Learning objectives
Complete the square and use it to find the vertex and range and to sketch the curve
Use the discriminant to determine the nature of roots
Solve quadratic inequalities
Solve simultaneous equations where one is linear and one quadratic, and equations that are quadratic in a function of x
Key definitions
Completed square form
a(x + p)² + q, whose vertex is at (−p, q).
Discriminant
b² − 4ac: positive gives two distinct real roots, zero a repeated root, negative no real roots.
Line of symmetry
The vertical line x = −p through the vertex of y = a(x + p)² + q; the curve is a mirror image of itself in this line.
Explanation
Completing the square converts ax² + bx + c into a(x + p)² + q. The minimum (a > 0) or maximum (a < 0) value is q, reached at x = −p — this gives the vertex, the line of symmetry and the range in one step, and it is the form examiners want for "express in the form" questions.
The discriminant decides how many times the parabola meets the x-axis. Questions about a line meeting a curve reduce to this: substitute, form a quadratic, and set b² − 4ac > 0, = 0 (tangent) or < 0 (no intersection). Quadratic inequalities are solved by finding the roots and sketching: the sign of a tells you whether the "between" or "outside" region is required.
Sketching a quadratic curve. The completed square gives the shape at once. For y = x² − 4x − 5 = (x − 2)² − 9, the curve is ∪-shaped because a > 0, its vertex (the minimum point) is (2, −9) and its line of symmetry is x = 2. Put x = 0 for the y-intercept, (0, −5), and solve y = 0 for the x-intercepts: (x − 5)(x + 1) = 0 gives (−1, 0) and (5, 0), which lie symmetrically either side of x = 2. If a < 0 the curve is ∩-shaped with a maximum point. When the discriminant is negative the curve does not meet the x-axis, so the sketch shows only the vertex and the y-intercept. Label the vertex and every intercept with its coordinates.
A linear and a quadratic equation together. Make x or y the subject of the linear equation and substitute into the quadratic one; this leaves a quadratic in a single variable. For x + y = 5 and x² + y² = 17, y = 5 − x gives x² + (5 − x)² = 17, so 2x² − 10x + 8 = 0, x² − 5x + 4 = 0 and x = 1 or x = 4. Substitute each value back into the linear equation to find its partner: the solutions are (1, 4) and (4, 1). Always give the solutions as pairs; two separate lists of x-values and y-values lose the final mark.
Equations that are quadratic in a function of x. If an equation contains only some function u of x and its square, substitute u and solve a quadratic. For 2x⁴ − 7x² + 3 = 0 let u = x²: 2u² − 7u + 3 = 0 gives (2u − 1)(u − 3) = 0, so x² = ½ or x² = 3 and x = ±1/√2 or x = ±√3. For x − 5√x + 6 = 0 let u = √x: u² − 5u + 6 = 0 gives u = 2 or u = 3, so x = 4 or x = 9. For 3 tan²θ − 4 tan θ + 1 = 0 let u = tan θ, which gives tan θ = 1 or tan θ = ⅓. Always return to x at the end, and reject any value that the function cannot take: √x cannot be negative, and neither can x².
Formulae
x = [−b ± √(b² − 4ac)] ÷ 2a
ax² + bx + c = a(x + b/2a)² + c − b²/4a
Two distinct real roots ⇔ b² − 4ac > 0
Vertex of y = a(x + p)² + q: (−p, q); line of symmetry x = −p
Hidden quadratic: substitute u = x², u = √x or u = tan x, solve for u, then return to x
Worked examples
q: Express 2x² − 12x + 7 in the form a(x + p)² + q and state the minimum value.
steps: Factor out 2 from the x terms: 2(x² − 6x) + 7
Complete the square inside: 2[(x − 3)² − 9] + 7
Expand: 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11
Minimum value −11 at x = 3.
In context
Projectile paths are parabolas; the vertex gives the maximum height.
Break-even analysis with quadratic revenue functions uses the discriminant to test whether profit is ever positive.
Common misconceptions
Forgetting to multiply the "−9" by the factor taken out.
Reversing the inequality region: for a > 0, "< 0" is the region between the roots.
Using ≥ instead of > for "two distinct real roots".
Students often stop at the values of u after substituting u = x² or u = √x. In fact the answer must be given in x, and a negative value of u = √x or u = x² has to be rejected.
Study tips
Always check the completed square by expanding it.
Sketch the parabola for inequalities — it prevents region errors.
Summary
Completed square gives vertex, symmetry, range.
Discriminant classifies roots and line–curve intersections.
Solve inequalities by roots plus a sketch.
Linear with quadratic: substitute, solve, then pair each x with its y.
Hidden quadratics: substitute u = x², √x or tan x, then return to x.
Check your understanding
Question 1: The discriminant of x² + 4x + 5 is…
1. 4
2. −4
3. 36
Question 2: The minimum value of (x − 2)² + 3 is…
1. 3
2. 2
3. −3