(x^2-x+1)^4+4x^2(x^2-x+1)^2=5x^4

Jasonbradley

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Jun 17, 2024

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Solving the Equation (x^2-x+1)^4+4x^2(x^2-x+1)^2=5x^4

This equation looks complicated, but we can solve it by using some clever algebraic manipulations.

Making a Substitution

Let's start by making a substitution to simplify the equation. Let y = x^2 - x + 1. Now our equation becomes:

y^4 + 4x^2y^2 = 5x^4

Rearranging and Factoring

We can rearrange this equation to look like a quadratic:

y^4 + 4x^2y^2 - 5x^4 = 0

Now we can factor the left-hand side:

(y^2 + 5x^2)(y^2 - x^2) = 0

Solving for y

This gives us two possible equations:

1. y^2 + 5x^2 = 0
2. y^2 - x^2 = 0

Let's analyze each equation:

• Equation 1: y^2 + 5x^2 = 0. Since both y^2 and 5x^2 are always non-negative, the only way this equation can hold is if both y^2 and 5x^2 are equal to zero. This means y = 0 and x = 0.

• Equation 2: y^2 - x^2 = 0. This can be factored as (y + x)(y - x) = 0. This leads to two possibilities:

  • y + x = 0 => y = -x
  • y - x = 0 => y = x

Substituting Back and Solving for x

Now we need to substitute back our original value for y (y = x^2 - x + 1) and solve for x:

• Case 1: y = 0

  • x^2 - x + 1 = 0. This quadratic equation has no real solutions.

• Case 2: y = -x

  • x^2 - x + 1 = -x. This simplifies to x^2 + 1 = 0, which has no real solutions.

• Case 3: y = x

  • x^2 - x + 1 = x. This simplifies to x^2 - 2x + 1 = 0. Factoring this gives us (x - 1)^2 = 0, leading to x = 1.

Conclusion

The only real solution to the equation (x^2-x+1)^4+4x^2(x^2-x+1)^2=5x^4 is x = 1.