Question

1. résous les équations polynômes suivantes :

a. $\mathbf{x^4 - 8x^2 + 16 = 0}$

Explanation:

Step1: Substitute \( y = x^2 \)

Let \( y = x^2 \), then the equation \( x^4 - 8x^2 + 16 = 0 \) becomes \( y^2 - 8y + 16 = 0 \).

Step2: Solve the quadratic equation

The quadratic equation \( y^2 - 8y + 16 = 0 \) can be factored as \( (y - 4)^2 = 0 \). Solving this, we get \( y = 4 \) (with multiplicity 2).

Step3: Substitute back \( y = x^2 \)

Since \( y = x^2 \) and \( y = 4 \), we have \( x^2 = 4 \).

Step4: Solve for \( x \)

Solving \( x^2 = 4 \), we get \( x = 2 \) or \( x = -2 \). Each of these solutions has multiplicity 2 because the original equation is a quartic (degree 4) and the quadratic in \( y \) had a double root, and the square root gives two solutions each with multiplicity 2. So the roots are \( x = 2 \) (multiplicity 2) and \( x = -2 \) (multiplicity 2).

Answer:

The solutions of the equation \( x^4 - 8x^2 + 16 = 0 \) are \( x = 2 \) (with multiplicity 2) and \( x = -2 \) (with multiplicity 2), or in set notation \( \{2, -2\} \) (where each root has multiplicity 2).