Mathematics guide, Foundations, step 2 of 33Previous step Next step

Step 2: Quadratics and Inequalities

By the EconTutorials team. 10-minute read with 6 practice questions.

Syllabus9709 P1 1.1 · IB SL 2.7

In 30 seconds

To solve $ ax^2 + bx + c = 0 $, factorise if you can, then use the zero-product rule: if $ pq = 0 $, then $ p = 0 $ or $ q = 0 $.
Completing the square writes a quadratic as $ a(x-h)^2 + k $. The vertex is $ (h, k) $, and the quadratic formula is this method done once in general.
The discriminant $ b^2-4ac $ counts the real roots: positive gives two, zero gives one repeated root, negative gives none. A line is a tangent to a curve when the resulting quadratic has discriminant zero.
Exam point: for a quadratic inequality, find the critical values and sketch. With a positive $ x^2 $ term, “$ > 0 $” is outside the roots and “$ < 0 $” is between them.

Video lesson coming soonA short animated lesson for this step will appear here.

Solving by factorising

A quadratic equation asks one question: which values of $ x $ make $ ax^2 + bx + c $ equal to zero? The quickest answer comes from a special property of zero. If two numbers multiply to give 6, either of them could be almost anything. But if they multiply to give 0, one of them must be 0. So once a quadratic is written as two brackets multiplied together, each bracket hands you a root.

Key termRoot: a value of $ x $ that makes an equation true. The real roots of $ f(x) = 0 $ are the x-coordinates where the graph of $ y = f(x) $ meets the x-axis.

Worked example

Solve $ 3x^2-10x-8 = 0 $.

Step 1Multiply $ a $ and $ c $: $ 3 \times (-8) = -24 $. The pair with product $ -24 $ and sum $ -10 $ is $ -12 $ and $ 2 $.

Step 2Split the middle term and group: $ 3x^2-12x + 2x-8 = 3x(x-4) + 2(x-4) = (3x + 2)(x-4) $.

Step 3Set each bracket to zero: $ 3x + 2 = 0 $ or $ x-4 = 0 $, so $ x = -\tfrac{2}{3} $ or $ x = 4 $.

Check one root: $ 3(16)-40-8 = 0 $.

Common mistakeDividing by $ x $. Faced with $ 2x^2 = 7x $, many students divide by $ x $ and write $ x = 3.5 $. That loses a root, because $ x $ might be 0. Rearrange instead: $ 2x^2-7x = 0 $, so $ x(2x-7) = 0 $, giving $ x = 0 $ or $ x = 3.5 $.

Completing the square and the vertex

Most quadratics do not factorise neatly. Completing the square works for every one, and it also reveals the turning point of the graph. The idea is to hide the x term inside a squared bracket. Since $ (x + 2)^2 = x^2 + 4x + 4 $, the expression $ x^2 + 4x $ is $ (x + 2)^2 $ with 4 too many. In general, halve the coefficient of $ x $ and subtract the square of that half.

\[ x^2 + bx = \left(x + \frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 \]

So $ x^2 + 4x-5 = (x + 2)^2-4-5 = (x + 2)^2-9 $. Now the minimum is easy to see. A square is never negative, so $ (x + 2)^2 \ge 0 $, and it equals 0 only at $ x = -2 $. The smallest value of the whole expression is therefore $ -9 $, at $ x = -2 $. The point $ (-2, -9) $ is the vertex, and $ a(x-h)^2 + k $ is called vertex form.

Solving also becomes easy: $ (x + 2)^2 = 9 $ gives $ x + 2 = \pm 3 $, so $ x = 1 $ or $ x = -5 $. For a sketch, mark the vertex, the roots and the y-intercept (put $ x = 0 $ to get $ -5 $). A positive $ x^2 $ coefficient means a $ \cup $ shape; a negative one means a $ \cap $.

y = x² + 4x − 5 = (x + 2)² − 9

y = x² + 4x − 5

Figure 1. Vertex form shows the lowest point (−2, −9) at once. The roots −5 and 1 sit 3 units either side of the line of symmetry x = −2.

Worked example

(a) Write $ 3x^2-12x + 7 $ in the form $ a(x + b)^2 + c $ and state its minimum value. (b) Find the maximum value of $ 8 + 6x-x^2 $.

Step 1(a) Take the 3 out of the first two terms only: $ 3(x^2-4x) + 7 $.

Step 2Complete the square inside: $ 3\left[(x-2)^2-4\right] + 7 = 3(x-2)^2-12 + 7 = 3(x-2)^2-5 $. The minimum is $ -5 $, when $ x = 2 $.

