Sec 2 Maths · Quadratic equations and graphs

Quadratic equations: word problems questions with answers

Practice questions on quadratic equations: word problems, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The length of a rectangle is $5$ cm more than its width. Its area is $104$ cm². Let the width be $x$ cm. Find the length of the rectangle.

Step 1 of 7

Which relationship gives the equation?

length × width = area

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Questions

Question 1Areas of rectangles and triangles
The length of a rectangle is $4$ cm more than its width. Its area is $32$ cm². Let the width be $x$ cm. Find the length of the rectangle.
Show the worked solution
1. Step 1: Which relationship gives the equation?length × width = area
2. Step 2: Form the equation $x(x + 4) = 32$
3. Step 3: Expand and make one side zero $x^{2} + 4x - 32 = 0$
4. Step 4: Factorise $(x - 4)(x + 8) = 0$
5. Step 5: Set each factor to zero $x - 4 = 0$ or $x + 8 = 0$ $x = 4$ or $x = -8$
6. Step 6: Reject the root that does not fit the problem $x = -8$ is rejected because a length cannot be negative. So $x = 4$ .
7. Step 7: Answer the question asked: the length, not xLength $= 4 + 4$ $= 8$ cm
Question 2Numbers
The product of two consecutive positive even numbers is $728$ . Let the smaller one be $x$ . Find the larger one.
Show the worked solution
1. Step 1: Which relationship gives the equation?smaller × larger = product
2. Step 2: Write the larger one in terms of x, then form the equation $x(x + 2) = 728$
3. Step 3: Expand and make one side zero $x^{2} + 2x - 728 = 0$
4. Step 4: Factorise $(x - 26)(x + 28) = 0$
5. Step 5: Set each factor to zero $x - 26 = 0$ or $x + 28 = 0$ $x = 26$ or $x = -28$
6. Step 6: Reject the root that does not fit the problem $x = -28$ is rejected because the integers are positive. So $x = 26$ .
7. Step 7: Answer the question asked: the larger one, not xThe larger one $= 26 + 2$ $= 28$
Question 3Frames, borders and paths
A rectangular garden measures $22$ m by $20$ m. A path of the same width, $x$ m, runs along the inside of all four edges, leaving a rectangular lawn of area $224$ m². Find the length of the lawn.
Show the worked solution
1. Step 1: Which relationship gives the equation?lawn length × lawn width = lawn area
2. Step 2: How much shorter is the lawn each way? Form the equation $(22 - 2x)(20 - 2x) = 224$
3. Step 3: Expand the brackets $4x^{2} - 84x + 440 = 224$
4. Step 4: Make one side zero, then divide by the common factor $4x^{2} - 84x + 216 = 0$ $x^{2} - 21x + 54 = 0$
5. Step 5: Factorise $(x - 3)(x - 18) = 0$
6. Step 6: Set each factor to zero $x - 3 = 0$ or $x - 18 = 0$ $x = 3$ or $x = 18$
7. Step 7: Reject the root that does not fit the problem $x = 18$ is rejected because it makes a side negative. So $x = 3$ .
8. Step 8: Answer the question asked: the length of the lawn, not xLength of the lawn $= 22 - 2(3)$ $= 16$ m
Question 4Right-angled triangles
The sides of a right-angled triangle are $x$ cm, $(x + 4)$ cm and $(x + 8)$ cm. Find the area of the triangle.
Show the worked solution
1. Step 1: Which side is the hypotenuse?The hypotenuse is the longest side, $x + 8$ .
2. Step 2: Which relationship gives the equation?hypotenuse² = sum of the squares of the other two sides
3. Step 3: Form the equation $x^2 + (x + 4)^2 = (x + 8)^2$
4. Step 4: Expand and make one side zero $x^2 + x^{2} + 8x + 16 = x^{2} + 16x + 64$ $x^{2} - 8x - 48 = 0$
5. Step 5: Factorise $(x - 12)(x + 4) = 0$
6. Step 6: Set each factor to zero $x - 12 = 0$ or $x + 4 = 0$ $x = 12$ or $x = -4$
7. Step 7: Reject the root that does not fit the problem $x = -4$ is rejected because a length cannot be negative. So $x = 12$ .
