, the quadratic formula confirms our previous results, demonstrating its power and versatility in solving quadratic equations.
Conclusion: Mastering Quadratic Equations
In this comprehensive guide, we have explored the intricacies of solving quadratic equations, focusing on two specific examples: x^2 - 10x + 25 = 0 and h^2 + 6h = 16. We delved into different methods, including factoring, completing the square, and the quadratic formula, showcasing their strengths and applicability. By understanding these techniques, you are now equipped with the skills to tackle a wide range of quadratic equations with confidence.
Remember, practice is key to mastering any mathematical concept. Work through various examples, explore different scenarios, and don't hesitate to revisit the methods we discussed. With consistent effort, you will develop a strong foundation in solving quadratic equations, a fundamental skill that will serve you well in your mathematical journey.