Step 3(b) Take out a minus sign: $ 8-(x^2-6x) = 8-\left[(x-3)^2-9\right] = 17-(x-3)^2 $.

Step 4A square is being subtracted, so the expression is largest when the square is 0. The maximum is 17, when $ x = 3 $.

In step 2, the $ -4 $ inside the bracket is multiplied by the 3 outside. Forgetting that multiplication is the most common slip in this method.

The quadratic formula and the discriminant

Complete the square on the general equation $ ax^2 + bx + c = 0 $ and you get a formula that solves every quadratic in one line:

\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]

Read it as two parts. The $ -\frac{b}{2a} $ is the x-coordinate of the vertex. The $ \pm $ part says how far each root sits on either side of it. The quantity under the root, $ b^2-4ac $, is the discriminant, and its sign decides how many real roots there are.

Discriminant	Roots	Graph of $ y = ax^2 + bx + c $
$ b^2-4ac > 0 $	two distinct real roots	crosses the x-axis twice
$ b^2-4ac = 0 $	one repeated root	touches the x-axis at the vertex
$ b^2-4ac < 0 $	no real roots	misses the x-axis

For example, $ 2x^2-3x-4 = 0 $ has $ b^2-4ac = 9 + 32 = 41 $, so $ x = \dfrac{3 \pm \sqrt{41}}{4} $, which is $ 2.35 $ or $ -0.851 $ to 3 significant figures.

Exam tipWrite down $ a $, $ b $ and $ c $ with their signs before substituting, and put negatives in brackets: $ b^2 = (-3)^2 = 9 $, not $ -9 $. If the question says “exact”, leave the surd and simplify it, for example $ \sqrt{48} = 4\sqrt{3} $. Discriminant questions on Cambridge 9709 Paper 1 almost always end with “find the values of $ k $” or “find the set of values of $ k $”.

The same idea tells you how a line meets a curve. Set the two expressions for $ y $ equal; the result is a quadratic, and its discriminant counts the meeting points. A line is a tangent to a parabola when that discriminant is zero.

Worked example

Find the value of $ c $ for which the line $ y = x + c $ is a tangent to the curve $ y = x^2-3x + 6 $, and find the point of contact.

Step 1Set the expressions equal: $ x^2-3x + 6 = x + c $, so $ x^2-4x + (6-c) = 0 $.

Step 2Tangent means discriminant zero: $ (-4)^2-4(1)(6-c) = 0 $, so $ 16-24 + 4c = 0 $ and $ c = 2 $.

Step 3Then $ x^2-4x + 4 = (x-2)^2 = 0 $, so $ x = 2 $ and $ y = 2 + 2 = 4 $.

The line $ y = x + 2 $ touches the curve at $ (2, 4) $. Figure 2 shows what happens for other values of $ c $.

Lines y = x + c meeting y = x² − 3x + 6

y = x² − 3x + 6c = 5: crosses twicec = 2: tangent at (2, 4)c = −1: misses

Figure 2. Substituting y = x + c gives x² − 4x + (6 − c) = 0 with discriminant 4c − 8. It is positive for c = 5 (two crossings), zero for c = 2 (the tangent) and negative for c = −1 (no meeting point).

Disguised quadratics and simultaneous equations

Some equations are quadratics in disguise. In $ 2^{2x}-6 \times 2^x + 8 = 0 $, notice that $ 2^{2x} = (2^x)^2 $. Let $ u = 2^x $; then $ u^2-6u + 8 = 0 $, so $ (u-2)(u-4) = 0 $ and $ u = 2 $ or $ 4 $. Translate back: $ 2^x = 2 $ gives $ x = 1 $, and $ 2^x = 4 $ gives $ x = 2 $. Do not stop at the values of $ u $: the question asked for $ x $. If a value of $ u = 2^x $ or $ u = \sqrt{x} $ comes out negative, reject it and say why.

When one equation is linear and the other is a quadratic, make one letter the subject of the linear equation and substitute it into the other. You get a quadratic, so expect up to two solution pairs.

Worked example

Solve the simultaneous equations $ x + 2y = 5 $ and $ x^2 + y^2 = 10 $.

Step 1From the linear equation, $ x = 5-2y $.

Step 2Substitute: $ (5-2y)^2 + y^2 = 10 $, so $ 25-20y + 5y^2 = 10 $, which gives $ 5y^2-20y + 15 = 0 $.

