Equations and Inequalities

Lessons

Senior 4 · Basic Mathematics

Equations and Inequalities

5 lessons

Lesson 5 of 5

Quadratic equations and their applications

25 min

Where buyers and sellers agree

At a vegetable market, a high price brings more tomatoes to the stalls (traders want to sell) but sends shoppers away; a low price does the opposite. The price at which the quantity offered equals the quantity bought is the equilibrium price. Economists find it by solving the supply equation and the demand equation together. When both are straight lines this is a linear equation, as in the last lesson; but supply often rises faster and faster as the price climbs, and then the equation becomes a quadratic. This lesson solves quadratic equations by factorising and with the discriminant, and then uses them for prices, revenue and interest.

Vegetables piled high on a stall at Kimironko market in Kigali: aubergines, green peppers, tomatoes, courgettes, beetroot and green beans. Photo: François Terrier, CC BY 2.0, via Wikimedia Commons.

Definition

Quadratic equation: an equation that can be written

ax^2 + bx + c = 0, \qquad a \ne 0 .

Its solutions are called roots. A quadratic equation has at most two real roots.

Example: $x^{2} - 5 x - 14 = 0$ has $a = 1$ , $b = - 5$ , $c = - 14$ , and roots 7 and $- 2$ .

Solving by factorising

The method rests on one fact, the zero-product property: if $A \times B = 0$ , then $A = 0$ or $B = 0$ . So write the equation as "something $= 0$ ", factorise, and set each factor to zero.

Worked example 1: four kinds of factorising

Solve (a) $x^{2} - 5 x - 14 = 0$ (b) $6 x^{2} - x - 2 = 0$ (c) $x^{2} = 9 x$ (d) $4 x^{2} - 25 = 0$ .

Solution

Two numbers with product $- 14$ and sum $- 5$ are $- 7$ and 2: $(x - 7) (x + 2) = 0$ , so $x = 7$ or $x = - 2$ .
$a c = - 12$ ; two numbers with product $- 12$ and sum $- 1$ are $- 4$ and 3. Split the middle term: $6 x^{2} - 4 x + 3 x - 2 = 2 x (3 x - 2) + (3 x - 2) = (3 x - 2) (2 x + 1) = 0$ , so $x = \frac{2}{3}$ or $x = -\frac{1}{2}$ .
$x^{2} - 9 x = 0$ , $x (x - 9) = 0$ , so $x = 0$ or $x = 9$ . (Dividing both sides by $x$ would lose the root 0.)
Difference of two squares: $(2 x - 5) (2 x + 5) = 0$ , so $x = \pm\frac{5}{2}$ .

The discriminant and the quadratic formula

Not every quadratic factorises neatly. Completing the square on $a x^{2} + b x + c = 0$ gives a formula that always works:

a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a} \quad\Longrightarrow\quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} .

Important

The discriminant $\Delta = b^2 - 4ac$ decides how many real roots there are:

Discriminant	Roots	Formula
$\Delta \gt 0$	two different real roots	$x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$
$\Delta = 0$	one (double) real root	$x = -\frac{b}{2a}$
$\Delta \lt 0$	no real root	( $\sqrt{\Delta}$ is not a real number)

Worked example 2: using the discriminant

(a) Solve $2 x^{2} - 3 x - 4 = 0$ to 2 decimal places. (b) How many real roots has $3 x^{2} - 7 x + 5 = 0$ ? (c) Find $k$ so that $x^{2} + k x + 9 = 0$ has a double root.

Solution

$\Delta = (-3)^2 - 4(2)(-4) = 9 + 32 = 41 \gt 0$ , so $x = \frac{3 \pm \sqrt{41}}{4}$ : $x \approx 2.35$ or $x \approx -0.85$ .
$\Delta = 49 - 60 = -11 \lt 0$ : no real roots.
A double root needs $\Delta = k^2 - 36 = 0$ , so $k = 6$ or $k = - 6$ . (For $k = 6$ : $(x + 3)^{2} = 0$ .)

Warning

Signs in the formula. With $b = - 3$ , $- b = 3$ and $b^{2} = 9$ (not $- 9$ ). Write $a$ , $b$ and $c$ with their signs before you substitute, and put negative numbers in brackets. Also make sure one side is 0 before you read off $a$ , $b$ , $c$ : for $x^{2} = 4 x + 5$ , use $x^{2} - 4 x - 5 = 0$ .

Applications: supply, demand, equilibrium and finance

The demand equation gives the price $P$ at which buyers take a quantity $Q$ ; the supply equation gives the price at which sellers offer $Q$ . The market is in equilibrium where the two prices are equal.

Worked example 3: equilibrium price and quantity

(a) On a tomato market, demand is $Q_{d} = 1200 - 2 P$ and supply is $Q_{s} = 3 P - 300$ (kg per day, $P$ in Frw per kg; illustrative). Find the equilibrium. (b) For another product, demand is $P = 900 - 20 Q$ and supply is $P = Q^{2} + 100$ , with $Q$ in hundreds of kg (illustrative). Find the equilibrium.

