Equations That Become Quadratics

Equations that turn into quadratics: show-that from a shape, clearing algebraic fractions, speed and cost stories, and choosing the answer that makes sense.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, IGCSE Additional Maths 0606, Edexcel IGCSE Maths 4MA1 · 33 real past-paper questions below

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Where Quadratics Hide

Some equations do not look quadratic on the page. After a few moves an $x^{2}$ appears, and then you solve it like any quadratic. Exams hide them in three places:

Steps

1
Shapes. Sides like (x + 3) and (2x − 1). Area or Pythagoras multiplies them, and x × x makes $x^{2}$ .
2
Fractions. x on the bottom, like $\frac{2}{x}$ . Clearing the bottoms multiplies x by x.
3
Stories. Speed, time or cost. Time = distance ÷ speed puts x on the bottom of a fraction.

Key idea

They all end the same way: everything on one side, = 0, then solve. And because x is often a length, speed or price, one of the two answers is usually thrown away.

Show That: From a Shape

“Show that” questions print the equation you must reach. Here is the rectangle kind, one move per line.

A rectangle is (x + 3) cm by (2x − 1) cm. Its area is 60 cm². Show that $2 x^{2} + 5 x - 63 = 0$ .

Worked example

1
$(x + 3) (2 x - 1) = 60$ — Area of a rectangle = length × width. The area is 60, so that is the equation.
2
$2 x^{2} - x + 6 x - 3 = 60$ — Expand the brackets: x × 2x, x × (−1), 3 × 2x and 3 × (−1). Four terms.
3
$2 x^{2} + 5 x - 3 = 60$ — Collect the x terms: −x + 6x = 5x.
4
$2 x^{2} + 5 x - 3 - 60 = 0$ — Take 60 from both sides, so the right side is 0.
5
$2 x^{2} + 5 x - 63 = 0$ — −3 − 60 = −63. This is the printed equation, so stop here.

Key idea

In a “show that”, the answer is already printed, so the marks are for your working. Write every line. Jumping straight to the printed equation scores nothing.

Common mistake

Writing 2x² + 5x − 3 + 60 = 0. Moving the 60 to the left changes its sign: −3 − 60 = −63.

Fractions with x on the Bottom

When x is on the bottom of a fraction, clear every bottom at once. Multiply every term by all the bottoms multiplied together.

Solve $\frac{2}{x} + \frac{3}{x + 2} = 1$ .

Worked example

1
$\frac{2}{x} + \frac{3}{x + 2} = 1$ — The question. The bottoms are x and (x + 2).
2
$\frac{2}{x} \times x (x + 2) + \frac{3}{x + 2} \times x (x + 2) = 1 \times x (x + 2)$ — Multiply every term by x(x + 2). That includes the 1 on the right.
3
$2 (x + 2) + 3 x = x (x + 2)$ — Cancel: in the first term the x cancels, in the second the (x + 2) cancels. No fractions left.
4
$2 x + 4 + 3 x = x^{2} + 2 x$ — Expand the brackets on both sides.
5
$5 x + 4 = x^{2} + 2 x$ — Collect the left side: 2x + 3x = 5x.
6
$0 = x^{2} - 3 x - 4$ — Take 5x and 4 from both sides. Move everything to the side with $x^{2}$ , so it stays positive.
7
$(x - 4) (x + 1) = 0$ — Factorise: −4 × 1 = −4 and −4 + 1 = −3.
8
$x = 4 or x = - 1$ — Each bracket equals 0. There is no story here, so keep both answers.

Common mistake

Forgetting the right side: writing 2(x + 2) + 3x = 1. The 1 is a term too, so it becomes x(x + 2).

Key idea

Check by putting an answer back in. x = 4 gives $\frac{2}{4} + \frac{3}{6} = \frac{1}{2} + \frac{1}{2} = 1$ . It works.

Fractions from a Story

Cost, speed and time questions put x on the bottom of a fraction. Write each amount as a fraction first, then clear the bottoms.

Sam spends $60 on tickets at $x each. Lee spends $60 on tickets at $(x + 3) each and gets 1 ticket fewer. Show that $x^{2} + 3 x - 180 = 0$ .

