Solving Systems Of Equations Algebraically Find X

In the realm of mathematics, solving systems of equations is a fundamental skill. It allows us to find the values of variables that satisfy multiple equations simultaneously. In this article, we will delve into the process of solving a system of equations algebraically, specifically focusing on finding the value(s) of x. We will explore the steps involved, the underlying concepts, and the potential solutions that can arise. The given system of equations involves two functions, f(x) and g(x), both expressed in terms of x. Our goal is to determine the value(s) of x for which f(x) equals g(x). This involves setting the two expressions equal to each other and then employing algebraic techniques to isolate x. The process may involve simplifying the equation, factoring, using the quadratic formula, or other methods, depending on the nature of the equation. It's crucial to understand the different types of solutions that can occur, such as real solutions, complex solutions, or no solutions at all. By mastering the art of solving systems of equations algebraically, we equip ourselves with a powerful tool for tackling a wide range of mathematical problems and real-world applications. This skill is not only essential in algebra but also forms the basis for more advanced mathematical concepts and applications in fields such as physics, engineering, and economics. This exploration will also enhance your analytical and problem-solving skills, which are valuable assets in various aspects of life.

We are given the following system of equations:

$$\begin{array}{l} f(x)=3 x^2-4 \\ g(x)=x^2+7 x-13 \end{array}$$

Our objective is to find the value(s) of x that satisfy both equations simultaneously. In other words, we need to find the x value(s) for which f(x) = g(x).

To solve the system of equations algebraically, we need to find the value(s) of x for which f(x) = g(x). This means we set the two expressions equal to each other and solve for x:

Step 1: Set the Equations Equal

We begin by equating the expressions for f(x) and g(x):

$$3x^2 - 4 = x^2 + 7x - 13$$

Step 2: Simplify the Equation

Next, we want to simplify the equation by moving all terms to one side. Subtract x², 7x, and add 13 to both sides of the equation:

$$3x^2 - x^2 - 7x - 4 + 13 = 0$$

This simplifies to:

$$2x^2 - 7x + 9 = 0$$

Step 3: Solve the Quadratic Equation

Now we have a quadratic equation in the form of ax² + bx + c = 0, where a = 2, b = -7, and c = 9. We can solve this quadratic equation using the quadratic formula:

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

Substitute the values of a, b, and c into the formula:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(9)}}{2(2)}$$

Step 4: Calculate the Discriminant

Calculate the discriminant (b² - 4ac):

$$(-7)^2 - 4(2)(9) = 49 - 72 = -23$$

Since the discriminant is negative (-23), the quadratic equation has no real solutions. This means the graphs of f(x) and g(x) do not intersect in the real number plane.

Step 5: Write the Solution

Since the discriminant is negative, there are no real solutions for x. Therefore, the correct answer is:

A. No solution

The fact that the discriminant (b² - 4ac) is negative in the quadratic formula indicates that the square root of a negative number would be involved in the solution. In the realm of real numbers, the square root of a negative number is undefined. This means that the quadratic equation 2x² - 7x + 9 = 0 has no real roots. Graphically, this implies that the parabola represented by the quadratic equation does not intersect the x-axis.

In the context of the original system of equations, the absence of real solutions means that the graphs of the two functions, f(x) = 3x² - 4 and g(x) = x² + 7x - 13, do not intersect. There is no real value of x for which the two functions have the same output.

However, it's important to note that the quadratic equation does have complex solutions. Complex numbers extend the real number system by including the imaginary unit i, where i² = -1. If we were to consider complex solutions, we would proceed with the quadratic formula and express the solutions in terms of i. The solutions would be complex conjugates, meaning they have the form a + bi and a - bi, where a and b are real numbers.

For instance, if we were to calculate the complex solutions for this problem, we would have:

$$x = \frac{7 \pm \sqrt{-23}}{4}$$

$$x = \frac{7 \pm i\sqrt{23}}{4}$$

So, the complex solutions would be:

$$x = \frac{7 + i\sqrt{23}}{4} \quad \text{and} \quad x = \frac{7 - i\sqrt{23}}{4}$$

These complex solutions indicate that the two functions f(x) and g(x) would intersect in the complex plane, but not in the real plane.

While the quadratic formula is a reliable method for solving quadratic equations, there are other techniques that can be used in certain situations. These methods include:

• Factoring: If the quadratic equation can be factored easily, factoring is often the quickest way to find the solutions. This involves expressing the quadratic expression as a product of two linear factors. The solutions are then the values of x that make each factor equal to zero.
• Completing the Square: This method involves manipulating the quadratic equation to create a perfect square trinomial on one side. The square root of both sides can then be taken, leading to the solutions.

In the case of the equation 2x² - 7x + 9 = 0, factoring is not straightforward, and completing the square would be more cumbersome than using the quadratic formula. Therefore, the quadratic formula is the most efficient method for solving this particular equation.

In this article, we have successfully solved the system of equations algebraically for x. By setting the two functions equal to each other, simplifying the equation, and applying the quadratic formula, we determined that there are no real solutions for x. This is because the discriminant of the quadratic equation was negative, indicating that the graphs of the two functions do not intersect in the real number plane. While there are no real solutions, we also explored the concept of complex solutions and how they would arise in this scenario. Understanding the nature of solutions, whether real or complex, is crucial in mathematics and its applications. This problem provides a solid example of how algebraic techniques can be used to solve systems of equations and interpret the results. The ability to solve quadratic equations is a cornerstone of algebra, with applications spanning various fields, including physics, engineering, and computer science. Mastering this skill opens doors to a deeper understanding of mathematical concepts and their real-world applications. Remember, the quadratic formula is not just a tool for finding solutions; it's a gateway to understanding the nature of those solutions and their implications.