Solving (x^2-7x)/2=15 A Step-by-Step Guide

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In the realm of mathematics, quadratic equations hold a significant position, appearing in various contexts, from basic algebra to advanced calculus. Understanding how to solve these equations is a fundamental skill for any aspiring mathematician or scientist. This comprehensive guide will delve into the step-by-step process of solving the quadratic equation $\frac{x^2-7x}{2}=15$, providing a clear and concise explanation for each step. We will explore the underlying concepts and techniques, ensuring a solid grasp of the subject matter.

Understanding Quadratic Equations

Before we embark on solving the equation, let's first define what a quadratic equation is. A quadratic equation is a polynomial equation of the second degree, meaning the highest power of the variable is 2. The general form of a quadratic equation is expressed as:

ax2+bx+c=0ax^2 + bx + c = 0

where a, b, and c are constants, and x represents the variable we aim to solve for. The coefficient a cannot be zero, otherwise, the equation would reduce to a linear equation.

In our given equation, $\frac{x^2-7x}{2}=15$, we can observe that it is indeed a quadratic equation. To make it more apparent, we can rewrite it in the standard form by performing a few algebraic manipulations. This involves multiplying both sides of the equation by 2 and then rearranging the terms to get a zero on one side.

Transforming the Equation into Standard Form

The first step in solving the equation $\frac{x^2-7x}{2}=15$ is to transform it into the standard quadratic form. This will make it easier to apply the various methods available for solving quadratic equations. To achieve this, we multiply both sides of the equation by 2:

2×x2−7x2=2×152 \times \frac{x^2-7x}{2} = 2 \times 15

This simplifies to:

x2−7x=30x^2 - 7x = 30

Next, we subtract 30 from both sides to set the equation equal to zero:

x2−7x−30=0x^2 - 7x - 30 = 0

Now, the equation is in the standard quadratic form, where a = 1, b = -7, and c = -30. This form allows us to readily identify the coefficients and apply methods such as factoring, completing the square, or using the quadratic formula.

Solving by Factoring

One of the most straightforward methods for solving quadratic equations is factoring. Factoring involves expressing the quadratic expression as a product of two linear expressions. This method is particularly effective when the quadratic equation has integer roots, making it easier to identify the factors.

To factor the equation $x^2 - 7x - 30 = 0$, we need to find two numbers that multiply to -30 (the constant term) and add up to -7 (the coefficient of the x term). Let's consider the factors of -30:

  • 1 and -30
  • -1 and 30
  • 2 and -15
  • -2 and 15
  • 3 and -10
  • -3 and 10
  • 5 and -6
  • -5 and 6

Among these pairs, the numbers 3 and -10 satisfy our conditions, as 3 \times -10 = -30 and 3 + (-10) = -7. Therefore, we can rewrite the quadratic expression as:

(x+3)(x−10)=0(x + 3)(x - 10) = 0

Now, according to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. Thus, we have two possible cases:

  1. x + 3 = 0, which implies x = -3
  2. x - 10 = 0, which implies x = 10

Therefore, the solutions to the quadratic equation are x = -3 and x = 10. These are the values of x that satisfy the original equation. To verify our solutions, we can substitute each value back into the original equation and check if the equation holds true.

Verifying the Solutions

To ensure the accuracy of our solutions, we must substitute them back into the original equation, $\frac{x^2-7x}{2}=15$, and verify that the equation holds true. Let's start with the first solution, x = -3:

(−3)2−7(−3)2=15\frac{(-3)^2 - 7(-3)}{2} = 15

9+212=15\frac{9 + 21}{2} = 15

302=15\frac{30}{2} = 15

15=1515 = 15

The equation holds true for x = -3, confirming that it is a valid solution. Now, let's verify the second solution, x = 10:

(10)2−7(10)2=15\frac{(10)^2 - 7(10)}{2} = 15

100−702=15\frac{100 - 70}{2} = 15

302=15\frac{30}{2} = 15

15=1515 = 15

The equation also holds true for x = 10, confirming that it is another valid solution. Therefore, we have successfully verified that both x = -3 and x = 10 are the solutions to the quadratic equation.

Alternative Methods for Solving Quadratic Equations

While factoring is a useful technique, it may not always be the most efficient method, especially for equations with non-integer roots or complex coefficients. In such cases, alternative methods like completing the square and the quadratic formula come into play. These methods provide a systematic approach to solving any quadratic equation, regardless of the nature of its roots.

Completing the Square

Completing the square is a technique that involves manipulating the quadratic equation to create a perfect square trinomial on one side. A perfect square trinomial is a trinomial that can be factored into the square of a binomial. By completing the square, we can transform the equation into a form that is easily solvable by taking the square root of both sides.

The Quadratic Formula

The quadratic formula is a powerful tool that provides a direct solution for any quadratic equation. It is derived from the method of completing the square and expresses the solutions in terms of the coefficients a, b, and c. The quadratic formula is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula guarantees a solution for any quadratic equation, regardless of the nature of its roots. The term inside the square root, b2 - 4ac, is known as the discriminant. The discriminant provides valuable information about the nature of the roots:

  • If b2 - 4ac > 0, the equation has two distinct real roots.
  • If b2 - 4ac = 0, the equation has one real root (a repeated root).
  • If b2 - 4ac < 0, the equation has two complex roots.

Conclusion

In conclusion, we have successfully solved the quadratic equation $\frac{x^2-7x}{2}=15$ using the factoring method. We first transformed the equation into the standard form, identified the coefficients, and then factored the quadratic expression. By applying the zero-product property, we found the solutions to be x = -3 and x = 10. We further verified these solutions by substituting them back into the original equation.

Furthermore, we explored alternative methods for solving quadratic equations, such as completing the square and the quadratic formula. These methods offer a more general approach and can be applied to solve any quadratic equation, regardless of the nature of its roots. The quadratic formula, in particular, provides a direct solution and is a valuable tool in mathematical problem-solving.

Understanding and mastering the techniques for solving quadratic equations is crucial for success in mathematics and related fields. This guide has provided a comprehensive overview of the process, equipping you with the knowledge and skills to tackle any quadratic equation that comes your way. Remember to practice these techniques regularly to solidify your understanding and build confidence in your problem-solving abilities.