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

Form a quadratic equation for each situation and solve it.

1. (a) The base of a triangle is 4 cm longer than its perpendicular height \(h\) cm. The area of the triangle is 21 cm². Find the height to 3 significant figures.

   • Area = ½ × base × height: \(\frac12 h(h + 4) = 21\).
   • \(h^{2} + 4 h - 42 = 0\)
   • From the calculator: \(h = −8.78\) or \(h = 4.78\).
   • A height cannot be negative, so \(h = 4.78\) cm.

2. (b) Two numbers differ by 2. Their product is 11. 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 \(x + 2\): \(x(x + 2) = 11\).
   • \(x^{2} + 2 x - 11 = 0\)
   • From the calculator: \(x = −4.46\) or \(x = 2.46\). Both work: the question does not say the numbers are positive.

Rearrange and solve a quadratic.

1. (a) Show that \((3x + 2)(x − 5) = 14\) 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: \((3x + 2)(x − 5) = 3 x^{2} - 13 x - 10\).
   • Subtract 14 from both sides: \(3 x^{2} - 13 x - 24 = 0\).
   • \(3 x^{2} - 13 x - 24 = 0\), so \(a = 3\), \(b = −13\), \(c = −24\).

2. (b) Hence solve \((3x + 2)(x − 5) = 14\). Give the solutions to 3 significant figures.

   • Polynomial solver: \(3 x^{2} - 13 x - 24 = 0\)
   • From the calculator: \(x = −1.40\) or \(x = 5.73\)

Turn a fraction equation into a cubic and solve it.

1. (a) Show that \(x^2 − 5 = \dfrac{3}{x + 2}\) 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 \((x + 2)\): \((x^2 − 5)(x + 2) = 3\).
   • Expand: \(x^{3} + 2 x^{2} - 5 x - 10 = 3\).
   • \(x^{3} + 2 x^{2} - 5 x - 13 = 0\), so \(a = 1\), \(b = 2\), \(c = −5\), \(d = −13\).

2. (b) Hence solve \(x^2 − 5 = \dfrac{3}{x + 2}\). Give the real solutions to 3 significant figures.

   • Polynomial solver: \(x^{3} + 2 x^{2} - 5 x - 13 = 0\)
   • From the calculator: \(x = 2.38\) (the cubic has only 1 real root).
   • Check that \(x \neq −2\): 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 990. 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 \(m\).
   • \((m − 1)m(m + 1) = 990\), so \(m^3 − m − 990 = 0\).
   • Polynomial solver: \(m = 10\) (the only real root). The mean is 10.
   • Check: 9 × 10 × 11 = 990.

2. (b) Two numbers differ by 1. Their cubes differ by 5. Find the possible values of the smaller number to 3 significant figures.

   • Let the numbers be \(x\) and \(x + 1\): \((x + 1)^3 − x^3 = 5\).
   • Expand: \(3 x^{2} + 3 x - 4 = 0\).
   • From the calculator: \(x = −1.76\) or \(x = 0.758\).

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

1. (a) The first term of an arithmetic sequence is 5 and the common difference is 3. The sum of the first \(n\) terms is 670. Find the value of \(n\).

   • \(S_n = \frac n2\bigl(2u_1 + (n − 1)d\bigr)\): \(\frac n2(2 × 5 + (n − 1) × 3) = 670\).
   • \(3 n^{2} + 7 n - 1340 = 0\).
   • From the calculator: \(n = −22.3\) or \(n = 20.0\). \(n\) must be a positive whole number, so \(n = 20\).

2. (b) At a meeting, every person shakes hands once with every other person. There are 105 handshakes in total. Find the number of people, \(n\).

   • Each of the \(n\) people shakes hands with \(n − 1\) others, and each handshake is counted twice: \(\frac{n(n − 1)}{2} = 105\).
   • \(n^{2} - n - 210 = 0\).
   • From the calculator: \(n = −14.0\) or \(n = 15.0\). A number of people cannot be negative, so \(n = 15\).

A cuboid has edges of length \(x\) cm, \((x + 3)\) cm and \((2x + 1)\) cm, as shown.

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

   • \(V = x(x + 3)(2x + 1)\)
   • \(V = 2 x^{3} + 7 x^{2} + 3 x\)

2. (b) The volume of the cuboid is 200 cm³. Find the value of \(x\) to 3 significant figures.

   • \(2 x^{3} + 7 x^{2} + 3 x = 200\), so \(2 x^{3} + 7 x^{2} + 3 x - 200 = 0\).
   • Polynomial solver: \(x = 3.64\) (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 150. Find the edge length of the cube to 3 significant figures.

   • Edge \(x\): volume \(x^3\), surface area \(6x^2\). \(x^3 + 6x^2 = 150\).
   • \(x^{3} + 6 x^{2} - 150 = 0\)
   • From the calculator: \(x = 3.89\).

A shop sells bicycles (2 wheels) and tricycles (3 wheels). It has 17 vehicles with 41 wheels in total. Let \(b\) be the number of bicycles and \(t\) the number of tricycles.

1. (a) Write down two equations in \(b\) and \(t\).

   • Vehicles: \(b + t = 17\).
   • Wheels: \(2b + 3t = 41\).

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

   • Simultaneous equation solver: \(b + t = 17\), \(2b + 3t = 41\).
   • From the calculator: \(b = 10\), \(t = 7\).
   • There are 7 tricycles.

Tickets for a concert are sold to adults and students. Let \(x\) be the number of adult tickets and \(y\) the number of student tickets sold. Adult tickets cost €14 and student tickets cost €9. 3 000 tickets are sold.

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

   • Every ticket is adult or student: \(x + y = 3000\).

2. (b) Find the cost of tickets for a group of 3 adults and 4 students.

