Solving and forming polynomial equations Problem set B Worked solutions

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

1. (a) \(x^2 + 7x + 5=0\)

   • Coefficients 1, 7, 5: x = −6.19 or −0.807

2. (b) \(36x^2 + 12x + 1=0\)

   • Coefficients 36, 12, 1: the solver gives one root, x = −0.167
   • It is repeated: the graph touches the x-axis

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

   • Coefficients 3, −4, 7: x = 0.667 ± 1.37i
   • Both roots contain i, so there are no real roots

Find the real roots of each equation.

1. (a) \(x^3 - 6x + 2=0\)

   • Coefficients 1, 0, −6, 2 (the x² term is missing: enter 0)
   • x = −2.60, 0.340 or 2.26

2. (b) \(2x^3 + x^2 + 3x + 7=0\)

   • Coefficients 2, 1, 3, 7: x = −1.34 and a complex pair
   • Only the real root counts

Find the real roots of each equation.

1. (a) \(x^4 - 8x^2 - 3x + 4=0\)

   • Coefficients 1, 0, −8, −3, 4 (the x³ term is missing: enter 0)
   • x = −2.48, −1, 0.552 or 2.93

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

   • Coefficients 2, −1, 0, 4, −3 (the x² term is missing: enter 0)
   • x = −1.32 or 0.712 and a complex pair
   • Only the real roots count

Each equation can be written as a polynomial equation.

1. (a) Show that \(2x+1=\frac{7}{x-3}\) can be written as \(2x^2 - 5x - 10=0\). Hence find the roots.

   • Multiply both sides by \((x-3)\): \((2x+1)(x-3)=7\)
   • Expand and subtract 7: \(2x^2 - 5x - 10=0\)
   • Coefficients 2, −5, −10: x = −1.31 or 3.81

2. (b) Show that \(x^2=5+3x^{-1}\) 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: \(x^3=5x+3\)
   • \(x^3-5x-3=0\)
   • Coefficients 1, 0, −5, −3: x = −1.83, −0.657 or 2.49

The base of a triangle is 4 cm longer than its perpendicular height. The area is 30 cm². Find the height of the triangle.

• Height x cm, base \((x+4)\) cm: \(\frac12x(x+4)=30\)
• \(x^2+4x-60=0\)
• Coefficients 1, 4, −60: x = −10 or 6
• A height is positive: 6 cm

A geometric sequence has first term 10 and the sum of the first three terms is 15.6. Find the possible values of the common ratio r.

• \(\frac{10(1-r^3)}{1-r}=15.6\), so \(15.6(1-r)=10(1-r^3)\)
• \(10r^3-15.6r+5.6=0\)
• Coefficients 10, 0, −15.6, 5.6: r = −1.4, 0.4 or 1
• r = 1 is not allowed (the formula needs r ≠ 1)
• r = −1.4 or 0.4

A right-angled triangle has sides x cm, \((x+6)\) cm and \((2x-3)\) cm, with \((2x-3)\) cm the longest side.

1. (a) Show that \(2x^2 - 24x - 27=0\).

   • \(x^2+(x+6)^2=(2x-3)^2\)
   • \(2x^2+12x+36=4x^2-12x+9\)
   • \(2x^2 - 24x - 27=0\)

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

   • Coefficients 2, −24, −27: x = −1.04 or 13.0
   • x = −1.04 gives negative lengths, so it is rejected
   • x = 13.0

A rectangular swimming pool measures 15 m by 6 m. A paved strip of width x m is laid outside the pool along one long side and one short side, as in the diagram. The area of the strip is 40 m².

1. (a) Show that \(x^2+21x-40=0\).

   • Whole rectangle: \((15+x)(6+x)\); pool: 90
   • Strip = \((15+x)(6+x)-90=x^2+21x\)
   • Set equal to 40: \(x^2+21x-40=0\)

2. (b) Find x.

   • Coefficients 1, 21, −40: x = −22.8 or 1.76
   • A width is positive: x = 1.76 m

A baker has a sheet of metal that measures 48 cm by 30 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open baking tray of volume 3000 cm³.

1. (a) Show that \(4x^3 - 156x^2 + 1440x - 3000=0\).

   • Tray: length \(48-2x\), width \(30-2x\), height x
   • \(x(48-2x)(30-2x)=3000\)
   • \(4x^3 - 156x^2 + 1440x - 3000=0\)

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

   • Coefficients 4, −156, 1440, −3000: x = 2.96, 9.57 or 26.5
   • x must be less than 15, so reject x = 26.5
   • x = 2.96 or 9.57

The volume of a cube, in cm³, and its surface area, in cm², are added together. The total is 400. Find the length of an edge.

• Edge x: volume \(x^3\), surface area \(6x^2\)
• \(x^3+6x^2=400\), so \(x^3+6x^2+0x-400=0\)
• Coefficients 1, 6, 0, −400: x = 5.82 and a complex pair
• The edge is 5.82 cm

A closed cylindrical gas canister has radius r cm and height h cm. Its volume is 250 cm³ and its total surface area is 240 cm².

1. (a) Show that \(h=\frac{120-\pi r^2}{\pi r}\).

   • Surface area: \(2\pi r^2+2\pi rh=240\)
   • \(\pi r^2+\pi rh=120\)
   • \(h=\frac{120-\pi r^2}{\pi r}\)

2. (b) Hence show that \(\pi r^3-120r+250=0\).

   • Volume: \(\pi r^2h=250\)
   • \(r(120-\pi r^2)=250\)
   • \(\pi r^3-120r+250=0\)

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

   • Coefficients \(\pi\), 0, −120, 250: r = −7.04, 2.49 or 4.55
   • A radius is positive: r = 2.49 or 4.55
   • Check h > 0 for each: both are valid

The first term of an arithmetic sequence is 5 and the common difference is 4. The sum of the first n terms is 945. Find n.

• \(S_n=\frac n2(2\times5+(n-1)4)=945\)
• \(4n^2+6n-1890=0\)
• Coefficients 4, 6, −1890: n = −22.5 or 21
• n is a positive whole number: n = 21

The difference between two consecutive perfect cubes is 271. Find the possible pairs of cubes.

• The cubes are \(n^3\) and \((n+1)^3\): \((n+1)^3-n^3=271\)
• \(3n^2+3n+1=271\), so \(3n^2+3n-270=0\)
• Coefficients 3, 3, −270: n = −10 or 9
• n = 9: 729 and 1000. n = −10: −1000 and −729

n posts stand at the corners of a regular polygon. A rope is tied between every pair of posts that are not next to each other. The diagram shows the ropes from one post of a heptagon.

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

   • Each post is joined to n − 3 others by a rope (not itself, not its two neighbours)
   • n posts give \(n(n-3)\), but each rope is counted twice
   • Number of ropes \(=\frac{n(n-3)}{2}\)

2. (b) There are 54 ropes. Find n.

   • \(\frac{n(n-3)}{2}=54\), so \(n^2-3n-108=0\)
   • Coefficients 1, −3, −108: n = −9 or 12
   • n is a positive whole number: n = 12