Solving the Quartic Equation x1x2x3x4 -1: Methods and Solutions

Solving the Quartic Equation (x_1x_2x_3x_4 -1): Methods and Solutions

In this article, we explore how to solve the given quartic equation (x_1x_2x_3x_4 -1). This problem involves advanced algebra and requires careful manipulation of polynomial expressions.

Introduction to the Problem

Consider the equation:

(x_1x_2x_3x_4 -1)

Algebraic Manipulation

One of the given equations can be rewritten as a matrix multiplication:

(left[x_1x_4right]cdotleft[x_2x_3right] -1)

Additionally, another related equation is:

(x^{25}x^4x^{25}x^6 -1)

Substitution and Simplification

Let us define a new variable (y x^{25}x^5). This simplifies our problem as follows:

(y-1y1 -1)

After further simplification:

(y^2 - 1 -1 implies y^2 0 implies y 0)

Substitute back the value of (y) into the original variable definition:

(x^{25}x^5 0)

From which we derive:

(x^{25}x -5)

Multiplying both sides by (x^{25}x^{25/4} -5^{25/4}), we get:

(left(frac{5}{2}right)^2 frac{25 - 20}{4} implies x^{5/2} pm frac{sqrt{5}}{2})

Therefore, the solutions are:

(x frac{-5 pm sqrt{5}}{2})

Another Approach using Polynomial Manipulation

Consider the equation:

(x_1x_4x_2x_3 -1)

We can see that:

(x_1x_2x_3x_4 -1)

Let us substitute

(d x^{5/2})

We now have:

(d - frac{3}{2}d - frac{1}{2}d^{1/2}d^{3/2} -1)

Which simplifies to:

(d^2 - frac{1}{4}d^2 - frac{9}{4} -1)

Further simplification gives:

(d^4 - frac{5}{2}d^2 frac{25}{16} 0)

This implies:

(d^2 - frac{5}{4} 0 implies d^2 frac{5}{4} implies d pm frac{sqrt{5}}{2})

Substituting back, we get:

(x^{5/2} pm frac{sqrt{5}}{2} implies x -frac{5}{2} pm frac{sqrt{5}}{2})

Thus, the solutions are:

(x -phisqrt{5} text{ and } x -frac{sqrt{5}}{phi})

Alternative Simplification and Solution

From the equation:

(x_1x_4x_2x_3 -1)

Divide both sides by (x_4x_2x_3), we get:

(x_1 -frac{1}{x_2x_3x_4})

Now, substituting (x_1) back into the original equation, we simplify to:

(x_1x_2x_3x_4 -1)

This equation can be simplified to:

(x^{25}x^4x^{25}x^6 -1)

Substituting (y x^{25}x^5), we get:

(y^2 - 2y - 1 0)
((y - 1)^2 0 implies y 1)

Thus, the solutions are:

(x frac{-5 pm sqrt{5}}{2})

Conclusion

The solutions to the quartic equation (x_1x_2x_3x_4 -1) are given by:

(x frac{-5 pm sqrt{5}}{2})

This demonstrates the power of algebraic manipulation and substitution in solving complex polynomial equations.

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