Solving and forming polynomial equations Problem set A Worked solutions

Use the polynomial solver to find the real roots of each equation.

1. (a) \(x^2 - 8x + 3=0\)

   • Coefficients 1, −8, 3: x = 0.394 or 7.61

2. (b) \(25x^2 - 30x + 9=0\)

   • Coefficients 25, −30, 9: the solver gives one root, x = 0.6
   • It is repeated: the graph touches the x-axis

3. (c) \(2x^2 - 3x + 5=0\)

   • Coefficients 2, −3, 5: x = 0.75 ± 1.39i
   • Both roots contain i, so there are no real roots

Find the real roots of each equation.

1. (a) \(x^3 - 5x + 1=0\)

   • Coefficients 1, 0, −5, 1 (the x² term is missing: enter 0)
   • x = −2.33, 0.202 or 2.13

2. (b) \(x^3 + 2x^2 + 3x + 5=0\)

   • Coefficients 1, 2, 3, 5: x = −1.84 and a complex pair
   • Only the real root counts

Find the real roots of each equation.

1. (a) \(x^4 - 7x^2 + 2x + 3=0\)

   • Coefficients 1, 0, −7, 2, 3 (the x³ term is missing: enter 0)
   • x = −2.71, −0.536, 0.872 or 2.37

2. (b) \(x^4 + 2x^3 - 3x - 5=0\)

   • Coefficients 1, 2, 0, −3, −5 (the x² term is missing: enter 0)
   • x = −1.90 or 1.39 and a complex pair
   • Only the real roots count

Each equation can be written as a polynomial equation.

1. (a) Show that \(3x-1=\frac{6}{x+2}\) can be written as \(3x^2 + 5x - 8=0\). Hence find the roots.

   • Multiply both sides by \((x+2)\): \((3x-1)(x+2)=6\)
   • Expand and subtract 6: \(3x^2 + 5x - 8=0\)
   • Coefficients 3, 5, −8: x = −2.67 or 1

2. (b) Show that \(x^2=4+2x^{-1}\) can be written in the form \(ax^3+bx^2+cx+d=0\), where a, b, c and d are whole numbers. Hence solve the equation.

   • Multiply every term by x: \(x^3=4x+2\)
   • \(x^3-4x-2=0\)
   • Coefficients 1, 0, −4, −2: x = −1.68, −0.539 or 2.21

The base of a triangle is 5 cm longer than its perpendicular height. The area is 42 cm². Find the height of the triangle.

• Height x cm, base \((x+5)\) cm: \(\frac12x(x+5)=42\)
• \(x^2+5x-84=0\)
• Coefficients 1, 5, −84: x = −12 or 7
• A height is positive: 7 cm

A geometric sequence has first term 8 and the sum of the first three terms is 14. Find the possible values of the common ratio r.

• \(\frac{8(1-r^3)}{1-r}=14\), so \(14(1-r)=8(1-r^3)\)
• \(8r^3-14r+6=0\)
• Coefficients 8, 0, −14, 6: r = −1.5, 0.5 or 1
• r = 1 is not allowed (the formula needs r ≠ 1)
• r = −1.5 or 0.5

A right-angled triangle has sides x cm, \((x+4)\) cm and \((2x-2)\) cm, with \((2x-2)\) cm the longest side.

1. (a) Show that \(2x^2 - 16x - 12=0\).

   • \(x^2+(x+4)^2=(2x-2)^2\)
   • \(2x^2+8x+16=4x^2-8x+4\)
   • \(2x^2 - 16x - 12=0\)

2. (b) Find x, and explain why one solution is rejected.

   • Coefficients 2, −16, −12: x = −0.690 or 8.69
   • x = −0.690 gives negative lengths, so it is rejected
   • x = 8.69

A rectangular lawn measures 12 m by 8 m. A path of width x m is laid outside the lawn along one long side and one short side, as in the diagram. The area of the path is 30 m².

1. (a) Show that \(x^2+20x-30=0\).

   • Whole rectangle: \((12+x)(8+x)\); lawn: 96
   • Path = \((12+x)(8+x)-96=x^2+20x\)
   • Set equal to 30: \(x^2+20x-30=0\)

2. (b) Find x.

   • Coefficients 1, 20, −30: x = −21.4 or 1.40
   • A width is positive: x = 1.40 m

A piece of card measures 36 cm by 24 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box of volume 1500 cm³.

1. (a) Show that \(4x^3 - 120x^2 + 864x - 1500=0\).

   • Box: length \(36-2x\), width \(24-2x\), height x
   • \(x(36-2x)(24-2x)=1500\)
   • \(4x^3 - 120x^2 + 864x - 1500=0\)

2. (b) Find the possible values of x.

   • Coefficients 4, −120, 864, −1500: x = 2.58, 7.17 or 20.2
   • x must be less than 12, so reject x = 20.2
   • x = 2.58 or 7.17

The volume of a cube, in cm³, and its surface area, in cm², are added together. The total is 300. Find the length of an edge.

• Edge x: volume \(x^3\), surface area \(6x^2\)
• \(x^3+6x^2=300\), so \(x^3+6x^2+0x-300=0\)
• Coefficients 1, 6, 0, −300: x = 5.18 and a complex pair
• The edge is 5.18 cm

A closed cylindrical tin has radius r cm and height h cm. Its volume is 300 cm³ and its total surface area is 300 cm².

1. (a) Show that \(h=\frac{150-\pi r^2}{\pi r}\).

   • Surface area: \(2\pi r^2+2\pi rh=300\)
   • \(\pi r^2+\pi rh=150\)
   • \(h=\frac{150-\pi r^2}{\pi r}\)

2. (b) Hence show that \(\pi r^3-150r+300=0\).

   • Volume: \(\pi r^2h=300\)
   • \(r(150-\pi r^2)=300\)
   • \(\pi r^3-150r+300=0\)

3. (c) Find the possible values of r.

   • Coefficients \(\pi\), 0, −150, 300: r = −7.75, 2.23 or 5.52
   • A radius is positive: r = 2.23 or 5.52
   • Check h > 0 for each: both are valid

The first term of an arithmetic sequence is 4 and the common difference is 3. The sum of the first n terms is 924. Find n.

• \(S_n=\frac n2(2\times4+(n-1)3)=924\)
• \(3n^2+5n-1848=0\)
• Coefficients 3, 5, −1848: n = −25.7 or 24
• n is a positive whole number: n = 24

The difference between two consecutive perfect cubes is 217. Find the possible pairs of cubes.

• The cubes are \(n^3\) and \((n+1)^3\): \((n+1)^3-n^3=217\)
• \(3n^2+3n+1=217\), so \(3n^2+3n-216=0\)
• Coefficients 3, 3, −216: n = −9 or 8
• n = 8: 512 and 729. n = −9: −729 and −512

A regular polygon has n sides. In the diagram, the dashed lines are the diagonals from one corner of a hexagon. A diagonal joins two corners that are not next to each other.

1. (a) Find an expression for the number of diagonals of the polygon.

   • Each corner joins to n − 3 others by a diagonal (not itself, not its two neighbours)
   • n corners give \(n(n-3)\), but each diagonal is counted twice
   • Number of diagonals \(=\frac{n(n-3)}{2}\)

2. (b) A polygon has 44 diagonals. Find n.

   • \(\frac{n(n-3)}{2}=44\), so \(n^2-3n-88=0\)
   • Coefficients 1, −3, −88: n = −8 or 11
   • n is a positive whole number: n = 11