Quadratic Functions and Equations

Master quadratic functions at A-Level: completing the square, the discriminant, sketching parabolas, and solving quadratic inequalities.

Quadratic functions and equations are among the most important topics in A-Level Mathematics. Building on the foundations laid at GCSE, the A-Level treatment requires a deeper understanding of the algebraic structure of quadratics, their graphical properties, and the connections between different forms of the same expression.

At AS Level, exam boards (AQA, Edexcel, OCR) expect you to factorise quadratics, complete the square, use the discriminant to analyse roots, sketch parabolas with key features labelled, and solve quadratic inequalities. These skills recur throughout pure mathematics — in differentiation, integration, coordinate geometry, and beyond.

A quadratic function has the general form f(x)=ax2+bx+cf(x) = ax^2 + bx + c where a0a \neq 0. Its graph is a parabola: opening upwards when a>0a > 0 and downwards when a<0a < 0. This guide covers all the techniques you need, with rigorous worked examples and exam-focused advice.

Core Concepts

The Quadratic Formula

The solutions (roots) of ax2+bx+c=0ax^2 + bx + c = 0 are given by:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula is derived by completing the square on the general quadratic and is guaranteed to find the roots whenever they exist. You should memorise this formula — it appears on the AQA and OCR formula sheets but not always on Edexcel's.

The Discriminant

The expression Δ=b24ac\Delta = b^2 - 4ac is called the discriminant. It determines the nature of the roots:

  • Δ>0\Delta > 0: two distinct real roots
  • Δ=0\Delta = 0: one repeated real root (the parabola touches the xx-axis)
  • Δ<0\Delta < 0: no real roots (the parabola does not cross the xx-axis)

The discriminant is a powerful tool for answering questions about whether a line intersects a curve, whether an equation has solutions, or whether a quadratic is always positive (or always negative).

Completing the Square

Any quadratic ax2+bx+cax^2 + bx + c can be written in the form a(x+p)2+qa(x + p)^2 + q. This is called completing the square and reveals the vertex (turning point) of the parabola.

For f(x)=x2+bx+cf(x) = x^2 + bx + c:

f(x)=(x+b2)2b24+cf(x) = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + c

For f(x)=ax2+bx+cf(x) = ax^2 + bx + c where a1a \neq 1, first factor out aa from the first two terms:

f(x)=a(x2+bax)+c=a(x+b2a)2b24a+cf(x) = a\left(x^2 + \frac{b}{a}x\right) + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c

The vertex is at (b2a,cb24a)\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right).

Completing the square also allows you to express the minimum (or maximum) value of a quadratic function, which is essential for optimisation problems.

Factorising Quadratics

A quadratic ax2+bx+cax^2 + bx + c can be factorised if and only if Δ0\Delta \geq 0. Methods include:

  • Inspection: Find two numbers that multiply to give acac and add to give bb.
  • Grouping: For a1a \neq 1, split the middle term using the two numbers found above.
  • Using roots: If the roots are α\alpha and β\beta, then ax2+bx+c=a(xα)(xβ)ax^2 + bx + c = a(x - \alpha)(x - \beta).

Sketching Parabolas

To sketch y=ax2+bx+cy = ax^2 + bx + c, identify:

  1. Shape: a>0a > 0 gives a U-shape; a<0a < 0 gives an inverted U.
  2. yy-intercept: Set x=0x = 0 to get y=cy = c.
  3. xx-intercepts (roots): Solve ax2+bx+c=0ax^2 + bx + c = 0 using factorisation, completing the square, or the quadratic formula.
  4. Vertex (turning point): Use completed square form or (b2a,f(b2a))\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right).
  5. Line of symmetry: x=b2ax = -\frac{b}{2a}.

Quadratic Inequalities

To solve an inequality such as ax2+bx+c>0ax^2 + bx + c > 0:

  1. Find the roots of ax2+bx+c=0ax^2 + bx + c = 0.
  2. Sketch the parabola (you only need a rough sketch).
  3. Read off the solution from the sketch.

For a>0a > 0:

  • ax2+bx+c>0ax^2 + bx + c > 0 is satisfied where the parabola is above the xx-axis (outside the roots): x<αx < \alpha or x>βx > \beta.
  • ax2+bx+c<0ax^2 + bx + c < 0 is satisfied where the parabola is below the xx-axis (between the roots): α<x<β\alpha < x < \beta.

