聯立二元一次方程

1.
(a)
y = 3x² + m ...... [1]
y = x + k ...... [2]

[1] = [2]:
3x² + m = x + k
3x² - x + (m - k) = 0

Since the equation has only one real root, determinant Δ = 0
(-1)² - 4(3)(m - k) = 0
1 - 12m + 12k = 0
12m = 12k + 1
m = (12k + 1)/12
or m = k + (1/12)

(b)
(α, -11/12) is a root of the simultaneous equations.
-11/12 = 3α² + m ...... [3]
-11/12 = α + k ...... [4]

From the results of (a): m = k + (1/4)
Hence, [3] becomes
-11/12 = 3α² + k + (1/12) ...... [5]

[5] - [4]:
3α² - α + (1/12) = 0
36α² - 12α + 1 = 0
(6α - 1)² = 0
α = 1/6 (double roots)

(c)
Put α = 1/6 into [3]:
-11/12 = 3(1/6)² + m
m = -1

Put α = 1/6 into [4]:
-11/12 = (1/6) + k
k = -13/12


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2.
Let d miles be the original distance and v miles/h be the original speed.
The original time taken = d/v h

If he had gone 5 more miles an hour, he would have saved an hour.
This means that when speed = (v + 5) miles/h, time taken = [(d/v) - 1] h
distance/speed = time taken
d/(v + 5) = (d/v) - 1 ..... [1]

If he had gone slower than he did by 3 miles an hour, he would have taken 1 hour longer.
This means that when speed = (v - 3) miles/h, time taken = [(d/v) + 1] h
d/(v - 3) = (d/v) + 1 ..... [2]

From [1]:
d/(v + 5) = (d - v)/v
vd = vd - v² + 5d - 5v
15d = 3v² + 15v ..... [3]

From [2]:
d/(v - 3) = (d + v)/v
vd = vd + v² - 3d - 3v
15d = 5v² - 15v ..... [4]

[4] = [3]:
5v² - 15v = 3v² + 15v
2v² - 30v = 0
2v(v - 15) = 0
v = 0 (rejected) or v = 15

Put v = 15 into [3]:
15d = 3(15)² + 15(15)
d = 60

The original distance is 60 miles.
The original speed is 15 miles/h.


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3.
Let v km/h be the speed in still water.

When upstream, speed = (v - 2) km/h
Time taken = distance/speed
Time taken for upstream = 3.5/(v - 2) h

When downstream, speed = (v + 2) km/h
Time taken for downstream = 3.5/(v + 2) h

(Time taken for upstream) + (Time taken for downstream) = Total time taken
[3.5/(v - 2)] + [3.5/(v + 2)] = (100/60)
(3.5v + 7 + 3.5v - 7) / (v² - 4) = 5/3
5v² - 21v - 20 = 0
(v - 5)(5v + 4) = 0
v = 5 or v = -4 (rejected)

Speed of the boat in still water is 5 km/h.