Step 3Divide by 5: $ y^2-4y + 3 = (y-1)(y-3) = 0 $, so $ y = 1 $ or $ y = 3 $.

Step 4Use the linear equation for each $ x $: $ y = 1 $ gives $ x = 3 $; $ y = 3 $ gives $ x = -1 $.

The solutions are $ (3, 1) $ and $ (-1, 3) $. Keep each $ x $ with its own $ y $.

Inequalities, including the modulus

An inequality asks for every value that works, so the answer is usually a range. Linear inequalities solve like equations with one exception: multiplying or dividing by a negative number reverses the sign, because negation flips order ($ 2 < 5 $ but $ -2 > -5 $). So $ 4-3x \ge 13 $ gives $ -3x \ge 9 $ and then $ x \le -3 $.

For a quadratic inequality, do not push symbols around. Use the graph:

Step	For $ x^2-2x-15 \le 0 $
1. Find the critical values	$ (x-5)(x + 3) = 0 $, so $ x = -3 $ and $ x = 5 $
2. Sketch	positive $ x^2 $ term, so a $ \cup $ through $ -3 $ and $ 5 $
3. Read off	on or below the axis between the roots: $ -3 \le x \le 5 $

In set notation that is $ \{x : -3 \le x \le 5\} $, and in interval notation $ [-3, 5] $. A square bracket includes the endpoint and a round one excludes it. An “outside” answer such as $ x < -3 $ or $ x > 5 $ is two separate pieces, written $ (-\infty, -3) \cup (5, \infty) $.

Worked example

Find the set of values of $ k $ for which $ 2x^2 + kx + 8 = 0 $ has two distinct real roots.

Step 1Two distinct roots means $ b^2-4ac > 0 $: $ k^2-64 > 0 $.

Step 2Factorise: $ (k-8)(k + 8) > 0 $. Critical values $ \pm 8 $; the graph in $ k $ is a $ \cup $.

Step 3Greater than zero means outside the roots: $ k < -8 $ or $ k > 8 $.

Never write $ k > \pm 8 $. It hides the negative half: $ k = -10 $ gives $ k^2 = 100 > 64 $, so it works.

An economics example: break-even and the profit zone. A firm faces the inverse demand curve $ P = 48-Q $, so its revenue is $ TR = PQ = 48Q-Q^2 $. Its costs are a fixed \$260 plus \$12 a unit: $ TC = 260 + 12Q $. Profit is $ \pi = TR-TC $, which simplifies and factorises as

\[ \pi = -Q^2 + 36Q-260 = -(Q-10)(Q-26) \]

The firm breaks even where $ \pi = 0 $, at $ Q = 10 $ and $ Q = 26 $. Profit is a $ \cap $, so it is positive between the roots: $ 10 < Q < 26 $. Completing the square gives $ \pi = 64-(Q-18)^2 $, so the best output is 18 units, with a profit of \$64. Revenue alone, $ TR = 576-(Q-24)^2 $, peaks later, at $ Q = 24 $: once costs count, the firm stops expanding sooner. No calculus was needed.

Revenue, cost and the profit zone $

TR = 48Q − Q²TC = 260 + 12Q

Figure 3. Revenue and cost meet at the break-even outputs Q = 10 and Q = 26. Between them revenue is above cost, so profit is positive (shaded).

The modulus. $ |x| $ is the size of $ x $ ignoring its sign, which is best read as a distance: $ |x-a| $ is the distance from $ x $ to $ a $ on the number line. So $ |x-a| < b $ means “within $ b $ of $ a $”, the interval $ a-b < x < a + b $, and $ |x-a| > b $ means the two outside pieces. For an equation $ |A| = c $ with $ c \ge 0 $, solve $ A = c $ and $ A = -c $. For example, $ |3x-2| < 7 $ gives $ -7 < 3x-2 < 7 $, so $ -\tfrac{5}{3} < x < 3 $. Modulus equations and inequalities are Cambridge 9709 Paper 2 and 3 content, and they appear in the IB HL extension; AP does not test them as a topic.

Common mistakeWriting $ |2x + 3| \ge 5 $ as $ -5 \le 2x + 3 \le 5 $. That is the “within” answer, and this inequality asks for points at least 5 away. The correct split is $ 2x + 3 \ge 5 $ or $ 2x + 3 \le -5 $, giving $ x \ge 1 $ or $ x \le -4 $.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Solve $ x^2 + 2x-24 = 0 $. Show worked answer

Two numbers with product $ -24 $ and sum 2: 6 and $ -4 $.
$ (x + 6)(x-4) = 0 $.
$ x = -6 $ or $ x = 4 $.

