Download CHAPTER 35 GRAPHICAL SOLUTION OF EQUATIONS

CHAPTER 35 GRAPHICAL SOLUTION OF EQUATIONS
EXERCISE 143 Page 369
1. Solve the simultaneous equations graphically: y = 3x – 2
y=–x+6
Since both equations represent straight-line graphs, only two coordinates plus one to check are
needed.
x
0
1
2
x
0
1
2
y = 3x – 2 – 2
1
4
y = –x + 6
6
5
4
Both graphs are shown plotted below.
The solution of the simultaneous equations is where the two graphs intersect
From the graphs the solution is x = 2, y = 4
2. Solve the simultaneous equations graphically: x + y = 2
3y – 2x = 1
x + y = 2 from which, y = –x + 2
3y – 2x = 1 from which, 3y = 2x + 1 and
y=
2
1
x+
3
3
Both graphs are shown plotted below.
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The solution of the simultaneous equations is where the two graphs intersect
From the graphs the solution is x = 1, y = 1
3. Solve the simultaneous equations graphically: y = 5 – x
x–y=2
Since both equations represent straight-line graphs, only two coordinates plus one to check are
needed.
x
y = –x + 5
0
1
2
x
0
1
5
4
3
y=x–2
–2
–1
2
0
Both graphs are shown plotted below.
The solution of the simultaneous equations is where the two graphs intersect
From the graphs the solution is x = 3.5, y = 1.5
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4. Solve the simultaneous equations graphically: 3x + 4y = 5
2x – 5y + 12 = 0
3x + 4y = 5
from which, 4y = –3x + 5
and
3
5
y = − x+
4
4
2x – 5y + 12 = 0 from which, 5y = 2x + 12 and y =
2
12
x+
5
5
Both graphs are shown plotted below.
The solution of the simultaneous equations is where the two graphs intersect
From the graphs the solution is x = –1, y = 2
5. Solve the simultaneous equations graphically: 1.4x – 7.06 = 3.2y
2.1x – 6.7y = 12.87
1.4x – 7.06 = 3.2y hence, y =
1.4
7.06
x−
3.2
3.2
i.e.
y = 0.4375x – 2.20625
x
0
1
y – 2.21 – 1.77
2.1x – 6.7y = 12.87
hence, y =
2.1
12.87
x−
6.7
6.7
i.e.
x
2
– 1.33
y = 0.31343x – 1.9209
0
1
y – 1.92 – 1.61
2
– 1.29
Both graphs are shown plotted below
The solution of the simultaneous equations is where the two graphs intersect
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From the graphs the solution is x = 2.3, y = –1.2
6. Solve the simultaneous equations graphically: 3x – 2y = 0
4x + y + 11 = 0
3x – 2y = 0
from which, y =
4x + y + 11 = 0
from which,
3
x
2
y = – 4x – 11
Both graphs are shown plotted below.
The solution of the simultaneous equations is where the two graphs intersect
From the graphs the solution is x = –2, y = –3
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7. The friction force F newtons and load L newtons are connected by a law of the form F = aL + b,
where a and b are constants. When F = 4 N, L = 6 N and when F = 2.4 N, L = 2 N. Determine
graphically the values of a and b.
Since F = aL + b, then 4 = 6a + b from which, b = –6a + 4
and
2.4 = 2a + b from which, b = –2a + 2.4
Both graphs are shown plotted below
The solution of the simultaneous equations is where the two graphs intersect
From the graphs the solution is a = 0.4, b = 1.6
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EXERCISE 144 Page 373
1. Sketch the following graphs and state the nature and coordinates of their turning points:
(a) y = 4x2
(b) y = 2x2 – 1
(c) y = –x2 + 3
(d) y = –
1 2
x –1
2
(a) A graph of y = 4x2 is shown below. The turning point is a minimum at (0, 0)
(b) A graph of y = 2x2 – 1 is shown below. The turning point is a minimum at (0, –1)
(c) A graph of y = –x2 + 3 is shown below. The turning point is a maximum at (0, 3)
1
(d) A graph of y = – x2 – 1 is shown below. The turning point is a maximum at (0, –1)
2
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2. Solve graphically the quadratic equation 4x2 – x – 1 = 0 by plotting the curve between the limits
x = –1 to x = 1. Give answers correct to 1 decimal place.
A graph of y= 4 x 2 − x − 1 is shown below
The points where y= 4 x 2 − x − 1 and y = 0 coincide give the solution of 4 x 2 − x − 1 =0 . This is seen
to be at x = –0.4 and x = 0.6
3. Solve graphically the quadratic equation x2 – 3x = 27 by plotting the curve between the limits
x = –5 to x = 8. Give answers correct to 1 decimal place.
A graph of y = x 2 − 3 x − 27 is shown below
The points where y = x 2 − 3 x − 27 and y = 0 coincide give the solution of x 2 − 3 x =
27 . This is seen
to be at x = –3.9 and x = 6.9
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4. Solve graphically the quadratic equation 2x2 – 6x – 9 = 0 by plotting the curve between the
limits x = –2 to x = 5. Give answers correct to 1 decimal place.
