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Quadratic functions

One parabola, three ways of writing it. Factorised shows the roots, completed-square shows the vertex, expanded shows the y-intercept. Underneath all three sits a single number, the discriminant, which knows how many times the curve meets the axis before you solve anything.

Year 12-13EDEXCEL 9MA0 2.3

Builds on Indices and surds.

IN THIS TOPIC

Read roots, vertex and intercept off the three written forms of a quadratic.
Complete the square when the leading coefficient is not 1.
Use the discriminant to count real roots.
Find the values of a parameter that force real, repeated or no roots.
Solve a quadratic in a function of the unknown by naming the substitution.

COMMON MISCONCEPTION

If b² − 4ac comes out negative, you made a mistake.

Three forms, one parabola

A quadratic function f(x) = ax² + bx + c draws a parabola, opening upward when a > 0 and downward when a < 0, and each way of writing it answers a different question at sight. The expanded form shows the y-intercept, c. The factorised form shows the roots. Completing the square shows the vertex, because

ax^{2} + bx + c = a(x + \frac{b}{2a})^{2} + (c − \frac{b^{2}}{4a})

puts all the x-dependence inside one squared bracket. A square is at its smallest when its bracket is zero, so the vertex sits at x = −b/(2a), and whatever constant is left outside the bracket is the minimum, or maximum, value itself.

Completing the square in practice

With a leading coefficient other than 1 there is one extra move, and it is where most of the marks in this topic go missing. Take the coefficient out of the x terms only. Complete the square inside. Multiply back out at the end.

WORKED EXAMPLE

A leading coefficient of 2

Express 2x² − 12x + 7 in the form a(x + p)² + q, and state the minimum point of the curve.

Take the 2 out of the x terms. 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7.

Distribute. 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11.

The bracket is zero at x = 3, so the minimum point is (3, −11). Expanding back is a ten-second check and worth every one of the ten seconds.

FIG. 1The completed square, drawn: y = 2(x − 3)² − 11 has its lowest point exactly at (3, −11), because the squared bracket cannot go below zero.

The discriminant

Feed a quadratic to the formula, which is on the must-learn list,

x = \frac{−b \pm \sqrt{b^{2} − 4ac}}{2a}

NOT IN THE BOOKLET — LEARN IT

and everything interesting happens under the root. The quantity b² − 4ac is the discriminant, and its sign is a complete census of the roots. Positive, two distinct real roots. Zero, one repeated root. Negative, no real roots at all. That last case is a healthy answer and not a symptom of a slip somewhere. The parabola simply never reaches the axis.

FIG. 2The three cases, computed: b² − 4ac positive and the curve crosses twice; zero and it touches once, at the vertex; negative and it misses the axis entirely.

WORKED EXAMPLE

Choosing k for a repeated root

Find the values of k for which x² + kx + 9 = 0 has a repeated root.

Repeated root means discriminant zero, so k² − 4 × 1 × 9 = 0 and k² = 36.

k = 6 or −6. Two different parabolas, each just touching the axis.

Give a quadratic a parameter and you have almost certainly been handed a discriminant question. Translate the condition on the roots into a condition on a sign, then compute.

GUIDED PRACTICE

A condition, not a value

Find the set of values of k for which x² + 3x + k = 0 has real roots, before opening the working.

Show the working

Real roots means b² − 4ac ≥ 0, so here 9 − 4k ≥ 0.

So k ≤ 9/4, or {k : k ≤ 9/4} in set notation. The ≥ matters. “Real roots” includes the repeated case, so the boundary value stays in.

The hidden quadratic

The examiners' favourite twist is a quadratic in a function of the unknown. Nothing about the equation looks quadratic in x until you name the right substitution. Powers, roots, and later on trigonometric and exponential expressions can all play the hidden variable, and last lesson's x − 6√x + 4 = 0 was one of them.

WORKED EXAMPLE

An exponential with a quadratic inside

Solve 4^x − 7(2^x) + 12 = 0.

Since 4^x = (2^x)² by the index laws, substitute u = 2^x to get u² − 7u + 12 = 0.

Factorise. (u − 3)(u − 4) = 0, so u = 3 or u = 4.

Return to x. 2^x = 3 gives x = log₂ 3, and 2^x = 4 gives x = 2. One exact logarithm and one integer, both valid because each value of u is positive.

The rhythm never changes. Spot the hidden square, name u, solve the visible quadratic, translate back, then check that each u makes sense for the substitution you chose.

INDEPENDENT PRACTICE

A quartic that is secretly a quadratic

Solve x⁴ − 5x² + 4 = 0.

Show the working

Substitute u = x². Then u² − 5u + 4 = (u − 1)(u − 4) = 0, so u = 1 or u = 4.

Translate back. x² = 1 gives x = ±1, and x² = 4 gives x = ±2. Four roots, at x = −2, −1, 1 and 2.

The translation step doubled the answers, because each positive u has two square roots. Solutions get gained and lost at the substitution, so that is the step to slow down for.

ASSESSMENT FOCUS

With a ≠ 1, take the coefficient out of the x terms only, complete the square inside the bracket, then distribute back out. Expand your answer to check.
Write the discriminant condition down before you compute it. “Real and distinct means b² − 4ac > 0” is itself a mark.
“Real roots” includes the repeated case, so ≥. “Distinct real roots” excludes it, so >. Read the wording twice; misreading it is the commonest lost mark on the topic.
For a hidden quadratic, write u = 2^x or u = x² down explicitly, and finish by translating every value of u back into x, discarding the impossible ones with a stated reason.
Match the tool to the question. Factorise when it is quick, complete the square when a vertex or an exact surd is wanted, and use the formula when nothing factorises.
A negative discriminant is an answer. Say “no real roots” and move on; candidates lose time re-checking correct arithmetic because they expect every quadratic to have solutions.

CHECK YOURSELF

Express 3x² + 6x − 2 in the form a(x + p)² + q, state the minimum point, and use the discriminant to say how many real roots 3x² + 6x − 2 = 0 has.

Show a hint

Take out the 3 first; the discriminant only needs a, b and c.

Show the answer

3(x² + 2x) − 2 = 3[(x + 1)² − 1] − 2 = 3(x + 1)² − 5.

Bracket zero at x = −1, leftover constant −5, so the minimum point is (−1, −5).

b² − 4ac = 36 + 24 = 60 > 0, so the equation has two distinct real roots, which agrees with a minimum sitting below the axis on an upward parabola.

The completed square names the vertex, and the leftover constant is the extreme value.

The discriminant counts the roots for you. Two, one, or none.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Or read them with their worked answers on the quadratic functions questions page.

CHECK YOUR PROGRESS

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Read roots, vertex and intercept off the three written forms of a quadratic.
Complete the square when the leading coefficient is not 1.
Use the discriminant to count real roots.
Find the values of a parameter that force real, repeated or no roots.
Solve a quadratic in a function of the unknown by naming the substitution.

Open the full revision checklist to see every objective in the course in one place.

Quadratic functions

Three forms, one parabola

Completing the square in practice

The discriminant

The hidden quadratic

Question workbook

Question workbook

Answer book