For a<0a < 0, the regions are reversed.

The Intersection of a Line and a Curve

To find where a line y=mx+dy = mx + d meets a curve y=ax2+bx+cy = ax^2 + bx + c, set them equal:

ax2+bx+c=mx+dax^2 + bx + c = mx + d

ax2+(bm)x+(cd)=0ax^2 + (b - m)x + (c - d) = 0

The discriminant of this equation tells you about the intersection:

  • Δ>0\Delta > 0: the line cuts the curve at two points
  • Δ=0\Delta = 0: the line is tangent to the curve
  • Δ<0\Delta < 0: the line does not meet the curve

Strategy Tips

Tip 1: Know When to Use Each Method

Factorisation is quickest when it works. The quadratic formula always works for real roots. Completing the square is best when you need the vertex or must express a minimum/maximum value.

Tip 2: Sketch Before Solving Inequalities

A quick sketch of the parabola makes inequalities straightforward. Without it, students often write the wrong regions or mix up strict and non-strict inequalities.

Tip 3: Use the Discriminant as a Condition

Many A-Level questions ask you to find values of a parameter kk such that an equation has certain properties (e.g., "no real roots" or "a repeated root"). Set up the discriminant condition and solve the resulting inequality or equation.

Tip 4: Present Completed Square Form Neatly

When completing the square, show each step clearly. Examiners award method marks for the intermediate stages, so even if you make an arithmetic error, you can still earn most of the marks.

Tip 5: Check with Substitution

After solving a quadratic equation, substitute your roots back into the original equation to verify they satisfy it. This takes only a few seconds and catches sign errors.

Worked Example: Example 1

Problem

Solve 2x25x3=02x^2 - 5x - 3 = 0 by factorisation.

Solution

We need two numbers that multiply to 2×(3)=62 \times (-3) = -6 and add to 5-5. These are 6-6 and 11.

Split the middle term: 2x26x+x3=02x^2 - 6x + x - 3 = 0

Group: 2x(x3)+1(x3)=02x(x - 3) + 1(x - 3) = 0

(2x+1)(x3)=0(2x + 1)(x - 3) = 0

So x=12x = -\frac{1}{2} or x=3x = 3.

Worked Example: Example 2

Problem

Write f(x)=x26x+11f(x) = x^2 - 6x + 11 in completed square form and hence state the minimum value of f(x)f(x).

Solution

f(x)=x26x+11=(x3)29+11=(x3)2+2f(x) = x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2

Since (x3)20(x - 3)^2 \geq 0 for all real xx, the minimum value of f(x)f(x) is 22, occurring when x=3x = 3.

The vertex of the parabola is at (3,2)(3, 2).

Worked Example: Example 3

Problem

Find the values of kk for which the equation x2+kx+9=0x^2 + kx + 9 = 0 has a repeated root.

Solution

For a repeated root, the discriminant equals zero:

Δ=k24(1)(9)=0\Delta = k^2 - 4(1)(9) = 0

k2=36k^2 = 36

k=±6k = \pm 6

Worked Example: Example 4

Problem

Solve the inequality x22x8<0x^2 - 2x - 8 < 0.

Solution

First, solve x22x8=0x^2 - 2x - 8 = 0:

(x4)(x+2)=0(x - 4)(x + 2) = 0

So the roots are x=2x = -2 and x=4x = 4.

Since the coefficient of x2x^2 is positive, the parabola is U-shaped. The quadratic is negative (below the xx-axis) between the roots.

Solution: 2<x<4-2 < x < 4.

Worked Example: Example 5

Problem

The line y=2x+ky = 2x + k is tangent to the curve y=x2+3y = x^2 + 3. Find the value of kk.

Solution

Set equal: x2+3=2x+kx^2 + 3 = 2x + k, giving x22x+(3k)=0x^2 - 2x + (3 - k) = 0.

