Solving Quadratic Equations

Chapter 5: Quadratic Equations and the Discriminant

Learning objectives

Solve quadratics by factoring, completing the square, and the quadratic formula
Compute and interpret the discriminant
Understand the relationship between roots and coefficients

Linear equations are a one-trick pony: isolate the variable. Quadratic equations are the first time you meet an equation where isolation alone does not work — $x^2$ and $x$ both contain the unknown, and you cannot pull them apart by simple subtraction. The four big ideas of this section (factoring, completing the square, the quadratic formula, and the discriminant) all amount to one strategy: rewrite the equation in a form where the variable IS isolated, namely as a perfect square equal to a number. A quadratic equation has the standard form $ax^2 + bx + c = 0$ where $a \neq 0$ .

The Quadratic Formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The expression $\Delta = b^2 - 4ac$ is the discriminant:

If $\Delta > 0$ : two distinct real roots. If $\Delta = 0$ : one repeated root. If $\Delta < 0$ : no real roots (two complex roots).

Completing the Square

ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}

Vieta's formulas relate roots $r_1, r_2$ to coefficients: $r_1 + r_2 = -\frac{b}{a}$ , $r_1 \cdot r_2 = \frac{c}{a}$ . This means you can find the sum and product of roots without solving.

Factoring: If you can find $r_1, r_2$ such that $r_1 + r_2 = -b/a$ and $r_1 r_2 = c/a$ , then $ax^2 + bx + c = a(x - r_1)(x - r_2)$ .

(Drag a, b, c in the widget. Watch the parabola morph, the roots slide along the x-axis, and the discriminant flip sign at the moment the parabola becomes tangent to the x-axis.)

Try it

In the widget, set $a =$ , $b =$ , $c =$ . Before setting these: what are the roots of $x^2 - 5x + 6 = 0$ ? Set the coefficients in the widget to verify, and confirm using Vieta's formulas.
Drag the coefficients until the discriminant equals zero exactly. What does the parabola do at that moment?
Solve $2x^2 - 5x + 3 = 0$ by the quadratic formula. Then verify with Vieta's formulas.
Drag the coefficients until the parabola has no real roots (discriminant $\Delta < 0$ ).

Try it in code

A trap to watch for

A common error: applying the quadratic formula to $(x - 1)^2 = 4$ . That is not in standard form yet. Expand first: $x^2 - 2x - 3 = 0$ , THEN read off $a = 1, b = -2, c = -3$ and apply the formula. Another sneakier trap: when $a < 0$ , students sometimes forget that $a$ stays negative throughout. The fix: always rewrite the equation as $ax^2 + bx + c = 0$ before extracting $a, b, c$ , and keep signs explicit.

What you now know

You can solve any quadratic by factoring (when easy), completing the square (always works), or the quadratic formula (when factoring is hard). The discriminant tells you the number of real roots before you commit to a method, and Vieta's formulas link roots to coefficients. The Intermezzo chapter that follows steps back from computation to discuss how mathematics is read, written, and reasoned about.

Quick check

References

Lang, S. (1971). Basic Mathematics. Springer. Chapter 4, §1 — quadratic equations, completing the square, and the discriminant.
Stewart, J. (2015). Calculus: Early Transcendentals, 8th ed. Cengage. Appendix C — the quadratic formula derived from completing the square step by step.
Garrity, T. (2002). All the Mathematics You Missed. Cambridge. Chapter 1 — quadratics as the bridge to algebraic geometry, with Vieta's formulas as the first symmetric-function identity.