8. Step 8: Answer the question asked: the area, not xArea $= \dfrac{1}{2} \times 12 \times (12 + 4)$ $= 96$ cm²
Question 5Journeys: speed and time
Mei cycles $24$ km from town A to town B at a steady speed of $x$ km/h. On the way back, Mei cycles $12$ km/h faster and takes 20 minutes less. Find the speed from A to B.
Show the worked solution
1. Step 1: Time = distance ÷ speed, for each journeyAt $x$ km/h: $\dfrac{24}{x}$ hoursAt $(x + 12)$ km/h: $\dfrac{24}{x + 12}$ hours
2. Step 2: Which relationship gives the equation?slower time − faster time = time difference
3. Step 3: Form the equation $\dfrac{24}{x} - \dfrac{24}{x + 12} = \dfrac{1}{3}$
4. Step 4: Multiply every term by the LCM of the denominators $72(x + 12) - 72x = x(x + 12)$
5. Step 5: Expand and make one side zero $x^{2} + 12x - 864 = 0$
6. Step 6: Factorise $(x - 24)(x + 36) = 0$
7. Step 7: Set each factor to zero $x - 24 = 0$ or $x + 36 = 0$ $x = 24$ or $x = -36$
8. Step 8: Reject the root that does not fit the problem $x = -36$ is rejected because a speed cannot be negative. So $x = 24$ .
9. Step 9: Answer the question askedSpeed from A to B $= 24$ km/h
Question 6Cost and quantity
Ravi spent $ $30$ on some pens, all at the same price. If each pen had cost $ $1$ less, Ravi could have bought $5$ more pens with the same money. Let $x$ be the number of pens Ravi bought. Find the price of each pen.
Show the worked solution
1. Step 1: Price of each pen = money spent ÷ number of pensNow: $\dfrac{30}{x}$ dollarsWith $5$ more: $\dfrac{30}{x + 5}$ dollars
2. Step 2: Which relationship gives the equation?higher price − lower price = difference
3. Step 3: Form the equation $\dfrac{30}{x} - \dfrac{30}{x + 5} = 1$
4. Step 4: Multiply every term by the LCM of the denominators $30(x + 5) - 30x = x(x + 5)$
5. Step 5: Expand and make one side zero $x^{2} + 5x - 150 = 0$
6. Step 6: Factorise $(x - 10)(x + 15) = 0$
7. Step 7: Set each factor to zero $x - 10 = 0$ or $x + 15 = 0$ $x = 10$ or $x = -15$
8. Step 8: Reject the root that does not fit the problem $x = -15$ is rejected because the number of people or things cannot be negative. So $x = 10$ .
9. Step 9: Answer the question asked: the price of each pen, not xPrice of each pen $= \dfrac{30}{10}$ $= 3$ dollars
Question 7Number grids
The numbers $1$ , $2$ , $3$ , … are written in rows of $12$ on a grid. Part of the grid is shown. A square is drawn around four numbers, as shown, and it can be moved to other positions on the grid. In one position, the product of the smallest and the largest numbers in the square is $90$ . Let the smallest number be $x$ . Form an equation in $x$ and solve it to find the smallest number in the square.
1 $\boxed{2}$ $\boxed{3}$ 4 5 6 7 8 9 10 11 12
13 $\boxed{14}$ $\boxed{15}$ 16 17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45 46 47 48
Show the worked solution
1. Step 1: Moving one row down adds the width of the grid. The largest number is $x + 13$
2. Step 2: Which relationship gives the equation?smallest × largest = product
3. Step 3: Form the equation $x(x + 13) = 90$
4. Step 4: Expand and make one side zero $x^{2} + 13x - 90 = 0$
5. Step 5: Factorise $(x - 5)(x + 18) = 0$
6. Step 6: Set each factor to zero $x - 5 = 0$ or $x + 18 = 0$ $x = 5$ or $x = -18$
7. Step 7: Reject the root that does not fit the problem $x = -18$ is rejected because the numbers on the grid are positive. So $x = 5$ .
8. Step 8: Answer the question askedSmallest number $= 5$

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Questions

Question 1Areas of rectangles and triangles

The length of a rectangle is $4$ cm more than its width. Its area is $32$ cm². Let the width be $x$ cm. Find the length of the rectangle.