A graph of y = 2 x 2 − 6 x − 9 is shown below
The points where y = 2 x 2 − 6 x − 9 and y = 0 coincide give the solution of 2 x 2 − 6 x − 9 =
0 . This is
seen to be at x = –1.1 and x = 4.1
5. Solve graphically the quadratic equation 2x(5x – 2) = 39.6 by plotting the curves between the
limits x = –2 to x = 3. Give answers correct to 1 decimal place.
2x(5x – 2) = 39.6
i.e.
10 x 2 − 4 x − 39.6 =
0
A graph of y = 10 x 2 − 4 x − 39.6 is shown below
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The points where y = 10 x 2 − 4 x − 39.6 and y = 0 coincide give the solution of 10 x 2 − 4 x − 39.6 = 0
This is seen to be at x = –1.8 and x = 2.2
6. Solve the quadratic equation 2x2 + 7x + 6 = 0 graphically, given that the solutions lie in the range
x = –3 to x = 1. Determine also the nature and coordinates of its turning point.
A graph of y = 2 x 2 + 7 x + 6 is shown below
The points where y = 2 x 2 + 7 x + 6 and y = 0 coincide give the solution of 2 x 2 + 7 x + 6 =
0 . This is
seen to be at x = –1.5 and x = –2
The turning point is a minimum and occurs at x = –1.75 and the minimum value is y = –0.1
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7. Solve graphically the quadratic equation 10x2 – 9x – 11.2 = 0, given that the roots lie between
x = –1 and x = 2
A graph of y = 10x2 – 9x – 11.2 is shown below
The points where y = 10x2 – 9x – 11.2 and y = 0 coincide give the solution of 10x2 – 9x – 11.2 = 0
This is seen to be at x = –0.7 and x = 1.6
8. Plot a graph of y = 3x2 and hence solve the equations:
(a) 3x2 – 8 = 0
(b) 3x2 – 2x – 1 = 0
A graph of y = 3 x 2 is shown below
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(a) 3 x 2 − 8 =
0 hence, 3 x 2 = 8 . The solution of 3 x 2 − 8 =
0 is determined from the intersection of
y = 3 x 2 and y = 8 and from the above graphs this occurs at x = –1.63 and x = 1.63
2
(b) 3 x 2 − 2 x − 1 =0 hence, 3 x=
2 x + 1 . The solution of 3 x 2 − 2 x − 1 =0 is determined from the
intersection of y = 3 x 2 and y = 2x + 1 and from the above graphs this occurs at x = –0.3 and
x=1
9. Plot the graphs y = 2x2 and y = 3 – 4x on the same axes and find the coordinates of the points of
intersection. Hence determine the roots of the equation 2x2 + 4x – 3 = 0
A graph of y = 2 x 2 is shown below
Also, a graph of y = 3 – 4x is shown on the same axes.
The coordinates of the points of intersection of the two graphs occurs at (–2.6, 13.2) and
(0.6, 0.8)
2 x2 + 4 x − 3 =
0 is equivalent to 2 x 2 = 3 − 4 x
The solution of 2 x 2 + 4 x − 3 =
0 is determined from the intersection of y = 2 x 2 and y = 3 – 4x and
from the above graphs, the roots of 2 x 2 + 4 x − 3 =
0 are x = –2.6 and x = 0.6
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10. Plot a graph of y = 10x2 – 13x – 30 for values of x between x = – and x = 3. Solve the equation
10x2 – 13x – 30 = 0 and from the graph determine:
(a) the value of y when x is 1.3, (b) the value of x when y is 10 and
(c) the roots of the equation 10x2 – 15x – 18 = 0
A graph of y = 10 x 2 − 13 x − 30 is shown below.
When y = 0, the solution of 10 x 2 − 13 x − 30 =
0 is x = –1.2 and x = 2.5
(a) From the graph, when x = 1.3, y = –30
(b) From the graph, when y = 10, x = –1.50 and x = 2.75
(c) 10 x 2 − 15 x − 18 =
0 is equivalent to 10 x 2 − 13 x − 30 = 2 x − 12
Thus the roots of the equation 10 x 2 − 15 x − 18 =
0 occurs at the intersection of the graphs
y = 10 x 2 − 13 x − 30 and y = 2x – 12, i.e. the roots are x = 2.3 and x = –0.8
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EXERCISE 145 Page 374
1. Determine graphically the values of x and y which simultaneously satisfy the equations:
y = 2(x2 – 2x – 4) and y + 4 = 3x
y = 2 ( x2 − 2x − 4) = 2x2 − 4x − 8
y + 4 = 3x i.e. y = 3x – 4
The two graphs are shown plotted below.
