Problems that lead to quadratic equations – Quadratic Equations (Senior 3 Mathematics) | Ganzaa

Quadratic Equations

Lessons

Senior 3 · Mathematics

Quadratic Equations

5 lessons

Lesson 5 of 5

Problems that lead to quadratic equations

25 min

From words to a quadratic, and back to words

Many real questions about areas, speeds, prices and moving objects lead to a quadratic equation. Solving the equation is only the middle of the work. Every problem in this lesson follows the same five steps:

Choose the unknown and write what it stands for, with its unit: "Let $x$ be the width of the path, in metres."
Form the equation from the facts in the question.
Solve it by the easiest method: factorising, the formula, or completing the square.
Test each root against the situation. A length cannot be negative; a number of people must be a whole number.
Answer the question in words, with units, and check it against the original facts.

Definition

Admissible root: a root of the equation that also makes sense in the real situation.

Notation: write "reject" next to a root that is not admissible, and say why.

Example: $w^{2} + 3 w - 40 = 0$ has roots 5 and $- 8$ ; for the width of a plot only $w = 5$ is admissible, because a width cannot be negative.

Worked example 1: consecutive numbers

The product of two consecutive even numbers is 168. Find the numbers.

Solution Let the smaller number be $n$ ; the next even number is $n + 2$ .

n(n + 2) = 168 \;\Rightarrow\; n^2 + 2n - 168 = 0 \;\Rightarrow\; (n + 14)(n - 12) = 0.

So $n = 12$ or $n = - 14$ . Both are admissible, since the question does not say "positive": the numbers are 12 and 14, or $- 14$ and $- 12$ . Check: $12 \times 14 = 168$ and $(- 14) (- 12) = 168$ . ✓

Worked example 2: a path round a school garden

A school garden in Nyamagabe measures 12 m by 8 m. A path of the same width goes all the way round it, and the garden and path together cover $140\ \mathrm{m^2}$ . How wide is the path?

The path adds $x$ metres on each side, so each outer side is $2 x$ longer than the garden.

Solution Let the width of the path be $x$ metres. The outer rectangle is $(12 + 2 x)$ m by $(8 + 2 x)$ m:

\begin{aligned} (12 + 2x)(8 + 2x) &= 140 \\ 96 + 40x + 4x^2 &= 140 \\ 4x^2 + 40x - 44 &= 0 \\ x^2 + 10x - 11 &= 0 \\ (x + 11)(x - 1) &= 0. \end{aligned}

$x = 1$ or $x = - 11$ ; reject $- 11$ , since a width cannot be negative. The path is 1 m wide. Check: $14 \times 10 = 140$ . ✓

Worked example 3: speed and time

A cyclist rides 60 km. If her average speed were $5\ \mathrm{km\,h^{-1}}$ greater, the ride would take 1 hour less. Find her average speed.

Solution Let her speed be $v\ \mathrm{km\,h^{-1}}$ . Time is distance divided by speed, so the times are $\frac{60}{v}$ and $\frac{60}{v + 5}$ hours, and

\frac{60}{v} - \frac{60}{v + 5} = 1.

Multiply every term by $v (v + 5)$ to clear the fractions:

\begin{aligned} 60(v + 5) - 60v &= v(v + 5) \\ 300 &= v^2 + 5v \\ v^2 + 5v - 300 &= 0 \\ (v + 20)(v - 15) &= 0. \end{aligned}

Reject $v = - 20$ . Her speed is $15\ \mathrm{km\,h^{-1}}$ . Check: 60 km takes 4 h at 15 and 3 h at 20, one hour less. ✓

Worked example 4: a ball kicked straight up

A ball is kicked straight up from the ground. A simple model that ignores air resistance gives its height after $t$ seconds as $h = 20 t - 5 t^{2}$ metres. (a) When is the ball 15 m high? (b) When does it land? (c) Does it ever reach 25 m?

Solution

(a) $20t - 5t^2 = 15 \Rightarrow 5t^2 - 20t + 15 = 0 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t - 1)(t - 3) = 0$ . The ball is at 15 m twice: at $t = 1$ s on the way up and at $t = 3$ s on the way down. Both roots are admissible.

(b) It lands when $h = 0$ : $20 t - 5 t^{2} = 5 t (4 - t) = 0$ , so $t = 0$ (the kick) or $t = 4$ . It lands after 4 s.

(c) $20t - 5t^2 = 25 \Rightarrow t^2 - 4t + 5 = 0$ , and $\Delta = 16 - 20 = -4 \lt 0$ . No real root, so the ball never reaches 25 m. Completing the square, $h = 20 - 5 (t - 2)^{2}$ : the greatest height is 20 m, at $t = 2$ s.

