Finding The Range Of X Given X + Y + Z = 4 And X² + Y² + Z² = 6
This article explores the range of possible values for x when given two equations: x + y + z = 4 and x² + y² + z² = 6, where x, y, and z are real numbers. We will delve into the mathematical concepts and techniques required to solve this problem, providing a comprehensive explanation suitable for anyone interested in mathematical problem-solving. Understanding the constraints imposed by these equations and how they interact is key to determining the bounds within which x can exist. This problem highlights the interplay between linear and quadratic equations and the geometric interpretations that can aid in finding solutions. Let's embark on this mathematical journey together to unravel the solution.
Problem Statement: Finding the Range of x
We are given that x, y, and z are real numbers that satisfy the following equations:
- x + y + z = 4
- x² + y² + z² = 6
The objective is to find the range of possible values for x. This means we want to determine the minimum and maximum values that x can take while still satisfying both equations. This involves a combination of algebraic manipulation and potentially geometric insights to understand the relationships between the variables.
Initial Observations and Strategic Approach
Our initial step in approaching this problem is to recognize that we have two equations with three unknowns. This typically means that there will be infinitely many solutions, but the second equation (x² + y² + z² = 6) introduces a constraint that limits the possible values. The key idea here is to eliminate one or two variables to obtain an expression involving only x, which we can then analyze to determine its range. We can achieve this by using the first equation to express y + z in terms of x and then substituting this into a modified form of the second equation. This process will lead us to a quadratic equation or inequality in x that we can solve. Another important strategy is to think about the geometric interpretation of these equations. x + y + z = 4 represents a plane in three-dimensional space, and x² + y² + z² = 6 represents a sphere centered at the origin with a radius of √6. The solutions to the system of equations are the points where the plane and the sphere intersect. Visualizing this intersection can give us a better understanding of the possible values of x. We will use these strategies to solve the problem step by step.
Step-by-Step Solution
To find the range of x, we will systematically manipulate the given equations. Here's the breakdown of the steps involved:
1. Express y + z in terms of x
From the equation x + y + z = 4, we can isolate y + z:
y + z = 4 - x (Equation 3)
This equation tells us that the sum of y and z is directly related to the value of x. As x changes, the sum y + z will also change accordingly.
2. Express y² + z² in terms of x
Next, we need to express y² + z² in terms of x. We start with the second given equation:
x² + y² + z² = 6
Rearrange the equation to isolate y² + z²:
y² + z² = 6 - x² (Equation 4)
This equation provides another crucial relationship. It shows that the sum of the squares of y and z is also dependent on the value of x. This, combined with Equation 3, will allow us to form a relationship that doesn't involve y or z individually.
3. Use the Identity (y + z)² = y² + z² + 2yz
We know that (y + z)² = y² + z² + 2yz. We can use this identity to relate the expressions we derived in the previous steps. From Equation 3, we have (y + z) = 4 - x, so
(y + z)² = (4 - x)²
Expanding the right side, we get:
(y + z)² = 16 - 8x + x² (Equation 5)
From Equation 4, we have y² + z² = 6 - x². Now we can substitute these expressions into the identity:
16 - 8x + x² = 6 - x² + 2yz
4. Express yz in terms of x
Now, let's isolate 2yz:
2yz = 16 - 8x + x² - (6 - x²)
Simplify:
2yz = 16 - 8x + x² - 6 + x²
2yz = 2x² - 8x + 10
Divide by 2:
yz = x² - 4x + 5 (Equation 6)
5. Form a Quadratic Equation with y and z as Roots
Now, consider a quadratic equation with y and z as its roots. Let this equation be:
t² - (y + z)t + yz = 0
We know y + z and yz in terms of x from Equations 3 and 6. Substitute the values:
t² - (4 - x)t + (x² - 4x + 5) = 0
This equation's roots are y and z. For y and z to be real numbers, the discriminant of this quadratic equation must be greater than or equal to zero.
6. Apply the Discriminant Condition
The discriminant (Δ) of a quadratic equation at² + bt + c = 0 is given by Δ = b² - 4ac. In our case, a = 1, b = -(4 - x), and c = x² - 4x + 5. So:
Δ = (-(4 - x))² - 4(1)(x² - 4x + 5)
For real roots, we need Δ ≥ 0:
(4 - x)² - 4(x² - 4x + 5) ≥ 0
7. Solve the Inequality
Expand and simplify the inequality:
16 - 8x + x² - 4x² + 16x - 20 ≥ 0
Combine like terms:
-3x² + 8x - 4 ≥ 0
Multiply by -1 (and reverse the inequality sign):
3x² - 8x + 4 ≤ 0
Factor the quadratic:
(3x - 2)(x - 2) ≤ 0
8. Determine the Range of x
To solve the inequality (3x - 2)(x - 2) ≤ 0, we need to find the intervals where the expression is negative or zero. The critical points are the roots of the equation (3x - 2)(x - 2) = 0, which are x = 2/3 and x = 2. We can test intervals to determine the solution:
- For x < 2/3, both (3x - 2) and (x - 2) are negative, so their product is positive.
- For 2/3 < x < 2, (3x - 2) is positive and (x - 2) is negative, so their product is negative.
- For x > 2, both (3x - 2) and (x - 2) are positive, so their product is positive.
Thus, the solution to the inequality is 2/3 ≤ x ≤ 2. This means that the range of possible values for x is the closed interval [2/3, 2]. This final range is the solution to our problem.
Final Answer: Range of x
Therefore, the range of x is [2/3, 2]. This corresponds to option (C) in the multiple-choice options provided. This means x can take any value between 2/3 and 2, inclusive, while still satisfying the given equations. The boundaries 2/3 and 2 represent the extreme values that x can take, considering the constraints imposed by both the linear and quadratic equations. Understanding this range provides a complete solution to the problem.
Importance of the Discriminant Condition
The use of the discriminant condition is a critical step in solving this problem. The discriminant, in essence, tells us about the nature of the roots of a quadratic equation. In the context of this problem, we needed y and z to be real numbers. The discriminant being non-negative (Δ ≥ 0) ensures that the roots (in this case, y and z) are real. Had we not applied this condition, we might have arrived at an incorrect range for x. This condition serves as a crucial filter, ensuring that our solution adheres to the constraints of the problem. The discriminant helps us transition from a general algebraic manipulation to a concrete condition that defines the feasible region for the values of x. Without this condition, we would not have a complete and accurate solution.
Conclusion: A Mathematical Exploration
In conclusion, we have successfully determined the range of x to be [2/3, 2] given the equations x + y + z = 4 and x² + y² + z² = 6. The solution involved a series of algebraic manipulations, including expressing variables in terms of each other, using the identity (y + z)² = y² + z² + 2yz, and applying the discriminant condition to ensure real roots. The problem illustrates the power of combining different algebraic techniques to solve systems of equations and inequalities. It also showcases the importance of understanding the constraints imposed by the equations, particularly the condition for real roots. This exploration provides a clear and comprehensive understanding of how to approach and solve similar mathematical problems. This kind of problem solving skill is invaluable in various fields beyond pure mathematics, including engineering, physics, and computer science.