Quadratic equations: word problems questions with answers

Questions

1. 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
2. Question 2Numbers

   A positive integer is multiplied by one more than twice itself. The result is $465$. Let the integer be $x$. Find the integer.

   Show the worked solution

   1. Step 1: Write 'one more than twice the integer' in terms of x, then form the equation$x(2x + 1) = 465$
   2. Step 2: Expand and make one side zero$2x^{2} + x - 465 = 0$
   3. Step 3: Factorise$(x - 15)(2x + 31) = 0$
   4. Step 4: Set each factor to zero$x - 15 = 0$ or $2x + 31 = 0$$x = 15$ or $x = -\dfrac{31}{2}$
   5. Step 5: Reject the root that does not fit the problem$x = -\dfrac{31}{2}$ is rejected because x must be a positive integer. So $x = 15$.
   6. Step 6: Answer the question askedThe integer $= 15$
3. Question 3Frames, borders and paths

   A rectangular garden measures $23$ m by $13$ m. A path of the same width, $x$ m, runs along the inside of all four edges, leaving a rectangular lawn of area $75$ 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$(23 - 2x)(13 - 2x) = 75$
   3. Step 3: Expand the brackets$4x^{2} - 72x + 299 = 75$
   4. Step 4: Make one side zero, then divide by the common factor$4x^{2} - 72x + 224 = 0$$x^{2} - 18x + 56 = 0$
   5. Step 5: Factorise$(x - 4)(x - 14) = 0$
   6. Step 6: Set each factor to zero$x - 4 = 0$ or $x - 14 = 0$$x = 4$ or $x = 14$
   7. Step 7: Reject the root that does not fit the problem$x = 14$ is rejected because it makes a side negative. So $x = 4$.
   8. Step 8: Answer the question asked: the length of the lawn, not xLength of the lawn $= 23 - 2(4)$$= 15$ m
4. Question 4Right-angled triangles

   The sides of a right-angled triangle are $x$ cm, $(x + 7)$ cm and $(x + 14)$ cm. Find the perimeter of the triangle.

   Show the worked solution

   1. Step 1: Which side is the hypotenuse?The hypotenuse is the longest side, $x + 14$.
   2. Step 2: Which relationship gives the equation?hypotenuse² = sum of the squares of the other two sides
   3. Step 3: Form the equation$x^2 + (x + 7)^2 = (x + 14)^2$
   4. Step 4: Expand and make one side zero$x^2 + x^{2} + 14x + 49 = x^{2} + 28x + 196$$x^{2} - 14x - 147 = 0$
   5. Step 5: Factorise$(x - 21)(x + 7) = 0$
   6. Step 6: Set each factor to zero$x - 21 = 0$ or $x + 7 = 0$$x = 21$ or $x = -7$
   7. Step 7: Reject the root that does not fit the problem$x = -7$ is rejected because a length cannot be negative. So $x = 21$.
   8. Step 8: Answer the question asked: the perimeter, not xPerimeter $= 21 + (21 + 7) + (21 + 14)$$= 84$ cm
5. Question 5Journeys: speed and time

   A car travels $112$ km at a steady speed of $x$ km/h. If its speed had been $2$ km/h less, the journey would have taken 1 hour longer. Find the slower speed.

   Show the worked solution

   1. Step 1: Time = distance ÷ speed, for each journeyAt $x$ km/h: $\dfrac{112}{x}$ hoursAt $(x - 2)$ km/h: $\dfrac{112}{x - 2}$ hours
   2. Step 2: Which relationship gives the equation?slower time − faster time = time difference
   3. Step 3: Form the equation$\dfrac{112}{x - 2} - \dfrac{112}{x} = 1$
   4. Step 4: Multiply every term by the LCM of the denominators$112x - 112(x - 2) = x(x - 2)$
   5. Step 5: Expand and make one side zero$x^{2} - 2x - 224 = 0$
   6. Step 6: Factorise$(x - 16)(x + 14) = 0$
   7. Step 7: Set each factor to zero$x - 16 = 0$ or $x + 14 = 0$$x = 16$ or $x = -14$
   8. Step 8: Reject the root that does not fit the problem$x = -14$ is rejected because a speed cannot be negative. So $x = 16$.
   9. Step 9: Answer the question asked: the slower speed, not xSlower speed $= 16 - 2$$= 14$ km/h
6. Question 6Cost and quantity

   A group of $x$ pupils shared the cost of a $$270$ gift equally. If $3$ more pupils had joined the group, each pupil would have paid $$3$ less. Find the amount each pupil paid.

