Classify and solve quadratic equations

Try

Solve x² − 6x + 5 = 0 by recognizing (x − 3)² − 4 = 0. The square must equal 4, so x − 3 equals 2 or −2. The roots are 5 and 1. Explain why keeping only the positive square-root branch loses a solution.

Predict

Compare (x − 3)² = 4, (x − 3)² = 0 and (x − 3)² = −4. Predict the number of distinct real roots in each: two, one and none. The final equation will require a larger number system.

Experiment

Sketch y = (x − 3)² and horizontal lines at heights 4, 0 and −4. Their intersections show the real solutions. A nonreal solution is not another hidden crossing on this real-coordinate graph.

Understand

For ax² + bx + c = 0 with a nonzero, a, b and c are real coefficients and x is the unknown. Complete the square in three steps:

1. Divide by a and move the constant: x² + (b/a)x = −c/a.
2. Add (b/(2a))² to both sides: x² + (b/a)x + b²/(4a²) = −c/a + b²/(4a²).
3. Factor the left and use a common denominator on the right: (x + b/(2a))² = (b² − 4ac)/(4a²).

Taking both square-root branches and isolating x gives the quadratic formula:

x = (−b ± √(b² − 4ac))/(2a).

The square root of 4a² is 2|a|. Because both signs are included, using 2a instead gives the same pair of roots, even when a is negative.

The symbol ± means that both the plus and minus expressions must be considered. The discriminant D = b² − 4ac determines the real-root cases. Positive D gives two distinct real roots, zero D gives one repeated root, and negative D gives no real roots. A positive discriminant need not be a perfect square; irrational roots still count as exact real solutions.

For 3x² + x − 2 = 0, D = 1 − 4(3)(−2) = 25. The roots are (−1 + 5)/6 = 2/3 and (−1 − 5)/6 = −1. Substitution verifies both.

Define i by i² = −1. For x² + 4 = 0, the complex roots are 2i and −2i. Their squares equal −4. More generally negative D produces a conjugate pair with opposite imaginary parts. These are solutions over the complex numbers, not real graph intersections.

If a = 0, the quadratic formula is invalid because it divides by 2a. Return to bx + c = 0 and classify it as linear, an identity or a contradiction.

Apply

The rational-root questions accept exact fractions. Irrational and complex answers use choices or a written rubric, not rounded decimal matching. Show coefficients with their signs before computing D, and check candidates in the original polynomial.

Continue

This finishes the first algebra foundations unit. It does not complete Algebra I or Algebra II: systems, function families, broader polynomial/rational work, exponential/logarithmic reasoning and cumulative modeling need further sequences. Revisit the domain and reversibility lessons whenever a new method produces candidates rather than guaranteed solutions.