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CUBIC EQUATIONS 133
(x + 3)3 = 16.8 Divide both sides of equation by 5
x + 3 = 2.56 Cube root of 16.8 is 2.56 to 2 dp
x = β0.44 Subtract 3 from both sides of equation
Example 4. Solve (x + 3)(x β 5)(5 β 2x) = 0.
Solution. If three brackets multiply together to make zero, then the equation is solved by equating each bracket to zero and obtaining three answers. Write
x + 3 = 0
x = β 3 Subtracting 3 from both sides of the equation
x β 5 = 0
x = 5 Adding 5 to both sides of the equation
5 β 2x = 0
β2x = β5 Subtracting 5 from both sides of the equation
x = 2.5 Dividing both sides of the equation by β2.
The three solutions are x = β3,5, 2.5
Example 5. Solve 4(x + 3)(x β 5)(5 β 2x) = 0.
(x + 3)(x β 5)(5 β 2x) = 0 Dividing both sides of the equation by 4
Then proceed as in the previous example to obtain the same three solutions as in Example 4.
Some cubic equations may have a squared bracket, as in the following example, in which case there are three solutions, but one is a repeated solution.
Example 6. Solve (x + 3)(x β 5)2 = 0.
(x + 3)(x β 5)(x β 5) = 0 Using an index law that b2 = b x b
The solutions are x = β3,5 (twice).
We say that x = 5 is a repeated solution, or a repeated root.
Example 7. David loves the outdoors and he made a sketch of a hillside with a small lake at the foot of the hill. He modeled the scene with an equation and drew its graph,
which is shown in the figure. All lengths are in meters. How long is the small lake if the equation of the graph is
y = Β±j(x +20)(.x - 50)0 - 80)?
Solution. The curve crosses the x-axis at x = A and x = B, and the length of the lake is B β A. Write
j^(x + 20)(x - 50)0 - 80) = 0 0 + 20)0 - 50)0 - 80) = 0
x + 20 = 0, x β 50 = 0, x β 80 = 0 x = β20, x = 50, x = 80 A = 50, B = 80
Length of lake = 30 meters
The curve crosses the x-axis when y = 0
Dividing both sides of the equation
Equating each bracket to zero
These are the three solutions
The curve crosses the x-axis at x = A andx = B
Length of lake = B β A
References: Cube Root, Cubic Equations, Cubic Graphs, Linear Equation.
Reference: Quadratic Graphs.
This is a hexahedron, which is a solid shape that has six rectangular faces that meet at right angles. The hexahedron is not regular; otherwise its faces would be squares and it would be a cube. The cuboid is also known as a rectangular block. The dimensions of the cuboid are length, width, and height, denoted by L, W, and H, respectively (see figure a).
Ceiling Left Floor Right
The net of the cuboid, which is made up of six rectangles, can take many arrangements; only one of them is drawn in figure a. These six rectangles are made up of three pairs of congruent rectangles. They can be thought of as floor and ceiling, front and back, and right and left.
The volume of the cuboid is given by
V = Length x width x height, cubic units
V = LWH Expressed in shortened form
The units of volume may be cubic centimeters (cm3), cubic meters (m3), and so on.
The surface area of the cuboid is the surface area of the net, which is made up of three pairs of rectangles:
A = 2 x L x Wr + 2xLx// + 2x W x H square units A = 2(LW + LH + WH) In factored form
The units of area may be square centimeters (cm2), square meters (m2), and so on.
Example. A shoebox with a lid is drawn in figure b, with the measurements in centimeters. The drawing is not to scale. Calculate the volume and surface area, and draw the net of the open box and the lid.
Volume = L x W x H Formula for the volume of a cuboid
Volume = 32x 17.5 x 11.2 Substituting L = 32, W = 17.5,
H = 11.2
Volume of shoebox = 6272 cm3 Using a calculator
136 CUMULATIVE FREQUENCY GRAPH
We will take the surface area of the shoebox to mean the surface area of the cardboard needed to make the box, not including the tabs used to glue the box together. Write
Area of box = 2 (LW + LH + WH) Formula for the area
of a cuboid
= 2(32 x 17.5 + 32 x 11.2 + 17.5 x 11.2) Substituting L = 32,
W = 17.5,
H = 11.2
= 2(560 + 358.4 + 196)
= 2228.8 cm2
The area of the surface of the shoebox is 2228.8 cm2.
The net of the shoebox is made up of the open box and the lid. Figure c is not drawn to scale; the key dimensions are marked on the figure.
References: Cube, Net.
CUMULATIVE FREQUENCY GRAPH
The cumulative frequency graph is explained using an example. It is especially useful for finding the median, upper quartile, lower quartile, and interquartile range of a set of continuous data.
Davidβs son William attends a small high school and all the students in his school are doing a statistics project about their heights. There are 100 students in the school and their heights are measured and recorded in a frequency table. Their heights in the table are entered in class intervals of 5 cm. The class interval β140-β means that the heights are from 140, to 145, including 140, but not including 145. Note that the final cumulative frequency of 100 is always the total frequency.