Show the worked solution

Step 1: Which relationship gives the equation?length × width = area
Step 2: Form the equation $x(x + 4) = 32$
Step 3: Expand and make one side zero $x^{2} + 4x - 32 = 0$
Step 4: Factorise $(x - 4)(x + 8) = 0$
Step 5: Set each factor to zero $x - 4 = 0$ or $x + 8 = 0$ $x = 4$ or $x = -8$
Step 6: Reject the root that does not fit the problem $x = -8$ is rejected because a length cannot be negative. So $x = 4$ .
Step 7: Answer the question asked: the length, not xLength $= 4 + 4$ $= 8$ cm

Question 2Numbers

The product of two consecutive positive even numbers is $728$ . Let the smaller one be $x$ . Find the larger one.

Show the worked solution

Step 1: Which relationship gives the equation?smaller × larger = product
Step 2: Write the larger one in terms of x, then form the equation $x(x + 2) = 728$
Step 3: Expand and make one side zero $x^{2} + 2x - 728 = 0$
Step 4: Factorise $(x - 26)(x + 28) = 0$
Step 5: Set each factor to zero $x - 26 = 0$ or $x + 28 = 0$ $x = 26$ or $x = -28$
Step 6: Reject the root that does not fit the problem $x = -28$ is rejected because the integers are positive. So $x = 26$ .
Step 7: Answer the question asked: the larger one, not xThe larger one $= 26 + 2$ $= 28$

Question 3Frames, borders and paths

A rectangular garden measures $22$ m by $20$ m. A path of the same width, $x$ m, runs along the inside of all four edges, leaving a rectangular lawn of area $224$ m². Find the length of the lawn.

Show the worked solution

Step 1: Which relationship gives the equation?lawn length × lawn width = lawn area
Step 2: How much shorter is the lawn each way? Form the equation $(22 - 2x)(20 - 2x) = 224$
Step 3: Expand the brackets $4x^{2} - 84x + 440 = 224$
Step 4: Make one side zero, then divide by the common factor $4x^{2} - 84x + 216 = 0$ $x^{2} - 21x + 54 = 0$
Step 5: Factorise $(x - 3)(x - 18) = 0$
Step 6: Set each factor to zero $x - 3 = 0$ or $x - 18 = 0$ $x = 3$ or $x = 18$
Step 7: Reject the root that does not fit the problem $x = 18$ is rejected because it makes a side negative. So $x = 3$ .
Step 8: Answer the question asked: the length of the lawn, not xLength of the lawn $= 22 - 2(3)$ $= 16$ m

Question 4Right-angled triangles

The sides of a right-angled triangle are $x$ cm, $(x + 4)$ cm and $(x + 8)$ cm. Find the area of the triangle.

Show the worked solution

Step 1: Which side is the hypotenuse?The hypotenuse is the longest side, $x + 8$ .
Step 2: Which relationship gives the equation?hypotenuse² = sum of the squares of the other two sides
Step 3: Form the equation $x^2 + (x + 4)^2 = (x + 8)^2$
Step 4: Expand and make one side zero $x^2 + x^{2} + 8x + 16 = x^{2} + 16x + 64$ $x^{2} - 8x - 48 = 0$
Step 5: Factorise $(x - 12)(x + 4) = 0$
Step 6: Set each factor to zero $x - 12 = 0$ or $x + 4 = 0$ $x = 12$ or $x = -4$
Step 7: Reject the root that does not fit the problem $x = -4$ is rejected because a length cannot be negative. So $x = 12$ .
Step 8: Answer the question asked: the area, not xArea $= \dfrac{1}{2} \times 12 \times (12 + 4)$ $= 96$ cm²

Question 5Journeys: speed and time

Mei cycles $24$ km from town A to town B at a steady speed of $x$ km/h. On the way back, Mei cycles $12$ km/h faster and takes 20 minutes less. Find the speed from A to B.