For tangency, Δ=0\Delta = 0:

(2)24(1)(3k)=0(-2)^2 - 4(1)(3 - k) = 0

412+4k=04 - 12 + 4k = 0

4k=84k = 8

k=2k = 2

Practice Problems

  1. Problem 1

    Solve 3x2+7x+2=03x^2 + 7x + 2 = 0 by factorisation. [Answer: x=13x = -\frac{1}{3} or x=2x = -2]

    Problem 2

    Write 2x2+8x+32x^2 + 8x + 3 in the form a(x+p)2+qa(x + p)^2 + q. [Answer: 2(x+2)252(x + 2)^2 - 5]

    Problem 3

    Find the range of values of kk for which x2+4x+k=0x^2 + 4x + k = 0 has two distinct real roots. [Answer: k<4k < 4]

    Problem 4

    Solve 2x2+3x502x^2 + 3x - 5 \geq 0. [Answer: x52x \leq -\frac{5}{2} or x1x \geq 1]

    Problem 5

    The curve y=x2+2x+cy = x^2 + 2x + c does not cross the xx-axis. Find the range of values of cc. [Answer: c>1c > 1]

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Common Mistakes

  • Sign errors when completing the square. The most common error is writing (x+3)2=x2+3x+9(x + 3)^2 = x^2 + 3x + 9 instead of the correct x2+6x+9x^2 + 6x + 9. Remember: (x+p)2=x2+2px+p2(x + p)^2 = x^2 + 2px + p^2.

  • Forgetting to reverse the inequality when multiplying by a negative. When a<0a < 0, dividing through changes the direction of the inequality.

  • Writing inequality solutions incorrectly. For x22x8<0x^2 - 2x - 8 < 0, the solution is 2<x<4-2 < x < 4 (a single interval), not x<2x < -2 and x<4x < 4.

  • Misusing the discriminant. Δ>0\Delta > 0 means two distinct roots, not "two positive roots". The sign of the roots depends on Vieta's formulae, not the discriminant alone.

  • Not factoring out the leading coefficient before completing the square. When a1a \neq 1, you must factor aa from the x2x^2 and xx terms first.

  • Confusing "no real roots" with "no roots". A quadratic with Δ<0\Delta < 0 has no real roots but does have two complex roots. At AS Level, you work with real numbers only, so "no real roots" is the correct phrase.

Frequently Asked Questions

When should I use the quadratic formula instead of factorising?

If the quadratic doesn't factorise easily (especially when the roots are irrational or the coefficients are large), use the formula. In exam conditions, if you can't spot factors within a minute, move to the formula.

What does the discriminant tell us geometrically?

The discriminant tells you how many times the parabola crosses the xx-axis. Two crossings (Δ>0\Delta > 0), one touching point (Δ=0\Delta = 0), or no crossings (Δ<0\Delta < 0).

How do I write the solution to a quadratic inequality?

Use interval notation or inequalities. For example, x<2x < -2 or x>4x > 4 (two separate regions), or 2<x<4-2 < x < 4 (one region between the roots). Never use "and" when you mean "or" for disjoint intervals.

Can I always complete the square?

Yes. Every quadratic can be written in completed square form. This works whether or not the quadratic has real roots.

How do completing the square and the quadratic formula relate?

The quadratic formula is derived by completing the square on ax2+bx+c=0ax^2 + bx + c = 0. The formula is simply the result of applying the completing the square technique to the general case.

Key Takeaways

  • Three forms, one function. A quadratic can be expressed as ax2+bx+cax^2 + bx + c (expanded), a(xα)(xβ)a(x - \alpha)(x - \beta) (factorised), or a(x+p)2+qa(x + p)^2 + q (completed square). Each form reveals different information.

  • The discriminant is a decision tool. b24acb^2 - 4ac instantly tells you about the nature of the roots and the behaviour of the parabola. Use it to answer "show that" and parameter questions.

  • Completing the square reveals the vertex. The turning point of y=a(x+p)2+qy = a(x + p)^2 + q is (p,q)(-p, q). This gives you the minimum or maximum value directly.

  • Sketch for inequalities. A rough parabola sketch turns inequality problems from confusing to straightforward. Always identify the roots first, then read off the solution.

  • Quadratics appear everywhere. This topic underpins coordinate geometry, calculus, mechanics, and more. Mastering it thoroughly saves time across the entire A-Level course.

  • Check your work. Substitute roots back into the equation, differentiate to verify completed square forms, and use the discriminant to confirm the number of roots.

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