   • 3 × €14 + 4 × €9 = €42 + €36 = €78.

3. (c) The total money from ticket sales is €35 700. Write down a second equation in \(x\) and \(y\).

   • Money from adults: \(14x\). Money from students: \(9y\).
   • \(14x + 9y = 35700\).

4. (d) Find the number of adult tickets and the number of student tickets sold.

   • Simultaneous equation solver: \(x + y = 3000\), \(14x + 9y = 35700\).
   • From the calculator: \(x = 1740\), \(y = 1260\).

Pipes are stacked in rows. Each row has one fewer than the row below, and the top row has 1. A stack has \(n\) in the bottom row and \(S\) in total.

1. (a) Show that a stack with 25 in the bottom row has 325 in total.

   • The rows are 1, 2, 3, …, 25: an arithmetic series with \(u_1 = 1\), \(d = 1\), \(n = 25\).
   • \(S_{25} = \frac{25}{2}(1 + 25) = 325\).

2. (b) A stack has 1 830 in total. Find the number in the bottom row.

   • \(\frac{n(n + 1)}{2} = 1830\), so \(n^2 + n − 3660 = 0\).
   • From the calculator: \(n = 60\) (the other root is negative).
   • There are 60 in the bottom 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 builder has 1 000. Explain why they cannot be stacked in this way with none left over.

   • \(n^2 + n − 2000 = 0\).
   • From the calculator: \(n = 44.2\).
   • \(n\) must be a whole number, so this number cannot make a full stack.

A rectangular lawn measures \((x + 5)\) m by \((x + 2)\) m. A square patio of side 2 m is built in one corner. The rest of the lawn 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: \((x + 5)(x + 2) = x^{2} + 7 x + 10\).
   • Take away the square: \(2^2 = 4\).
   • \(A = x^{2} + 7 x + 6\).

2. (b) The area of the shaded region is 100 m². Find \(x\) to 3 significant figures.

   • \(x^{2} + 7 x + 6 = 100\), so \(x^{2} + 7 x - 94 = 0\).
   • From the calculator: \(x = 6.81\) (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 \(= 2\bigl((x + 5) + (x + 2)\bigr) = 4 x + 14\).
   • With \(x = 6.81\): perimeter \(= 41.2\) m.

Use the sequence formulae and the calculator.

1. (a) In an arithmetic sequence, \(S_{30} = 1 515\) and \(u_{30} = 94\). Find \(u_1\) and \(d\).

   • \(u_{30} = u_1 + 29d = 94\).
   • \(S_{30} = \frac{30}{2}(u_1 + u_{30}) = 1 515\), so \(u_1 + 94 = 101\), giving \(u_1 = 7\).
   • Then \(29d = 94 − (7) = 87\), so \(d = 3\).

2. (b) A geometric sequence has first term 9 and the sum of its first three terms is 13. Find the possible values of the common ratio \(r\), to 3 significant figures.

   • \(u_1 + u_1r + u_1r^2 = 9(1 + r + r^2) = 13\).
   • \(9r^2 + 9r - 4 = 0\).
   • From the calculator: \(r = −1.33\) or \(r = 0.333\).

In triangle ABC, \(AB = x\), \(AC = (x + 4)\), \(BC = 2x\) and angle \(BAC = 60°\).

1. (a) Show that \(3 x^{2} - 4 x - 16 = 0\).

   • Cosine rule: \(BC^2 = AB^2 + AC^2 − 2(AB)(AC)\cos 60°\).
   • \((2x)^2 = x^2 + (x + 4)^2 − 2x(x + 4)\cos 60°\), with \(\cos 60° = \frac{1}{2}\).
   • \(4 x^{2} = x^{2} + 4 x + 16\), so \(3 x^{2} - 4 x - 16 = 0\).

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

   • Polynomial solver: \(3 x^{2} - 4 x - 16 = 0\)
   • From the calculator: \(x = 3.07\) (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 \((3, 15)\) and has gradient 11 at that point. Find the value of \(y\) when \(x = −2\).

   • The curve passes through \((3, 15)\): \(9a + 3b = 15\).
   • Gradient: \(\frac{dy}{dx} = 2ax + b = 11\) at \(x = 3\): \(6a + b = 11\).
   • Simultaneous equation solver: \(a = 2\), \(b = −1\).
   • \(y = 2x^2 + (-1)x\), so at \(x = -2\): \(y = 10\).

2. (b) Solve the simultaneous equations \(x^3 + y^3 + z^3 = −4,\ 2x^3 − y^3 + z^3 = 15,\ x^3 + 2y^3 − z^3 = −14\).

   • Let \(a = x^3\), \(b = y^3\), \(c = z^3\). The equations are linear in \(a\), \(b\), \(c\).
   • Simultaneous equation solver: \(a = 3\), \(b = -8\), \(c = 1\).
   • Take cube roots: \(x = \sqrt[3]{3} = 1.44\), \(y = −2.00\), \(z = 1.00\).

Every student in a year group of 40 plays hockey or tennis or both. The number who play hockey is 5 more than the number who play tennis. The number who play both is 14 less than the number who play only one of the two sports.

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

   • Everyone is in at least one set: \(h + t + b = 40\).
   • The number who do hockey is \(h + b\) and the number who do tennis is \(t + b\). They differ by 5: \(h − t = 5\).
   • The number who do both is 14 less than the number who do only one: \(b = h + t − 14\).

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

   • Write as \(h + t + b = 40\), \(h − t = 5\), \(−h − t + b = −14\).
   • Simultaneous equation solver: \(h = 16\), \(t = 11\), \(b = 13\).
   • \(P(\text{both}) = \frac{13}{40} = \frac{13}{40} = 0.325\).