EasyQ2. Solve $ 5-3x > 17 $. Show worked answer

Subtract 5: $ -3x > 12 $.
Divide by $ -3 $ and reverse the sign: $ x < -4 $.

MediumQ3. Write $ 2x^2 + 12x + 11 $ in the form $ a(x + b)^2 + c $, and state the minimum value. Show worked answer

$ 2(x^2 + 6x) + 11 = 2\left[(x + 3)^2-9\right] + 11 $.
$ = 2(x + 3)^2-18 + 11 = 2(x + 3)^2-7 $.
Minimum value $ -7 $, when $ x = -3 $.

MediumQ4. Find the values of $ k $ for which $ x^2-(k + 2)x + 9 = 0 $ has equal roots. Show worked answer

Equal roots: $ b^2-4ac = 0 $, so $ (k + 2)^2-36 = 0 $.
$ k + 2 = \pm 6 $.
$ k = 4 $ or $ k = -8 $.

MediumQ5. Solve $ 3x^2 + 5x-2 > 0 $. Show worked answer

Critical values: $ (3x-1)(x + 2) = 0 $, so $ x = \tfrac{1}{3} $ and $ x = -2 $.
The graph is a $ \cup $, and we want it above the axis: outside the roots.
$ x < -2 $ or $ x > \tfrac{1}{3} $.

ExamQ6. A curve has equation $ y = 2x^2-4x + 7 $ and a line has equation $ y = kx + 5 $, where $ k $ is a constant. (a) Show that the x-coordinates of any points of intersection satisfy $ 2x^2-(4 + k)x + 2 = 0 $. [1] (b) Find the values of $ k $ for which the line is a tangent to the curve. [3] (c) For the negative value of $ k $, find the coordinates of the point of contact. [2] (d) Find the set of values of $ k $ for which the line meets the curve at two distinct points. [2] Show worked answer

(a) $ 2x^2-4x + 7 = kx + 5 $, so $ 2x^2-4x-kx + 2 = 0 $, that is $ 2x^2-(4 + k)x + 2 = 0 $. [B1]
(b) Tangent when $ b^2-4ac = 0 $: $ (4 + k)^2-16 = 0 $. [M1 A1]
$ 4 + k = \pm 4 $, so $ k = 0 $ or $ k = -8 $. [A1]
(c) $ k = -8 $: $ 2x^2 + 4x + 2 = 2(x + 1)^2 = 0 $, so $ x = -1 $. [M1]
$ y = -8(-1) + 5 = 13 $; the point is $ (-1, 13) $. (Check on the curve: $ 2 + 4 + 7 = 13 $.) [A1]
(d) Two points when $ (4 + k)^2-16 > 0 $, so $ k(k + 8) > 0 $. [M1]
Outside the critical values: $ k < -8 $ or $ k > 0 $. [A1]
Total: 8 marks.

Check yourself

1. How many real roots does $ 2x^2 + 3x + 5 = 0 $ have?

The discriminant is $ 9-40 = -31 < 0 $, so there are no real roots: the graph sits entirely above the x-axis.

2. What is the solution of $ x^2 < 9 $?

$ x^2-9 = (x-3)(x + 3) < 0 $ holds between the roots. $ x < 3 $ wrongly includes values such as $ -4 $, where $ x^2 = 16 $.

3. What is the vertex of $ y = (x-4)^2 + 1 $?

The square is zero at $ x = 4 $, leaving $ y = 1 $. Vertex form $ (x-h)^2 + k $ has vertex $ (h, k) $.

Your score: 0 out of 3

The short version

If a product is zero, one factor is zero. Factorise, set each bracket to zero, and never divide by $ x $.
Completing the square gives $ a(x-h)^2 + k $: the vertex is $ (h, k) $, a minimum if $ a > 0 $ and a maximum if $ a < 0 $.
$ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $. The discriminant $ b^2-4ac $ is positive, zero or negative for two, one or no real roots; a line is a tangent when it is zero.
For a line and a curve, substitute from the linear equation. For quadratic inequalities, find the critical values and sketch: “$ > 0 $” is outside, “$ < 0 $” is between.
$ |x-a| < b $ means $ a-b < x < a + b $. In economics, break-even solves profit $ = 0 $, and the profit zone is a quadratic inequality.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 2: Quadratics and Inequalities

Solving by factorising

Completing the square and the vertex

The quadratic formula and the discriminant

Disguised quadratics and simultaneous equations

Inequalities, including the modulus

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

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