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Topic: Solving for the Roots of a Quadratic Equation ax² + bx + c = 0

Problem Description
A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). Solving for its solutions (roots) is a fundamental problem in algebra, requiring the determination of the nature of the roots (real or complex) based on the value of the discriminant, followed by the application of the quadratic formula to calculate their specific values.

Solution Process

  1. Discriminant Analysis
    First, calculate the discriminant \(\Delta\) (Delta):

\[ \Delta = b^2 - 4ac \]

  • If \(\Delta > 0\), the equation has two distinct real roots;
  • If \(\Delta = 0\), the equation has two equal real roots (a repeated root);
  • If \(\Delta < 0\), the equation has two conjugate complex roots.
  1. Case-based Solution
    • Case 1: \(\Delta \geq 0\) (Real Roots)
      Directly use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]

 If $ \Delta = 0 $, the formula simplifies to $ x = -\frac{b}{2a} $.
  • Case 2: \(\Delta < 0\) (Complex Roots)
    Express \(\sqrt{\Delta}\) in imaginary form:

\[ \sqrt{\Delta} = i\sqrt{|\Delta|} \quad (i \text{ is the imaginary unit, } i^2 = -1) \]

 Substitute into the quadratic formula:

\[ x = \frac{-b \pm i\sqrt{|\Delta|}}{2a} \]

  1. Example Verification

    • Example 1: \(2x^2 - 4x + 2 = 0\)
      \(\Delta = (-4)^2 - 4 \times 2 \times 2 = 0\) → repeated root \(x = \frac{4}{4} = 1\).

    • Example 2: \(x^2 + 2x + 5 = 0\)
      \(\Delta = 4 - 20 = -16\) → complex roots \(x = \frac{-2 \pm 4i}{2} = -1 \pm 2i\).

Key Points Summary

  • The discriminant determines the type of roots and must be calculated first;
  • Complex roots always appear in conjugate pairs; pay attention to the sign of the imaginary part during calculation.