When should I use the quadratic formula instead of factorising?

When the question asks for answers to 2 decimal places or 3 significant figures, the roots are not whole numbers, so the expression will not factorise. Go straight to the formula.

Why do I lose marks for rounding too early?

Rounding the square root before dividing changes the final digit. Keep the full calculator value until the last line, then round once.

Do I have to reject a negative answer?

Only when the context forbids it. A length, a time or a number of people cannot be negative, so say so in a sentence and give the positive value.

4024 Quadratic Equations Questions with Worked Solutions

Quadratic equations turn up on almost every 4024 paper, usually as a short 2 or 3 mark question plus a longer word problem. The four questions below cover each kind the examiners set. Try each one on paper before you open the solution.

Question 1: solve by factorising [2 marks]

Solve $x^{2} - 5 x - 14 = 0$ .

Show the worked solution

Look for two numbers that multiply to $- 14$ and add to $- 5$ . They are $- 7$ and $+ 2$ .

(x - 7) (x + 2) = 0

A product is zero only when one of its factors is zero, so

x = 7 or x = - 2

Check: $7^{2} - 5 (7) - 14 = 49 - 35 - 14 = 0$ .

Question 2: the formula, to 2 decimal places [3 marks]

Solve $3 x^{2} + 4 x - 2 = 0$ . Give your answers correct to 2 decimal places.

Show the worked solution

"2 decimal places" tells you it will not factorise. Take $a = 3$ , $b = 4$ , $c = - 2$ , keeping the sign of $c$ .

x = \frac{- b \pm b ^{2} - 4 a c}{2 a} = \frac{- 4 \pm 16 + 24}{6} = \frac{- 4 \pm 40}{6}

$40 = 6.3245 \dots$ Keep that full value.

x = \frac{- 4 + 6.3245 \dots}{6} = 0.387 \dots x = \frac{- 4 - 6.3245 \dots}{6} = - 1.720 \dots

x = 0.39 or x = - 1.72

The usual slip is $b^{2} - 4 a c = 16 - 24$ . Here $c$ is negative, so $- 4 a c = - 4 (3) (- 2) = + 24$ .

Question 3: completing the square [4 marks]

(a) Write $x^{2} - 8 x + 5$ in the form $(x - p)^{2} - q$ .

(b) Hence solve $x^{2} - 8 x + 5 = 0$ , leaving your answers in surd form.

Show the worked solution

(a) Halve the coefficient of $x$ : $p = 4$ . Squaring the bracket adds $4^{2} = 16$ , so take it away again.

x^{2} - 8 x + 5 = (x - 4)^{2} - 16 + 5 = (x - 4)^{2} - 11

So $p = 4$ and $q = 11$ .

(b) "Hence" means use part (a).

(x - 4)^{2} = 11 ⟹ x - 4 = \pm 11 ⟹ x = 4 \pm 11

Leave it as a surd. A decimal here loses the mark.

Question 4: a word problem [4 marks]

A rectangle is $x$ cm wide and $(x + 3)$ cm long. Its area is $40 cm^{2}$ .

(a) Show that $x^{2} + 3 x - 40 = 0$ .

(b) Solve the equation and find the perimeter of the rectangle.

Show the worked solution

(a) Area is length times width:

x (x + 3) = 40 ⟹ x^{2} + 3 x - 40 = 0

(b) Two numbers that multiply to $- 40$ and add to $+ 3$ : $+ 8$ and $- 5$ .

(x + 8) (x - 5) = 0 ⟹ x = - 8 or x = 5

A width cannot be negative, so $x = 5$ . The rectangle is $5$ cm by $8$ cm.

Perimeter = 2 (5 + 8) = 26 cm

Write the sentence rejecting $x = - 8$ . It is part of the answer, not decoration.

Try it: see any quadratic solved

Type your own quadratic below, try it first, then reveal the working.

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The three mistakes that cost the most marks

Dropping the sign of $c$ inside $b^{2} - 4 a c$ . Write the substitution out in brackets every time.
Rounding the square root early. Round once, on the last line.
Dividing both sides by $x$ . In $x^{2} = 5 x$ that throws away the root $x = 0$ . Move everything to one side and factorise: $x (x - 5) = 0$ .

O Level Maths 4024: quadratic equations exam questions

Question 1: solve by factorising [2 marks]

Question 2: the formula, to 2 decimal places [3 marks]

Question 3: completing the square [4 marks]

Question 4: a word problem [4 marks]

Try it: see any quadratic solved

The three mistakes that cost the most marks

Questions students ask