The height–time graph is a parabola. The dashed line $h = 15$ meets it at $t = 1$ and $t = 3$ , the two roots of part (a).

Worked example 5: a question of price

A trader in Musanze spent 36,000 Frw on trays of eggs. If each tray had cost 600 Frw less, she could have bought 2 more trays with the same money. How many trays did she buy? (Illustrative prices.)

Solution Let $n$ be the number of trays. Each tray cost $\frac{36000}{n}$ Frw, and at the lower price she could buy $n + 2$ trays, so

\frac{36000}{n} - 600 = \frac{36000}{n + 2}.

Multiply by $n (n + 2)$ : $36000 (n + 2) - 600 n (n + 2) = 36000 n$ . The $36000 n$ terms cancel, leaving $72000 - 600 n^{2} - 1200 n = 0$ . Divide by $- 600$ :

n^2 + 2n - 120 = 0 \;\Rightarrow\; (n + 12)(n - 10) = 0.

Reject $n = - 12$ . She bought 10 trays at 3,600 Frw each. Check: at 3,000 Frw each, 36,000 Frw buys 12 trays, 2 more. ✓

Warning

The equation does not know the story. Algebra gives every root; only you can decide which roots fit. Reject negative lengths, speeds and counts, fractions of people or trays, and times before the start. But do not reject a root just because it is negative when the question allows it, as in Worked example 1.

Tip

Sketch, label, then write "Let …". A quick diagram with the unknown marked on it (as in the garden path) prevents the most common error: adding $x$ to only one side instead of both.

Math Liberty (formerly MathWOEs): Solving "Rectangular Area" word problem with a quadratic equation (11 min). Two worked examples: watch how the unknown is defined, how the area equation is formed and factorised, and how the negative root is rejected.

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Practice

Follow the five steps. The last three are harder.

The product of two consecutive positive integers is 132. Find them.
A rectangle has a perimeter of 34 cm and an area of $60\ \mathrm{cm^2}$ . Find its length and width.
The sides of a right-angled triangle are $x$ , $x + 7$ and $x + 8$ centimetres, the longest being the hypotenuse. Find $x$ .
A classroom floor is 4 m longer than it is wide and has an area of $60\ \mathrm{m^2}$ . Find its dimensions.
A ball's height is $h = 25 t - 5 t^{2}$ metres after $t$ seconds. When is it 20 m high, and when does it land?
A number plus its reciprocal equals $\frac{29}{10}$ . Find the number.
Friends share the 60,000 Frw hire of a minibus equally. If 2 more friends had come, each would have paid 1,500 Frw less. How many friends were there?
A path of uniform width round a lawn 20 m by 15 m has an area of $74\ \mathrm{m^2}$ . Find the width of the path.

Show answersHide answers

$n (n + 1) = 132$ , $(n + 12) (n - 11) = 0$ : 11 and 12.
$l + w = 17$ , so $w (17 - w) = 60$ , $w^{2} - 17 w + 60 = 0$ , $(w - 5) (w - 12) = 0$ : 12 cm by 5 cm.
$x^{2} + (x + 7)^{2} = (x + 8)^{2}$ gives $x^{2} - 2 x - 15 = 0$ , $(x - 5) (x + 3) = 0$ : $x = 5$ (sides 5, 12, 13 cm).
$w (w + 4) = 60$ , $(w + 10) (w - 6) = 0$ : 6 m by 10 m.
$t^{2} - 5 t + 4 = 0$ : at $t = 1$ s and $t = 4$ s. It lands when $5 t (5 - t) = 0$ : after 5 s.
$10 x^{2} - 29 x + 10 = 0$ , $\Delta = 841 - 400 = 441 = 21^2$ : $x = \frac{29 \pm 21}{20}$ , so $x = \frac{5}{2}$ or $x = \frac{2}{5}$ .
$\frac{60000}{n} - \frac{60000}{n + 2} = 1500$ gives $n^{2} + 2 n - 80 = 0$ , $(n + 10) (n - 8) = 0$ : 8 friends (7,500 Frw each, or 6,000 Frw each for 10).
$(20 + 2 x) (15 + 2 x) - 300 = 74$ gives $2 x^{2} + 35 x - 37 = 0$ , $(2 x + 37) (x - 1) = 0$ : the path is 1 m wide.

1From words to a quadratic, and back to words

From words to a quadratic, and back to words