Mixed practice: equations Problem set B One question at a time

Form a quadratic equation for each situation and solve it.

1. (a) The base of a triangle is 6 cm longer than its perpendicular height \(h\) cm. The area of the triangle is 17 cm². Find the height to 3 significant figures.

   • Area = ½ × base × height: \(\frac12 h(h + 6) = 17\).
   • \(h^{2} + 6 h - 34 = 0\)
   • From the calculator: \(h = −9.56\) or \(h = 3.56\).
   • A height cannot be negative, so \(h = 3.56\) cm.

2. (b) Two numbers differ by 4. Their product is 9. The smaller number is \(x\). Write this as a quadratic equation \(ax^2 + bx + c = 0\) and find the possible values of \(x\) to 3 significant figures.

   • The numbers are \(x\) and \(x + 4\): \(x(x + 4) = 9\).
   • \(x^{2} + 4 x - 9 = 0\)
   • From the calculator: \(x = −5.61\) or \(x = 1.61\). Both work: the question does not say the numbers are positive.

Rearrange and solve a quadratic.

1. (a) Show that \((2x − 3)(x + 4) = 16\) can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\) and \(c\) are integers. Write down \(a\), \(b\) and \(c\).

   • Expand: \((2x − 3)(x + 4) = 2 x^{2} + 5 x - 12\).
   • Subtract 16 from both sides: \(2 x^{2} + 5 x - 28 = 0\).
   • \(2 x^{2} + 5 x - 28 = 0\), so \(a = 2\), \(b = 5\), \(c = −28\).

2. (b) Hence solve \((2x − 3)(x + 4) = 16\). Give the solutions to 3 significant figures.

   • Polynomial solver: \(2 x^{2} + 5 x - 28 = 0\)
   • From the calculator: \(x = −5.19\) or \(x = 2.69\)

Turn a fraction equation into a cubic and solve it.

1. (a) Show that \(x^2 − 6 = \dfrac{4}{x + 3}\) can be written in the form \(ax^3 + bx^2 + cx + d = 0\), where \(a\), \(b\), \(c\) and \(d\) are integers. Write down \(a\), \(b\), \(c\) and \(d\).

   • Multiply both sides by \((x + 3)\): \((x^2 − 6)(x + 3) = 4\).
   • Expand: \(x^{3} + 3 x^{2} - 6 x - 18 = 4\).
   • \(x^{3} + 3 x^{2} - 6 x - 22 = 0\), so \(a = 1\), \(b = 3\), \(c = −6\), \(d = −22\).

2. (b) Hence solve \(x^2 − 6 = \dfrac{4}{x + 3}\). Give the real solutions to 3 significant figures.

   • Polynomial solver: \(x^{3} + 3 x^{2} - 6 x - 22 = 0\)
   • From the calculator: \(x = 2.59\) (the cubic has only 1 real root).
   • Check that \(x \neq −3\): it is not, so the root is allowed.

Form a polynomial equation and solve it with the calculator.

1. (a) The product of three consecutive integers is 2184. Find the mean of the three integers.

   • Let the middle integer be \(m\). The integers are \(m − 1\), \(m\), \(m + 1\), and their mean is \(m\).
   • \((m − 1)m(m + 1) = 2184\), so \(m^3 − m − 2184 = 0\).
   • Polynomial solver: \(m = 13\) (the only real root). The mean is 13.
   • Check: 12 × 13 × 14 = 2184.

2. (b) Two numbers differ by 1. Their cubes differ by 8. Find the possible values of the smaller number to 3 significant figures.

   • Let the numbers be \(x\) and \(x + 1\): \((x + 1)^3 − x^3 = 8\).
   • Expand: \(3 x^{2} + 3 x - 7 = 0\).
   • From the calculator: \(x = −2.11\) or \(x = 1.11\).

Form a quadratic in \(n\) and solve it.

1. (a) The first term of an arithmetic sequence is 4 and the common difference is 5. The sum of the first \(n\) terms is 837. Find the value of \(n\).

   • \(S_n = \frac n2\bigl(2u_1 + (n − 1)d\bigr)\): \(\frac n2(2 × 4 + (n − 1) × 5) = 837\).
   • \(5 n^{2} + 3 n - 1674 = 0\).
   • From the calculator: \(n = −18.6\) or \(n = 18.0\). \(n\) must be a positive whole number, so \(n = 18\).

2. (b) In a chess tournament, every player plays every other player once. There are 136 games in total. Find the number of players, \(n\).

