Section 3.cq

Complex Numbers & Quadratics

Complex numbers and quadratics are heavily tested on the final — expect 3–4 questions on completing the square, the quadratic formula, and complex arithmetic. If you can nail vertex form and the discriminant, you're in good shape for this chunk of the exam.

⚡ Quick Summary

A quadratic function creates a U-shaped curve called a parabola — think of the path a ball traces when you toss it in the air. The highest (or lowest) point of that curve is the vertex, which tells you the maximum or minimum value. If the U opens upward, the vertex is the bottom; if it opens downward, the vertex is the top.

Standard Form of a Complex Number

z = a + bi, \quad i^2 = -1

Complex Conjugate Product

(a + bi)(a - bi) = a^2 + b^2

Multiply by the conjugate to clear i from a denominator

📋 Before You Start

Tap any item if you need a refresher:

Plain-English version

Complex numbers were invented to answer a question that used to have no answer: "What number, multiplied by itself, gives a negative result?" Mathematicians created $i$ (where $i^2 = -1$ ) to fill that gap. A complex number like $3 + 2i$ is just a pair: a regular part (3) and an imaginary part ( $2i$ ). You do arithmetic with them almost exactly like regular algebra — the only trick is replacing $i^2$ with $-1$ whenever it appears.

The quadratic formula is a universal key that unlocks any quadratic equation. The piece under the square root, called the discriminant, is like a preview: if it's positive you get two normal answers, if it's zero you get one repeated answer, and if it's negative the answers involve $i$ (complex numbers).

Completing the square is a technique that rewrites a quadratic into a form where you can read the vertex directly. It's like rearranging a messy room so everything is in its place — once you see the pattern, the vertex (and everything else) becomes obvious.

Key Formulas

Standard Form of a Complex Number

z = a + bi, \quad i^2 = -1

Complex Conjugate Product

(a + bi)(a - bi) = a^2 + b^2

Multiply by the conjugate to clear i from a denominator

Vertex Form of a Quadratic

f(x) = a(x - h)^2 + k

Vertex at (h, k); axis of symmetry x = h

Quadratic Formula

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

Discriminant

\Delta = b^2 - 4ac

Δ > 0 → two real roots; Δ = 0 → one repeated root; Δ < 0 → two complex roots

Key Takeaways

✓Complex numbers: $i^2 = -1$ ; add/subtract by combining like terms; multiply with FOIL and replace $i^2$ with $-1$
✓To divide complex numbers, multiply top and bottom by the conjugate of the denominator
✓Vertex form $a(x - h)^2 + k$ gives vertex $(h, k)$ — complete the square to convert from standard form
✓Discriminant $b^2 - 4ac$ : positive → 2 real roots, zero → 1 repeated root, negative → 2 complex roots

⚠️ Common Mistakes

✗Forgetting the negative sign on $b$ in the quadratic formula — when $b = -5$ , you get $-(-5) = 5$ , not $-5$
✗Writing $i^2 = 1$ instead of $i^2 = -1$ — the whole point of $i$ is that its square is negative one
✗When completing the square, adding a number inside the parentheses but forgetting to balance the other side — if you add $9$ inside, add $9$ (times the leading coefficient) to the other side too
✗Reading the discriminant $b^2 - 4ac$ as the answer instead of taking its square root — $\sqrt{b^2 - 4ac}$ goes in the quadratic formula, not $b^2 - 4ac$ itself

Next: Section 3.poly Builds on: 2.lin

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2.lin Linear Functions

3.poly Polynomials