Show the worked solution

Step 1: Time = distance ÷ speed, for each journeyAt $x$ km/h: $\dfrac{24}{x}$ hoursAt $(x + 12)$ km/h: $\dfrac{24}{x + 12}$ hours
Step 2: Which relationship gives the equation?slower time − faster time = time difference
Step 3: Form the equation $\dfrac{24}{x} - \dfrac{24}{x + 12} = \dfrac{1}{3}$
Step 4: Multiply every term by the LCM of the denominators $72(x + 12) - 72x = x(x + 12)$
Step 5: Expand and make one side zero $x^{2} + 12x - 864 = 0$
Step 6: Factorise $(x - 24)(x + 36) = 0$
Step 7: Set each factor to zero $x - 24 = 0$ or $x + 36 = 0$ $x = 24$ or $x = -36$
Step 8: Reject the root that does not fit the problem $x = -36$ is rejected because a speed cannot be negative. So $x = 24$ .
Step 9: Answer the question askedSpeed from A to B $= 24$ km/h

Question 6Cost and quantity

Ravi spent $ $30$ on some pens, all at the same price. If each pen had cost $ $1$ less, Ravi could have bought $5$ more pens with the same money. Let $x$ be the number of pens Ravi bought. Find the price of each pen.

Show the worked solution

Step 1: Price of each pen = money spent ÷ number of pensNow: $\dfrac{30}{x}$ dollarsWith $5$ more: $\dfrac{30}{x + 5}$ dollars
Step 2: Which relationship gives the equation?higher price − lower price = difference
Step 3: Form the equation $\dfrac{30}{x} - \dfrac{30}{x + 5} = 1$
Step 4: Multiply every term by the LCM of the denominators $30(x + 5) - 30x = x(x + 5)$
Step 5: Expand and make one side zero $x^{2} + 5x - 150 = 0$
Step 6: Factorise $(x - 10)(x + 15) = 0$
Step 7: Set each factor to zero $x - 10 = 0$ or $x + 15 = 0$ $x = 10$ or $x = -15$
Step 8: Reject the root that does not fit the problem $x = -15$ is rejected because the number of people or things cannot be negative. So $x = 10$ .
Step 9: Answer the question asked: the price of each pen, not xPrice of each pen $= \dfrac{30}{10}$ $= 3$ dollars

Question 7Number grids

The numbers $1$ , $2$ , $3$ , … are written in rows of $12$ on a grid. Part of the grid is shown. A square is drawn around four numbers, as shown, and it can be moved to other positions on the grid. In one position, the product of the smallest and the largest numbers in the square is $90$ . Let the smallest number be $x$ . Form an equation in $x$ and solve it to find the smallest number in the square.

1	$\boxed{2}$	$\boxed{3}$	4	5	6	7	8	9	10	11	12
13	$\boxed{14}$	$\boxed{15}$	16	17	18	19	20	21	22	23	24
25	26	27	28	29	30	31	32	33	34	35	36
37	38	39	40	41	42	43	44	45	46	47	48

Show the worked solution

Step 1: Moving one row down adds the width of the grid. The largest number is $x + 13$
Step 2: Which relationship gives the equation?smallest × largest = product
Step 3: Form the equation $x(x + 13) = 90$
Step 4: Expand and make one side zero $x^{2} + 13x - 90 = 0$
Step 5: Factorise $(x - 5)(x + 18) = 0$
Step 6: Set each factor to zero $x - 5 = 0$ or $x + 18 = 0$ $x = 5$ or $x = -18$
Step 7: Reject the root that does not fit the problem $x = -18$ is rejected because the numbers on the grid are positive. So $x = 5$ .
Step 8: Answer the question askedSmallest number $= 5$