A Level Mathematics: Quadratics — Revision Notes

Condensed for the final weeks. For the full explanation, use the Quadratics study guide.

Completed square form

a x^2 + b x + c  =  a(x + p)^2 + q

Reading it off immediately:

• Vertex at (−p, q).
• Line of symmetry x = −p.
• If a > 0, q is the minimum value; if a < 0, it is the maximum.

Method when a ≠ 1: factor a out of the x² and x terms only, complete the square inside, then multiply back.

2x^2 - 12x + 5  =  2(x^2 - 6x) + 5  =  2[(x-3)^2 - 9] + 5  =  2(x-3)^2 - 13

Vertex (3, −13), minimum value −13. Forgetting to multiply the −9 by 2 is the classic slip.

Solving quadratics

Three connected methods, all examinable: factorising (fastest when the quadratic factorises neatly into two linear brackets), completing the square (gives the vertex directly, useful for turning-point questions), and the quadratic formula, which always works regardless of whether the expression factorises:

x = (-b +/- sqrt(b^2 - 4ac)) / (2a)

The quadratic formula always works, even when factorising fails — use it whenever factors are not obvious rather than spending time hunting for them.

Worked example. Solve 3x² + 5x − 2 = 0 using the quadratic formula.

x = (-5 +/- sqrt(25 + 24)) / 6 = (-5 +/- 7) / 6
x = 1/3  or  x = -2

Checking mentally that a = 3, b = 5, c = −2 are substituted correctly avoids the most common slip — a sign error inside the square root.

The discriminant

b^2 - 4ac  >  0    two distinct real roots
b^2 - 4ac  =  0    one repeated root  (curve TOUCHES the x-axis)
b^2 - 4ac  <  0    no real roots       (curve never meets the x-axis)

Almost every “find the values of k” question is a discriminant question in disguise, since introducing an unknown constant k into a quadratic and asking about its roots is really asking about the discriminant. The standard patterns:

• Tangent to a line → substitute, form a quadratic, set the discriminant to zero.
• Two intersections → discriminant > 0.
• Never touches / no intersection → discriminant < 0.
• Always positive → a > 0 and discriminant < 0. Both conditions are needed; giving only the discriminant loses a mark.

Quadratic inequalities

Always sketch. Solving algebraically without a sketch produces wrong inequality directions more often than not, since the direction of the inequality depends on whether the parabola is above or below the axis in the region of interest.

1. Rearrange so one side is zero.
2. Factorise to find the critical values.
3. Sketch the parabola.
4. Read off the region.

(x - 2)(x - 5) < 0    ->    2 < x < 5        (below the axis, BETWEEN the roots)
(x - 2)(x - 5) > 0    ->    x < 2  or  x > 5  (above the axis, OUTSIDE the roots)

For an upward parabola: “less than zero” gives one interval between the roots; “greater than zero” gives two separate intervals. Writing 5 < x < 2 is impossible — if the answer looks like that, it should be two intervals joined by “or”.

Disguised quadratics

If an equation contains a term and its square, substitute.

x^4 - 5x^2 + 4 = 0        let u = x^2      ->  u^2 - 5u + 4 = 0
x - 7 sqrt(x) + 12 = 0    let u = sqrt(x)  ->  u^2 - 7u + 12 = 0
2^(2x) - 5(2^x) + 4 = 0   let u = 2^x      ->  u^2 - 5u + 4 = 0

Two things must follow: substitute back to find x, and reject impossible values — √x cannot be negative, and 2^x is always positive for real x. Marks are routinely lost by stopping at u, or by keeping a negative root for u = √x.

Exam traps

• Not multiplying q by a when completing the square with a ≠ 1.
• Sign error on the vertex: a(x + p)² + q has vertex at (−p, q).
• Solving inequalities without sketching.
• Writing an impossible double inequality instead of two intervals.
• Giving only the discriminant condition for “always positive”.
• Forgetting to substitute back, or keeping invalid roots.

Self-test

1. Write 2x² − 12x + 5 in completed square form and state the minimum point.
2. What does b² − 4ac = 0 mean geometrically?
3. Solve x² − 7x + 10 > 0.
4. What two conditions make ax² + bx + c positive for all x?
5. Solve x − 7√x + 12 = 0.
6. Solve 3x² + 5x − 2 = 0 using the quadratic formula.

Answers: 1. 2(x − 3)² − 13, minimum at (3, −13). 2. The curve touches the x-axis at exactly one point — a repeated root, so the line is a tangent. 3. Roots 2 and 5; the parabola is above the axis outside them, so x < 2 or x > 5. 4. a > 0 and b² − 4ac < 0. 5. Let u = √x: u² − 7u + 12 = 0, so u = 3 or 4, both valid as they are positive; x = 9 or x = 16. 6. x = (−5 ± √(25+24)) ÷ 6 = (−5 ± 7) ÷ 6, giving x = 1/3 or x = −2.