Solving Sets with Given Conditions: Analyzing a2b2c2 23 and a2b4c 22

When dealing with algebraic problems where multiple variables are involved, it's pivotal to break down complex expressions into manageable steps. This article provides a detailed exploration of the problem of finding sets of (a, b, c) that satisfy the equations a2b2c2 23 and a2b4c 22. By employing a variety of methods, including substitution and geometric interpretation, we aim to determine the feasibility and solution sets of these equations.

Algebraic Substitution and Equation Transformation

The given equations are:

a2b4c 22

This equation can be rearranged to express a in terms of b and c:

a 22 - 2b - 4c

We then substitute this expression for a into the second equation:

a2b2c2 23

This results in a complex polynomial:

(22 - 2b - 4c)2b2c2 23

Expanding the left-hand side gives:

5b4c4 - 88b3c3 - 176b2c2 576b2c2 - 88bc 576b2c2 - 176b2c2 - 88bc 576b2c2 - 176b2c2 23

Further simplification provides:

5b4c4 - 88b3c3 - 288b2c2 176b2 - 176bc 484 23

After subtraction, the equation is:

5b4c4 - 88b3c3 - 288b2c2 176b2 - 176bc 461 0

This high-degree polynomial has no real roots as determined by Wolfram Alpha, indicating no viable solution set for this approach.

Geometric Interpretation

This problem can be viewed in a geometric context. The first equation represents a sphere:

a2b2c2 23

With a midpoint at the origin (0, 0, 0) and a radius of:

r sqrt{23}

The second equation represents a plane:

a2b4c 22

This plane is orthogonal to the vector n (1, 2, 4), and its distance from the origin is:

dOP 22 / sqrt{21}

Comparing the radius of the sphere and the distance from the origin to the plane, we find:

dOP 22 / sqrt{21}

Since:

22sqrt{21}

We find that the sphere is too small to intersect the plane. Thus, no solution exists.

Alternative Approach: Substitution and Trial Method

By arbitrarily assigning a value to one of the variables, we can attempt to find solutions by trial and error. Let's take a 0 and solve the two equations for b and c:

a2b4c 22 becomes:

0 22, which is false.

This approach confirms the problem is inconsistent and no solutions exist for the arbitrary value of a 0.

Substituting other values for a such as a 2 and a 4 leads to similar contradictions. Therefore, the problem has no solutions.

Conclusion

The problem of finding sets of (a, b, c) that satisfy the given equations is non-existent. Through algebraic substitution and geometric interpretation, we have shown that these equations do not intersect in any meaningful way, leading to the conclusion that no solutions exist.