   • Each of the \(n\) players plays \(n − 1\) others, and each game is counted twice: \(\frac{n(n − 1)}{2} = 136\).
   • \(n^{2} - n - 272 = 0\)
   • From the calculator: \(n = −16.0\) or \(n = 17.0\). A number of players cannot be negative, so \(n = 17\).

A storage crate is a cuboid with edges of length \(x\) cm, \((x + 4)\) cm and \((3x + 1)\) cm, as shown.

1. (a) Write down an expression for the volume \(V\) of the cuboid. Expand and simplify it.

   • \(V = x(x + 4)(3x + 1)\)
   • \(V = 3 x^{3} + 13 x^{2} + 4 x\)

2. (b) The volume of the cuboid is 160 cm³. Find the value of \(x\) to 3 significant figures.

   • \(3 x^{3} + 13 x^{2} + 4 x = 160\), so \(3 x^{3} + 13 x^{2} + 4 x - 160 = 0\).
   • Polynomial solver: \(x = 2.67\) (the other roots are not positive real lengths).

3. (c) The numerical values of the volume and the surface area of a cube are added together. The total is 200. Find the edge length of the cube to 3 significant figures.

   • Edge \(x\): volume \(x^3\), surface area \(6x^2\). \(x^3 + 6x^2 = 200\).
   • \(x^{3} + 6 x^{2} - 200 = 0\)
   • From the calculator: \(x = 4.39\).

A car park holds cars (4 wheels) and motorbikes (2 wheels). There are 25 vehicles with 78 wheels in total. Let \(c\) be the number of cars and \(m\) the number of motorbikes.

1. (a) Write down two equations in \(c\) and \(m\).

   • Vehicles: \(c + m = 25\).
   • Wheels: \(4c + 2m = 78\).

2. (b) Solve them with the calculator. How many cars are there?

   • Simultaneous equation solver: \(c + m = 25\), \(4c + 2m = 78\).
   • From the calculator: \(c = 14\), \(m = 11\).
   • There are 14 cars.

A market stall sells small boxes and large boxes of strawberries. Let \(x\) be the number of large boxes and \(y\) the number of small boxes sold. A large box costs €16 and a small box costs €8. 2 500 boxes are sold.

1. (a) Write down an equation in \(x\) and \(y\) for the number of boxes sold.

   • Every box is large or small: \(x + y = 2500\).

2. (b) Find the cost of 2 large boxes and 5 small boxes.

   • 2 × €16 + 5 × €8 = €32 + €40 = €72.

3. (c) The total money from sales is €31 600. Write down a second equation in \(x\) and \(y\).

   • Money from large boxes: \(16x\). Money from small boxes: \(8y\).
   • \(16x + 8y = 31600\).

4. (d) Find the number of large boxes and the number of small boxes sold.

   • Simultaneous equation solver: \(x + y = 2500\), \(16x + 8y = 31600\).
   • From the calculator: \(x = 1450\), \(y = 1050\).

A forester plants trees in rows to make a triangle. The first row has 1 tree and each row has one more tree than the row before. A plantation has \(n\) trees in the last row and \(S\) in total.

1. (a) Show that a plantation with 30 trees in the last row has 465 trees in total.

   • The rows are 1, 2, 3, …, 30: an arithmetic series with \(u_1 = 1\), \(d = 1\), \(n = 30\).
   • \(S_{30} = \frac{30}{2}(1 + 30) = 465\).

2. (b) A plantation has 1 275 trees in total. Find the number of trees in the last row.

   • \(\frac{n(n + 1)}{2} = 1275\), so \(n^2 + n − 2550 = 0\).
   • From the calculator: \(n = 50\) (the other root is negative).
   • There are 50 trees in the last row.

3. (c) Show that \(n^2 + n − 2S = 0\).

   • \(S = \frac n2(1 + n)\), so \(2S = n + n^2\).
   • Rearrange: \(n^2 + n − 2S = 0\).

4. (d) A forester has 600 trees. Explain why they cannot be planted in this way with none left over.

   • \(n^2 + n − 1200 = 0\).
   • From the calculator: \(n = 34.1\).
   • \(n\) must be a whole number, so 600 trees cannot make a full triangle.

A rectangular sheet of card measures \((x + 7)\) cm by \((x + 1)\) cm. A square of side 1.5 cm is cut from one corner. The card that is left is the shaded region.

1. (a) Write down an expression for the area \(A\) of the shaded region in terms of \(x\). Expand and simplify it.

   • Whole rectangle: \((x + 7)(x + 1) = x^{2} + 8 x + 7\).
   • Take away the square: \(1.5^2 = 2.25\).
   • \(A = x^{2} + 8 x + 4.75\).