Worked example

1
$Sam: \frac{60}{x} Lee: \frac{60}{x + 3}$ — Number of tickets = money ÷ price of one ticket.
2
$\frac{60}{x} - \frac{60}{x + 3} = 1$ — Lee has 1 fewer, so Sam's number minus Lee's number is 1.
3
$60 (x + 3) - 60 x = x (x + 3)$ — Multiply every term by x(x + 3), including the 1. The bottoms cancel.
4
$60 x + 180 - 60 x = x^{2} + 3 x$ — Expand both sides.
5
$180 = x^{2} + 3 x$ — 60x − 60x = 0, so the x terms on the left cancel out.
6
$x^{2} + 3 x - 180 = 0$ — Take 180 from both sides. This is the printed equation.

Memory trick

Time = distance ÷ speed. Number bought = money spent ÷ price of one.

Common mistake

Adding the fractions: $\frac{60}{x} + \frac{60}{x + 3} = 1$ . “1 fewer” is a difference, so subtract: bigger number − smaller number = 1.

Solve, Then Choose the Real One

Solving gives two answers. The story decides which one is real. Then answer the question that was actually asked.

Carry on: solve $x^{2} + 3 x - 180 = 0$ and find how many tickets Lee buys.

Worked example

1
$x^{2} + 3 x - 180 = 0$ — The equation from the last card. x is the price of Sam's tickets.
2
$(x + 15) (x - 12) = 0$ — Two numbers that multiply to −180 and add to +3: 15 and −12.
3
$x = - 15 or x = 12$ — Each bracket equals 0.
4
$x = - 15 x = 12$ — Reject −15: a price cannot be negative. So Sam's tickets cost $12.
5
$x + 3 = 12 + 3 = 15$ — The question is about Lee, whose tickets cost $(x + 3). So $15 each.
6
$60 \div 15 = 4$ — Lee buys 4 tickets. Check: Sam buys 60 ÷ 12 = 5, and 5 − 1 = 4.

Common mistake

Stopping at x = 12. The question asked how many tickets Lee buys, not what x is. Read the last line of the question again.

Key idea

Throw away an answer that makes a length, speed, price or time negative, or a number of people or tickets that is not a whole number. Say why in words: “a price cannot be negative”.

When It Won't Factorise

Stories often give an equation that does not factorise. If the question says “to 2 decimal places”, use the formula, then still choose the real answer.

A cyclist rides 30 km at x km/h, then 20 km at (x − 5) km/h. The ride takes 3 hours, which gives $3 x^{2} - 65 x + 150 = 0$ . Find x.

Worked example

1
$a = 3, b = - 65, c = 150$ — Read off a, b and c. Keep the minus on b.
2
$b^{2} - 4 a c = (- 65)^{2} - 4 \times 3 \times 150$ — Work out the part under the square root first. Put b in brackets.
3
$= 4225 - 1800 = 2425$ — (−65)² = +4225: a negative number squared is positive. 4 × 3 × 150 = 1800.
4
$x = \frac{- ( - 65 ) \pm 2425}{2 \times 3} = \frac{65 \pm 2425}{6}$ — −b = −(−65) = +65. The bottom is 2a = 6.
5
$x = \frac{65 + 2425}{6} = 19.04$ — √2425 = 49.244…, so (65 + 49.244) ÷ 6 = 19.04 to 2 d.p.
6
$x = \frac{65 - 2425}{6} = 2.63$ — (65 − 49.244) ÷ 6 = 2.626…, which is 2.63 to 2 d.p.
7
$x - 5 = 2.63 - 5 = - 2.37$ — Reject 2.63: the second speed x − 5 would be negative, and a speed cannot be negative.
8
$x = 19.04$ — So the first speed is 19.04 km/h.

Common mistake

Typing −65² into the calculator. It gives −4225, because it squares 65 first. Type (−65)² with the brackets.

Key idea

Both answers here were positive, but 2.63 still failed: check every length or speed in the story, not just x.

Which Move? Read the Question

The wording tells you the first move.

You see	You do
Sides like (x + 3), and an area or volume	Area = length × width, expand, make it = 0
A right-angled triangle with sides in x	a² + b² = c², expand, make it = 0
x on the bottom of a fraction	Multiply every term by all the bottoms
A speed of x km/h, or a price of $x	Time = distance ÷ speed; number = money ÷ price
“Show that … = 0”	Write every line and stop at the printed equation
“To 2 decimal places”	Use the quadratic formula
“Find the speed”, “find the length”	Reject any negative answer, then answer the question

Key idea

Before you finish, read the question again: does it want x, or something worked out from x?

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