Solution

$1200 - 2 P = 3 P - 300$ , so $5 P = 1500$ , $P = 300$ Frw per kg and $Q = 1200 - 600 = 600$ kg per day.
$Q^{2} + 100 = 900 - 20 Q$ , so $Q^{2} + 20 Q - 800 = 0$ and $(Q + 40) (Q - 20) = 0$ . A quantity cannot be negative, so reject $Q = - 40$ : $Q = 20$ (2 000 kg) and $P = 900 - 20 (20) = 500$ Frw per kg. Check with supply: $400 + 100 = 500$ . ✓

Worked example 4: revenue and the discriminant

With the demand $Q = 1200 - 2 P$ from (a), a trader's daily revenue is $R = P Q = P (1200 - 2 P)$ . For which prices is the revenue (i) 160 000 Frw, (ii) 180 000 Frw, (iii) 200 000 Frw?

Solution

$P (1200 - 2 P) = R$ gives $2 P^{2} - 1200 P + R = 0$ , that is $P^2 - 600P + \frac{R}{2} = 0$ , with $\Delta = 360\,000 - 2R$ .

$R = 160\,000$ : $\Delta = 40\,000$ , $\sqrt{\Delta} = 200$ , so $P = \frac{600 \pm 200}{2}$ : 400 or 200 Frw.
$R = 180\,000$ : $\Delta = 0$ , one price: $P = 300$ Frw.
$R = 200\,000$ : $\Delta \lt 0$ : impossible. No price brings in 200 000 Frw.

The discriminant has found the greatest possible revenue, 180 000 Frw at 300 Frw per kg: that is the only revenue reached at exactly one price.

Worked example 5: an interest rate from a quadratic

500 000 Frw invested at a yearly rate $r$ , compounded yearly, grows to 583 200 Frw in 2 years. Find $r$ .

Solution

500\,000(1 + r)^2 = 583\,200 \;\Longrightarrow\; (1 + r)^2 = 1.1664 \;\Longrightarrow\; 1 + r = 1.08 \ \ (\text{or } -1.08) .

A growth factor cannot be negative, so $1 + r = 1.08$ and $r = 0.08$ : 8% a year.

Tip

In real problems a quadratic often gives one root that makes no sense: a negative quantity, a negative price or a growth factor below zero. Always ask "which root fits the story?" and say why you reject the other.

Economicsfun: How to Calculate Equilibrium Price and Quantity. Watch how setting quantity demanded equal to quantity supplied gives one equation in the price.

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Practice

Try each one before opening the answers. They get harder as you go.

Solve by factorising: (a) $x^{2} + 3 x - 10 = 0$ (b) $3 x^{2} - 12 = 0$ (c) $2 x^{2} + 7 x + 3 = 0$ .
Solve $x^{2} - 5 x - 6 = 0$ , and say which roots are valid if $x$ is a number of days.
Find the discriminant and the number of real roots of (a) $x^{2} + 4 x + 4 = 0$ (b) $2 x^{2} - x + 3 = 0$ (c) $x^{2} - 3 x - 1 = 0$ .
Solve $x^{2} - 3 x - 1 = 0$ to 2 decimal places.
Find $m$ so that $x^{2} - 8 x + m = 0$ has a double root, and give the root.
Demand: $Q_{d} = 500 - 4 P$ ; supply: $Q_{s} = 6 P - 100$ . Find the equilibrium price and quantity.
Demand: $P = 200 - Q$ ; supply: $P = \frac{Q^2}{4} + 80$ . Find the equilibrium ( $Q \ge 0$ ).

Show answersHide answers

(a) $(x + 5) (x - 2) = 0$ : $- 5$ , 2 (b) $3 (x - 2) (x + 2) = 0$ : $\pm 2$ (c) $(2 x + 1) (x + 3) = 0$ : $-\frac{1}{2}$ , $- 3$ .
$(x - 6) (x + 1) = 0$ : 6 or $- 1$ ; only $x = 6$ days is valid.
(a) $\Delta = 0$ : one root (b) $\Delta = 1 - 24 = -23$ : none (c) $\Delta = 13$ : two.
$x = \frac{3 \pm \sqrt{13}}{2}$ : $3.30$ or $- 0.30$ .
$64 - 4 m = 0$ , $m = 16$ ; root $x = 4$ .
$500 - 4 P = 6 P - 100$ , $P = 60$ , $Q = 260$ .
$\frac{Q^2}{4} + 80 = 200 - Q$ , $Q^{2} + 4 Q - 480 = 0$ , $(Q + 24) (Q - 20) = 0$ : $Q = 20$ , $P = 180$ .

Quick check

Key points

Write $a x^{2} + b x + c = 0$ , factorise, and use $AB = 0 \Rightarrow A = 0$ or $B = 0$ .
$\Delta = b^2 - 4ac$ : two roots if $\Delta \gt 0$ , one if $\Delta = 0$ , none if $\Delta \lt 0$ ; $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ .
Equilibrium: set the supply price equal to the demand price (or the quantities equal) and solve.
Reject roots that make no sense in the story, and say why.

1Where buyers and sellers agree

2Solving by factorising

3The discriminant and the quadratic formula

4Applications: supply, demand, equilibrium and finance

Where buyers and sellers agree

Solving by factorising

The discriminant and the quadratic formula

Applications: supply, demand, equilibrium and finance