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Quadratic inequalities

When is it above or below?

  • In 3 of 20 recent exams
  • Paper 1 or 2
  • 2 to 4 marks
How it is set in the exam

Paper 1 (no calculator), 2–4 marks a part, as a stand-alone question or the interval step of a longer one. Typical (3 of 20 exam versions, 4 parts, all Paper 1 (no calculator)). On Paper 2 (calculator) it would sit inside a context ("for how long…", Example 3.2), but no such part has been set since 2021.

Spot it in the exam
Exam phrases to spot (3)
  • “solve …>…\ldots > \ldots”

  • “find the values of xx for which…”

  • “for what values of xx…”

In disguise (6)
  • "f(x)>g(x)f(x) > g(x)" or "the graph of ff is above the graph of gg": subtract first
  • "above / below a height of…", "more than / less than…" in a context
  • "for how long…": the answer is the difference between two roots
  • "ff is increasing" when f′(x)f'(x) is a quadratic (Where is f increasing or decreasing?)
  • two conditions at once, such as where ff is negative and increasing: the sign gives the interval between (or outside) the roots, and increasing or decreasing is one side of the vertex (x>hx > h is the increasing side when a>0a > 0). The answer is the overlap, e.g. from the vertex's x-coordinate to the larger root. f′(x)>0f'(x) > 0 also scores, but the vertex is quicker when an earlier part has found it.
  • where the log of a quadratic is defined, e.g. ln⁡(8+2x−x2)\ln\left(8 + 2x - x^2\right): its argument must be positive (Logarithm graphs)
Looks like it, but isn't (2)
  • an inequality in kk that comes from Δ\Delta. That is Find k from the number of roots, where the inequality is one step.
  • a graph that is not a parabola, above or below a level, read off a given sketch or found on the GDC: Solve graphically (a sinusoidal model against a level: Above or below a level)
Worked example

Two functions, one interval

Paper 1 (no calculator)
Question

Let f(x)=x2−3xf(x) = x^2 - 3x and g(x)=x+5g(x) = x + 5. Solve f(x)<g(x)f(x) < g(x). [4]

Try it yourself first, then compare.

Now try a question

Questions of this type are on their way. Try another one meanwhile: write your working on paper or on screen and see where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Steps
  1. Rearrange so one side is 00. Never divide by an expression containing xx.

  2. Solve the equation =0= 0: factorise on , or the GDC on .

  3. Sketch the parabola through the roots, opening up if a>0a > 0 and down if a<0a < 0.

  4. Read off the region. Include the roots for ≤\le or ≥\ge. Write an "outside" answer as two inequalities.

  5. In a context, check the domain (for example t≥0t \ge 0), and answer the question asked (an interval, or a length of time).

Common mistakes
  • Dividing by xx

    (or by anything that could be negative or zero). Rearrange and factorise instead.

  • Reading the wrong region

    always sketch. With a<0a < 0 the "between" and "outside" rules swap.

  • Dropping the endpoints for ≤\le / ≥\ge

    , or including them for << / >>.

More mistakes (2)
  • Giving the roots instead of the answer

    in "for how long" questions. The answer is one number: the difference.

  • Rounding before subtracting

    1.98−0.875=1.1051.98 - 0.875 = 1.105 is borderline, and 1.98−0.88=1.101.98 - 0.88 = 1.10 is wrong. Use the unrounded values from the GDC (keep full-accuracy values).

Next typeRational functions