Finding The Intersection Point Value Of X In A System Of Equations

In mathematics, determining the intersection points of different curves is a fundamental problem. This article delves into finding the intersection point of a parabola and a line. Specifically, we will address the problem where we are given a system of equations representing a parabola and a line, and we know they intersect at exactly one point. Our goal is to find the x-coordinate of this intersection point. This problem combines concepts from algebra and coordinate geometry, making it a great exercise in mathematical problem-solving.

We are given the following system of equations:

1. y = 3x^2 - 5x + 4
2. y = 7x - a

where a is a constant. The graphs of these equations intersect at exactly one point (x, y) in the xy-plane. The question we need to answer is: What is the value of x?

Before diving into the solution, let's break down what the problem is asking. We have a parabola defined by the quadratic equation y = 3x^2 - 5x + 4 and a line defined by the linear equation y = 7x - a. The fact that the graphs intersect at exactly one point tells us that the line is tangent to the parabola. This means the line touches the parabola at only one point, and there is only one solution to the system of equations. This is a crucial piece of information that will guide our solution approach.

Step 1: Equate the Expressions for y

Since both equations are equal to y, we can set them equal to each other:

3x^2 - 5x + 4 = 7x - a

This step combines the two equations into a single equation in terms of x. This is a standard technique when solving systems of equations.

Step 2: Rearrange into a Quadratic Equation

Now, we want to rearrange the equation into the standard form of a quadratic equation, which is Ax^2 + Bx + C = 0. To do this, we subtract 7x from both sides and add a to both sides:

3x^2 - 5x + 4 - 7x + a = 0

Combine like terms:

3x^2 - 12x + (4 + a) = 0

Now we have a quadratic equation in the standard form, where A = 3, B = -12, and C = 4 + a.

Step 3: Apply the Discriminant Condition

The discriminant of a quadratic equation Ax^2 + Bx + C = 0 is given by Δ = B^2 - 4AC. The discriminant tells us about the nature of the roots of the quadratic equation:

• If Δ > 0, the equation has two distinct real roots.
• If Δ = 0, the equation has exactly one real root (a repeated root).
• If Δ < 0, the equation has no real roots.

Since the problem states that the graphs intersect at exactly one point, the quadratic equation must have exactly one real root. This means the discriminant must be equal to zero:

Δ = B^2 - 4AC = 0

Substitute the values of A, B, and C:

(-12)^2 - 4(3)(4 + a) = 0

Step 4: Solve for a

Simplify the equation:

144 - 12(4 + a) = 0

144 - 48 - 12a = 0

96 - 12a = 0

12a = 96

a = 8

So, we have found the value of the constant a.

Step 5: Substitute a Back into the Quadratic Equation

Now that we know a = 8, we can substitute it back into the quadratic equation:

3x^2 - 12x + (4 + 8) = 0

3x^2 - 12x + 12 = 0

Step 6: Solve for x

We can simplify the quadratic equation by dividing all terms by 3:

x^2 - 4x + 4 = 0

This is a perfect square trinomial, which can be factored as:

(x - 2)^2 = 0

Taking the square root of both sides:

x - 2 = 0

x = 2

Therefore, the x-coordinate of the intersection point is 2.

The value of x is 2.

It's important to understand why students might choose the incorrect answer options. Let's look at the common mistakes that could lead to choosing the wrong answers:

• Option A (8): This is the value of a that we found. Students might mistakenly think that they have solved for x when they found a. It's crucial to remember what the question is asking for and to complete all the steps.
• Option B (-8): This could result from an error in solving for a. For example, if a student makes a mistake in the sign when rearranging the equation or when dividing, they might end up with a negative value for a. They might then stop there, assuming they've solved for x.
• Option C (6): This option is less likely but could result from an algebraic error when solving the quadratic equation for x after substituting the value of a. For example, a student might incorrectly factor the quadratic or make a mistake when applying the quadratic formula (although factoring is more straightforward in this case).

In this article, we have walked through the process of finding the intersection point of a parabola and a line. We started with a system of equations, used the condition for tangency (discriminant equals zero), solved for the constant a, and then solved for the x-coordinate of the intersection point. The key to solving this type of problem is to understand the relationship between the equations and the geometric interpretation of their intersection. This problem exemplifies the application of algebraic techniques to solve geometric problems, a fundamental concept in mathematics. The correct answer is D. 2.

This article focuses on the keywords intersection point, parabola, line, system of equations, discriminant, and quadratic equation. These terms are crucial for students studying algebra and coordinate geometry. The title is optimized for search engines to attract students and educators seeking help with these topics. The article provides a detailed explanation of the problem-solving process, making it a valuable resource for understanding the concepts and techniques involved.

By providing a clear and step-by-step solution, along with an analysis of common errors, this article aims to help students master this type of problem and build a strong foundation in mathematics.