Solve the following simultaneous equations.
1. x^2 – 2y^2 = 3x + 4y = 1
2. If the simultaneous equations
y = x^2 – k
y = 2x
have only one solution, find the value of k.
題(1)及題(2)的答案分別是x = 3, y = -2和-1。
兩條關於Solving Equations的簡單數學問題
2009-03-22 11:19 pm
回答 (2)
2009-03-22 11:56 pm
✔ 最佳答案
1.x2 - 2y2 = 1 … (1)
3x + 4y = 1 … (2)
x = (1 - 4y)/3 … (3)
Put (3) into (1):
[(1 - 4y)/3]2 - 2y2 = 1
(1 - 8y + 16y2)/9 - 2y2 = 1
1 - 8y + 16y2 - 18y2 = 9
2y2 + 8y + 8 = 0
2(y + 2)2 = 0
y = -2 (double roots)
Put y = -2 into (3):
x = [1 - 4(-2)]/3
x = 3
Ans: x = 3, y = -2
2.
y = x2 - k … (1)
y = 2x … (2)
Put (2) into (1):
2x = x2 - k
x2 - 2x - k = 0
The equation has only one solution. Then,
Δ = 0
(-2)2 - 4(1)(-k) = 0
4 + 4k = 0
k = -1
=
2009-03-23 3:32 am
1.
x2 - 2y2 = 1 … (1)
3x + 4y = 1 … (2)
x = (1 - 4y)/3 … (3)
Put (3) into (1):
[(1 - 4y)/3]2 - 2y2 = 1
(1 - 8y + 16y2)/9 - 2y2 = 1
1 - 8y + 16y2 - 18y2 = 9
2y2 + 8y + 8 = 0
2(y + 2)2 = 0
y = -2 (double roots)
Put y = -2 into (3):
x = [1 - 4(-2)]/3
x = 3
Ans: x = 3, y = -2
2.
y = x2 - k … (1)
y = 2x … (2)
Put (2) into (1):
2x = x2 - k
x2 - 2x - k = 0
The equation has only one solution. Then,
Δ = 0
(-2)2 - 4(1)(-k) = 0
4 + 4k = 0
k = -1
x2 - 2y2 = 1 … (1)
3x + 4y = 1 … (2)
x = (1 - 4y)/3 … (3)
Put (3) into (1):
[(1 - 4y)/3]2 - 2y2 = 1
(1 - 8y + 16y2)/9 - 2y2 = 1
1 - 8y + 16y2 - 18y2 = 9
2y2 + 8y + 8 = 0
2(y + 2)2 = 0
y = -2 (double roots)
Put y = -2 into (3):
x = [1 - 4(-2)]/3
x = 3
Ans: x = 3, y = -2
2.
y = x2 - k … (1)
y = 2x … (2)
Put (2) into (1):
2x = x2 - k
x2 - 2x - k = 0
The equation has only one solution. Then,
Δ = 0
(-2)2 - 4(1)(-k) = 0
4 + 4k = 0
k = -1
參考: me
收錄日期: 2021-04-29 18:17:22
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