10. Quadratics & Polynomials
High SchoolCurved growth begins with the quadratic — the squared term that bends a line into a parabola. This chapter covers solving quadratics three ways, the discriminant, and the wider world of polynomials.
The quadratic formula and the discriminant
A quadratic has the form ax² + bx + c = 0 with a ≠ 0. It can be solved by factoring, by completing the square, or universally by the quadratic formula:
The quantity under the root, Δ = b² − 4ac, is the discriminant. It reveals the nature of the roots before you solve:
- Δ > 0 — two distinct real roots (parabola crosses the x-axis twice).
- Δ = 0 — one repeated real root (parabola touches the axis).
- Δ < 0 — no real roots (parabola misses the axis entirely).
The graph is a parabola with vertex at x = −b/(2a). A polynomial of degree n has at most n real roots, and its end behaviour is set by the leading term.
Where the formula comes from
The quadratic formula is not magic — it is completing the square done once, in general. Take ax² + bx + c = 0, divide by a, and rewrite the left side as a perfect square plus a correction. Solving that square root step produces exactly the ±√(b² − 4ac) over 2a that you memorise. Every quadratic you will ever solve is one instance of that single derivation.
The ± is why parabolas are symmetric: the two roots sit equal distances either side of the vertex line x = −b/(2a). And the discriminant is the "reality check": a negative under the square root means the answer needs imaginary numbers, so on the real plane the curve simply never reaches the axis.
Worked example: solve and complete the square
Solve 2x² − 4x − 6 = 0, then find the vertex.
- Divide through by 2: x² − 2x − 3 = 0.
- Factor: two numbers multiplying to −3, adding to −2 are −3 and +1 → (x − 3)(x + 1) = 0.
- Roots: x = 3 or x = −1.
- Vertex x-coordinate is midway: (3 + (−1))/2 = 1. Substitute: y = 1 − 2 − 3 = −4. Vertex (1, −4).
- Check with the discriminant of the original: Δ = (−4)² − 4(2)(−6) = 16 + 48 = 64 > 0 — two real roots, as found. ✓
Answer: roots x = 3, −1; vertex (1, −4).
Test your understanding
Solve x² + 4x + 5 = 0. What does the answer tell you?
Discriminant Δ = 16 − 20 = −4 < 0, so there are no real roots — the parabola never touches the x-axis (its vertex sits above it). Using complex numbers, x = (−4 ± √−4)/2 = −2 ± i, a pair of complex conjugates.
For a quadratic, how do the roots relate to b and c (when a = 1)?
By Vieta's formulas: if the roots are p and q, then p + q = −b and pq = c. For x² − 5x + 6, the roots 2 and 3 sum to 5 (= −(−5)) and multiply to 6. This gives a fast check and a fast way to build a quadratic from its roots.
Why can a cubic never have exactly two real roots (counting distinct values)?
Complex roots of a real polynomial always come in conjugate pairs. A cubic has 3 roots in total, and pairs are even, so the number of non-real roots is 0 or 2 — leaving 3 or 1 real roots. A continuous cubic also runs from −∞ to +∞, so it must cross the axis at least once. Hence a cubic has 1 or 3 real roots, never exactly 2 distinct ones (a repeated root can make it look like 2 values, but that is a double root).
Complete the square for x² + 6x + 1.
Half of 6 is 3; 3² = 9. Write x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8. This form instantly gives the vertex (−3, −8) and shows the minimum value is −8.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1Solve x^2 - 5x + 6 = 0.
- Factor into (x - 2)(x - 3) = 0
- Set each factor to zero
- x = 2 or x = 3
Example 2Solve 2x^2 + 3x - 2 = 0 with the formula.
- a=2, b=3, c=-2, discriminant = 9 + 16 = 25
- x = (-3 +/- 5)/4
- x = 1/2 or x = -2
Now you try
Work each one out first, then tap to reveal the worked answer.