2. (b) The area of the shaded region is 60 cm². Find \(x\) to 3 significant figures.

   • \(x^{2} + 8 x + 4.75 = 60\), so \(x^{2} + 8 x - 55.25 = 0\).
   • From the calculator: \(x = 4.44\) (the other root is negative).

3. (c) Find the perimeter of the shaded region to 3 significant figures.

   • Removing a corner square leaves the perimeter of the rectangle unchanged.
   • Perimeter \(= 2\bigl((x + 7) + (x + 1)\bigr) = 4 x + 16\).
   • With \(x = 4.44\): perimeter \(= 33.8\) cm.

Use the sequence formulae and the calculator.

1. (a) In an arithmetic sequence, \(S_{24} = 1 284\) and \(u_{24} = 111\). Find \(u_1\) and \(d\).

   • \(u_{24} = u_1 + 23d = 111\).
   • \(S_{24} = \frac{24}{2}(u_1 + u_{24}) = 1 284\), so \(u_1 + 111 = 107\), giving \(u_1 = −4\).
   • Then \(23d = 111 − (-4) = 115\), so \(d = 5\).

2. (b) A geometric sequence has first term 5 and the sum of its first three terms is 7.2. Find the possible values of the common ratio \(r\), to 3 significant figures.

   • \(u_1 + u_1r + u_1r^2 = 5(1 + r + r^2) = 7.2\).
   • \(5r^2 + 5r - 2.2 = 0\).
   • From the calculator: \(r = −1.33\) or \(r = 0.331\).

In triangle ABC, \(AB = x\), \(AC = (x + 2)\), \(BC = 3x\) and angle \(BAC = 120°\).

1. (a) Show that \(3 x^{2} - 3 x - 2 = 0\).

   • Cosine rule: \(BC^2 = AB^2 + AC^2 − 2(AB)(AC)\cos 120°\).
   • \((3x)^2 = x^2 + (x + 2)^2 − 2x(x + 2)\cos 120°\), with \(\cos 120° = - \frac{1}{2}\).
   • \(9 x^{2} = 3 x^{2} + 6 x + 4\), so \(6 x^{2} - 6 x - 4 = 0\). Divide by 2: \(3 x^{2} - 3 x - 2 = 0\).

2. (b) Hence find the value of \(x\) to 3 significant figures.

   • Polynomial solver: \(3 x^{2} - 3 x - 2 = 0\)
   • From the calculator: \(x = 1.46\) (the other root is negative, so rejected).

Form equations and solve them with the calculator.

1. (a) The curve \(y = ax^2 + bx\) passes through the point \((2, -2)\) and has gradient 5 at that point. Find the value of \(y\) when \(x = 4\).

   • The curve passes through \((2, -2)\): \(4a + 2b = -2\).
   • Gradient: \(\frac{dy}{dx} = 2ax + b = 5\) at \(x = 2\): \(4a + b = 5\).
   • Simultaneous equation solver: \(a = 3\), \(b = −7\).
   • \(y = 3x^2 + (-7)x\), so at \(x = 4\): \(y = 20\).

2. (b) Solve the simultaneous equations \(x^3 + y^3 + z^3 = 21,\ 3x^3 + y^3 − z^3 = 13,\ x^3 − 2y^3 + 3z^3 = −62\).

   • Let \(a = x^3\), \(b = y^3\), \(c = z^3\). The equations are linear in \(a\), \(b\), \(c\).
   • Simultaneous equation solver: \(a = -5\), \(b = 27\), \(c = -1\).
   • Take cube roots: \(x = \sqrt[3]{-5} = −1.71\), \(y = 3.00\), \(z = −1.00\).

Every member of a club of 60 does swimming or cycling or both. The number who swim is 4 more than the number who cycle. The number who do both is 16 less than the number who do only one of the two activities.

1. (a) Let \(h\) be the number who do swimming only, \(t\) the number who do cycling only and \(b\) the number who do both. Write down three equations in \(h\), \(t\) and \(b\).

   • Everyone is in at least one set: \(h + t + b = 60\).
   • The number who do swimming is \(h + b\) and the number who do cycling is \(t + b\). They differ by 4: \(h − t = 4\).
   • The number who do both is 16 less than the number who do only one: \(b = h + t − 16\).

2. (b) Solve the equations with the calculator. A person is chosen at random. Find the probability that they do both.

   • Write as \(h + t + b = 60\), \(h − t = 4\), \(−h − t + b = −16\).
   • Simultaneous equation solver: \(h = 21\), \(t = 17\), \(b = 22\).
   • \(P(\text{both}) = \frac{22}{60} = \frac{11}{30} = 0.367\).