The points of intersection provide the solutions to the simultaneous equations, i.e. x = 4, y = 8 and
x = –0.5, y = –5.5
2. Plot the graph of y = 4x2 – 8x – 21 for values of x from –2 to +4. Use the graph to find the roots of
the following equations:
(a) 4x2 – 8x – 21 = 0
(b) 4x2 – 8x – 16 = 0
(c) 4x2 – 6x – 18 = 0
A graph of y = 4 x 2 − 8 x − 21 is shown plotted below
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(a) From the graph, the roots of 4 x 2 − 8 x − 21 =
0 are x = –1.5 and x = 3.5
(b) 4 x 2 − 8 x − 16 =
0 is equivalent to 4 x 2 − 8 x − 21 =
−5
Hence, the intersection of the graphs y = 4 x 2 − 8 x − 21 and y = –5 gives the roots of the
equation 4 x 2 − 8 x − 16 =
0 , i.e. x = –1.24 and x = 3.24
(c) 4 x 2 − 6 x − 18 =
−2 x − 3
0 is equivalent to 4 x 2 − 8 x − 21 =
Hence, the intersection of the graphs y = 4 x 2 − 8 x − 21 and y = –2x – 3 gives the roots of the
equation 4 x 2 − 6 x − 18 =
0 , i.e. x = –1.5 and x = 3.0
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EXERCISE 146 Page 375
1. Plot the graph y = 4x3 + 4x2 – 11x – 6 between x = –3 and x = 2 and use the graph to solve the
cubic equation 4x3 + 4x2 – 11x – 6 = 0
A graph of y = 4 x3 + 4 x 2 − 11x − 6 is shown plotted below.
The solution of 4 x3 + 4 x 2 − 11x − 6 =
0 is determined from the intersection of
y = 4 x3 + 4 x 2 − 11x − 6 and y = 0 and from the above graph, the roots of 4 x3 + 4 x 2 − 11x − 6 =
0
are x = –2.0, x = –0.5 and x = 1.5
2. By plotting a graph of y = x3 – 2x2 – 5x + 6 between x = –3 and x = 4 solve the equation
x3 – 2x2 – 5x + 6 = 0. Determine also the coordinates of the turning points and distinguish between
them.
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A graph of y = x 3 − 2 x 2 − 5 x + 6 is shown plotted below.
The solution of x 3 − 2 x 2 − 5 x + 6 =
0 is determined from the intersection of
y = x 3 − 2 x 2 − 5 x + 6 and y = 0 and from the above graph, the roots of x3 − 2 x 2 − 5 x + 6 =
0
are x = –2, x = 1 and x = 3
The turning points are: minimum at (2.1, –4.1) and maximum at (–0.8, 8.2)
3. Solve graphically the cubic equation: x3 – 1 = 0 correct to 2 significant figures.
Let =
y x3 − 1
A table of values is shown below
x
–2
–1
0 1 2
3
y
–9 –2 –1 0 7 26
A graph of =
y x 3 − 1 is shown below and it can be seen that there is only one root for the equation
x3 − 1 =0 , i.e. the root is at x = 1
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4. Solve graphically the cubic equation: x3 – x2 – 5x + 2 = 0 correct to 2 significant figures.
A graph of y = x 3 − x 2 − 5 x + 2 is shown below
The solution of x 3 − x 2 − 5 x + 2 =
0 is determined from the intersection of
y = x 3 − x 2 − 5 x + 2 and y = 0 and from the above graph, the roots of x 3 − x 2 − 5 x + 2 =
0
are x = –2.0, x = 0.4 and x = 2.6
5. Solve graphically the cubic equation: x3 – 2x2 = 2x – 2 correct to 2 significant figures
x3 – 2x2 = 2x – 2 hence, x3 – 2x2 – 2x + 2 = 0
A graph of y = x3 – 2x2 – 2x + 2 is shown below
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The solution of x3 – 2x2 – 2x + 2 = 0 is: x = –1.2, 0.70 and 2.5
6. Solve graphically the cubic equation: 2x3 – x2 – 9.08x + 8.28 = 0 correct to 2 significant figures.
A graph of y = 2x3 – x2 – 9.08x + 8.28 is shown below
The solution of 2x3 – x2 – 9.08x + 8.28 = 0 is: x = –2.3, 1.0 and 1.8
7. Show that the cubic equation 8x3 + 36x2 + 54x + 27 = 0 has only one real root and determine its
value.
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Let y = 8 x 3 + 36 x 2 + 54 x + 27
A table of values is shown below
x
–3
–2
–1
0
1
2
y
–27 –1
1
27 125 343
A graph of y = 8 x 3 + 36 x 2 + 54 x + 27
The solution of 8 x3 + 36 x 2 + 54 x + 27 =
0 is determined from the intersection of
y = 8 x 3 + 36 x 2 + 54 x + 27 and y = 0 and from the above graph, the one and only root of
8 x3 + 36 x 2 + 54 x + 27 =
0 is x = –1.5 (It can also be seen from the above table of values that there
is only one root to the equation)
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