   Show the worked solution

   1. Step 1: Each pupil's share = cost ÷ number of pupilsNow: $\dfrac{270}{x}$ dollarsWith $3$ more: $\dfrac{270}{x + 3}$ dollars
   2. Step 2: Which relationship gives the equation?bigger share − smaller share = difference
   3. Step 3: Form the equation$\dfrac{270}{x} - \dfrac{270}{x + 3} = 3$
   4. Step 4: Multiply every term by the LCM of the denominators$270(x + 3) - 270x = 3x(x + 3)$
   5. Step 5: Expand, make one side zero and divide by the common factor$3x^{2} + 9x - 810 = 0$$x^{2} + 3x - 270 = 0$
   6. Step 6: Factorise$(x - 15)(x + 18) = 0$
   7. Step 7: Set each factor to zero$x - 15 = 0$ or $x + 18 = 0$$x = 15$ or $x = -18$
   8. Step 8: Reject the root that does not fit the problem$x = -18$ is rejected because the number of people or things cannot be negative. So $x = 15$.
   9. Step 9: Answer the question asked: the amount each pupil paid, not xEach pupil paid $= \dfrac{270}{15}$$= 18$ dollars
7. Question 7Areas, by the formula

   The length of a rectangle is $9$ cm more than its width. Its area is $79$ cm². Let the width be $x$ cm. Find the length of the rectangle, correct to 2 decimal places.

   Show the worked solution

   1. Step 1: Which relationship gives the equation?length × width = area
   2. Step 2: Form the equation$x(x + 9) = 79$
   3. Step 3: Expand and make one side zero$x^{2} + 9x - 79 = 0$
   4. Step 4: Read off a, b and c$a = 1,\ b = 9,\ c = -79$
   5. Step 5: Substitute into the formula$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$x = \dfrac{-9 \pm \sqrt{(9)^2 - 4(1)(-79)}}{2(1)}$$x = \dfrac{-9 \pm \sqrt{397}}{2}$
   6. Step 6: Work out both values, then round$x = 5.4624\ldots$ or $x = -14.4624\ldots$$x = 5.46$ or $x = -14.46$ (2 d.p.)
   7. Step 7: Reject the root that does not fit the problem$x = -14.46$ is rejected because a length cannot be negative. So $x = 5.46$.
   8. Step 8: Answer the question asked: the length, not xLength $= 5.4624\ldots + 9$$= 14.46$ cm (2 d.p.)
8. Question 8Numbers, by the formula

   The sum of a positive number and its square is $108$. Let the number be $x$. Find the number, correct to 2 decimal places.

   Show the worked solution

   1. Step 1: The number plus its square$x + x^2 = 108$
   2. Step 2: Make one side zero$x^{2} + x - 108 = 0$
   3. Step 3: Read off a, b and c$a = 1,\ b = 1,\ c = -108$
   4. Step 4: Substitute into the formula$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$x = \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-108)}}{2(1)}$$x = \dfrac{-1 \pm \sqrt{433}}{2}$
   5. Step 5: Work out both values, then round$x = 9.9043\ldots$ or $x = -10.9043\ldots$$x = 9.90$ or $x = -10.90$ (2 d.p.)
   6. Step 6: Reject the root that does not fit the problem$x = -10.90$ is rejected because the number is positive. So $x = 9.90$.
   7. Step 7: Answer the question askedThe number $= 9.90$ (2 d.p.)
9. Question 9Number 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 $= 5$
10. Question 10Working together: rates of work

    Two taps fill an empty tank. Tap A on its own fills the tank in $x$ hours. Tap B on its own takes $4$ hours longer than tap A. With both taps turned on, the tank is filled in 5 hours 50 minutes. Form an equation in $x$ and solve it to find the time, in hours, tap B takes on its own.

    Show the worked solution

    1. Step 1: In one hour, the fraction of the job each does (1 ÷ the time it takes)Tap A: $\dfrac{1}{x}$Tap B: $\dfrac{1}{x + 4}$
    2. Step 2: Which relationship gives the equation?first part + second part = part done together
    3. Step 3: Form the equation (together, in one hour, they do 1 ÷ the time together)$\dfrac{1}{x} + \dfrac{1}{x + 4} = \dfrac{6}{35}$
    4. Step 4: Multiply every term by the LCM of the denominators$35(x + 4) + 35x = 6x(x + 4)$
    5. Step 5: Expand, make one side zero and divide by the common factor$6x^{2} - 46x - 140 = 0$$3x^{2} - 23x - 70 = 0$
    6. Step 6: Factorise$(x - 10)(3x + 7) = 0$
    7. Step 7: Set each factor to zero$x - 10 = 0$ or $3x + 7 = 0$$x = 10$ or $x = -\dfrac{7}{3}$
    8. Step 8: Reject the root that does not fit the problem$x = -\dfrac{7}{3}$ is rejected because a time cannot be negative. So $x = 10$.
    9. Step 9: Answer the question asked: the time for tap b, not xTime for tap B $= 10 + 4$$= 14$ hours