What is the solution set for the equation x² – 5x + 1 = 0?

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The equation you provided is a quadratic equation in the standard form of ax² + bx + c = 0. Here, a = 1, b = -5, and c = 1. To find the solution set of this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b² – 4ac)) / (2a)

Now, let’s calculate the values step by step:

  1. First, we need to calculate the discriminant (Δ):

    2. Δ = b² – 4ac = (-5)² – 4(1)(1) = 25 – 4 = 21

  2. Since the discriminant is positive, it indicates that there are two distinct real roots.
  3. Now, we can substitute back into the quadratic formula:

    4. x = (5 ± √21) / 2

  4. This gives us two solutions:

    5. 1. x₁ = (5 + √21) / 2

    2. x₂ = (5 – √21) / 2

Thus, the solution set for the equation x² – 5x + 1 = 0 is:

{(5 + √21) / 2, (5 – √21) / 2}

In decimal form, these roots approximately evaluate to:

1. x₁ ≈ 4.79

2. x₂ ≈ 0.21

In conclusion, the solution set of the equation x² – 5x + 1 = 0 comprises two decimal values: {4.79, 0.21}.

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