Solving $x^2 - 10x + 25 = 0$ And 1/3 = M + (m-1)/m A Math Tutorial
This article delves into solving two distinct mathematical problems. First, we will explore the quadratic equation and determine its roots using various methods, including factoring and the quadratic formula. Understanding quadratic equations is fundamental in algebra, as they appear in numerous contexts, from physics to engineering. We will break down the steps involved in solving this particular equation, highlighting key concepts and techniques that are applicable to a wide range of similar problems. Secondly, we will tackle the algebraic equation . This equation involves fractions and requires careful manipulation to isolate the variable m. We will discuss the process of clearing fractions, simplifying expressions, and ultimately finding the solution(s) for m. This exercise will reinforce skills in algebraic manipulation and problem-solving, which are crucial for advanced mathematical studies.
Solving the Quadratic Equation
Quadratic equations are polynomial equations of the second degree, generally expressed in the form ax² + bx + c = 0, where a, b, and c are constants, and x is the variable. The equation fits this form, with a = 1, b = -10, and c = 25. There are several methods to solve quadratic equations, each with its own advantages depending on the specific equation. Let's explore two common methods: factoring and the quadratic formula.
Factoring the Quadratic Equation
Factoring is a method that involves expressing the quadratic expression as a product of two binomials. This method is particularly effective when the quadratic equation can be easily factored. In our case, can be recognized as a perfect square trinomial. A perfect square trinomial is a trinomial that can be written as the square of a binomial. The general form of a perfect square trinomial is a² ± 2ab + b², which can be factored as (a ± b)². Comparing to this form, we can see that x² corresponds to a², 25 corresponds to b², and -10x corresponds to -2ab. Specifically, a = x and b = 5, since 5² = 25. Therefore, -2ab = -2 * x * 5 = -10x, which confirms that our equation is indeed a perfect square trinomial. Consequently, we can factor as (x - 5)². Now, setting the factored expression equal to zero, we have (x - 5)² = 0. Taking the square root of both sides, we get x - 5 = 0. Solving for x, we find x = 5. This means that the quadratic equation has one real root, x = 5. This root is a repeated root, also known as a double root, because the factor (x - 5) appears twice in the factored form of the equation. Factoring is a powerful technique for solving quadratic equations, particularly when the equation is easily factorable. It allows us to transform the equation into a simpler form, making it straightforward to find the roots. In this case, recognizing the perfect square trinomial greatly simplified the process of finding the solution.
Using the Quadratic Formula
When factoring is not straightforward or the equation is not easily factorable, the quadratic formula provides a general method for solving quadratic equations of the form ax² + bx + c = 0. The quadratic formula is given by: x = (-b ± √(b² - 4ac)) / (2a). This formula can be used to find the roots of any quadratic equation, regardless of whether it can be factored. Applying the quadratic formula to our equation, , we have a = 1, b = -10, and c = 25. Substituting these values into the formula, we get: x = (-(-10) ± √((-10)² - 4 * 1 * 25)) / (2 * 1). Simplifying the expression inside the square root, we have: (-10)² - 4 * 1 * 25 = 100 - 100 = 0. Therefore, the square root term becomes √0, which is 0. Substituting this back into the formula, we get: x = (10 ± 0) / 2. This simplifies to x = 10 / 2 = 5. Thus, the quadratic formula also yields the solution x = 5. The fact that the discriminant (the expression inside the square root, b² - 4ac) is zero indicates that the quadratic equation has exactly one real root, which is a repeated root. This confirms the result we obtained through factoring. The quadratic formula is a versatile tool that guarantees a solution for any quadratic equation. It is especially useful when dealing with equations that have complex roots or are difficult to factor. While factoring is often quicker when applicable, the quadratic formula provides a reliable method for finding the roots in all cases.
Solving the Algebraic Equation
Now, let's shift our focus to the algebraic equation . This equation involves fractions, so our first step will be to clear the fractions to simplify the equation. Clearing fractions typically involves multiplying both sides of the equation by the least common denominator (LCD) of the fractions present. In this case, we have fractions with denominators 3 and m. Therefore, the least common denominator is 3m. Multiplying both sides of the equation by 3m, we get: 3m * (1/3) = 3m * (m + (m-1)/m). On the left side, the 3's cancel out, leaving us with m. On the right side, we need to distribute 3m to both terms: 3m * m + 3m * ((m-1)/m). This simplifies to 3m² + 3(m-1). So, our equation now becomes: m = 3m² + 3(m-1). Next, we need to simplify and rearrange the equation into a standard form. Distributing the 3 in the term 3(m-1), we get 3m - 3. Substituting this back into the equation, we have: m = 3m² + 3m - 3. To bring all terms to one side and set the equation to zero, we subtract m from both sides: 0 = 3m² + 3m - 3 - m. Combining like terms, we get: 0 = 3m² + 2m - 3. This is a quadratic equation in terms of m. We can solve this equation using the quadratic formula.
Applying the Quadratic Formula
The quadratic formula, as mentioned earlier, is a general method for solving quadratic equations of the form ax² + bx + c = 0. In our case, the equation is 3m² + 2m - 3 = 0, so a = 3, b = 2, and c = -3. Substituting these values into the quadratic formula, m = (-b ± √(b² - 4ac)) / (2a), we get: m = (-2 ± √(2² - 4 * 3 * (-3))) / (2 * 3). Now, let's simplify the expression inside the square root: 2² - 4 * 3 * (-3) = 4 + 36 = 40. So, we have: m = (-2 ± √40) / 6. We can simplify √40 by factoring out the largest perfect square: √40 = √(4 * 10) = 2√10. Substituting this back into the equation, we get: m = (-2 ± 2√10) / 6. We can further simplify this by dividing both the numerator and the denominator by 2: m = (-1 ± √10) / 3. This gives us two possible solutions for m: m₁ = (-1 + √10) / 3 and m₂ = (-1 - √10) / 3. These are the exact solutions for the equation. To get approximate decimal values, we can use a calculator. √10 is approximately 3.162. Therefore, m₁ ≈ (-1 + 3.162) / 3 ≈ 2.162 / 3 ≈ 0.721 and m₂ ≈ (-1 - 3.162) / 3 ≈ -4.162 / 3 ≈ -1.387. Thus, the solutions for the equation are approximately m ≈ 0.721 and m ≈ -1.387. In summary, solving this equation involved clearing fractions by multiplying by the least common denominator, simplifying the resulting equation, and then applying the quadratic formula to find the roots.
Conclusion
In this article, we have successfully solved two distinct mathematical problems. The first was the quadratic equation , which we solved by both factoring and using the quadratic formula, yielding the single repeated root x = 5. This exercise demonstrated the application of key algebraic techniques for solving quadratic equations, highlighting the importance of recognizing perfect square trinomials and the versatility of the quadratic formula. The second problem involved the algebraic equation , which we tackled by clearing fractions and then applying the quadratic formula to the resulting quadratic equation in m. This process led us to two solutions: m ≈ 0.721 and m ≈ -1.387. This problem underscored the importance of algebraic manipulation skills, such as clearing fractions and simplifying expressions, and reinforced the application of the quadratic formula in a different context. These examples illustrate the interconnectedness of mathematical concepts and the importance of mastering fundamental techniques to solve a variety of problems. By understanding and applying these methods, one can confidently approach and solve a wide range of mathematical challenges. Both problems required a careful and systematic approach, emphasizing the need for precision and attention to detail in mathematical problem-solving. The ability to solve quadratic equations and manipulate algebraic expressions is essential for further studies in mathematics and related fields, making these exercises valuable for students and professionals alike.