A quadation has roots at -1 and -5. The quadratic equation "Q1" would intercept with a linear equation L1:y=2x-4 at two points A and B .(remind all your answer in surd form)
i)Calculate the interception points.
ii)Calculate the distance between point A and B.
iii)Calculate the vertex of the quadratic equation "Q1"
問一條拋物線同直線既問題
2007-08-28 7:28 am
回答 (2)
2007-08-31 1:43 am
✔ 最佳答案
i)y=(x+1)(x-5)
=x^2 - 4x -5
2x-4=x^2 - 4x -5
x^2 - 6x - 1=0
x=3+√10 or 3-√10
So,
the interception points are (3+√10,2+2√10) and (3-√10,2-2√10)
ii)
the distance between pointA and pointB
=√[(3+√10-3+√10)^2+(2+2√10-2+2√10)^2]
=√(40+160)
=√200
=10√2(units)
iii)
Because there's only one vertex,
so
y=x^2 - 4x - 5.......1
x^2 - 4x - 5 - y=0
D=(-4)^2 - 4(-5-y)=0
16+20+4y=0
y= - 9
Substitude y= - 9 into equation 1
- 9=x^2 - 4x -5
x^2-4x+4=0
x=2
So,the vertex of the quadratic equation "Q1" is (2, - 9)
2007-08-28 6:59 pm
y=(x+1)(x+5)
=x^2+6x+5----(1)
i) y=2x-4-----(2)
Put (2) into (1):
2x-4 = x^2+6x+5
x^2 +4x +9=0
Discriminant <0, Is there any intersection points between them? Or mistakes in the question?
Please check!
2007-08-29 17:22:10 補充：
y=(x+1)(x-5)=x^2-4x-5---(1),put (2)into(1):x^2-6x-1=0, x=3+10^1/2 or 3-10^1/2, put them into (2), intersection points are (3+10^1/2, 2+2(10)^1/2) and (3-10^1/2, 2-2(10)^1/2)
=x^2+6x+5----(1)
i) y=2x-4-----(2)
Put (2) into (1):
2x-4 = x^2+6x+5
x^2 +4x +9=0
Discriminant <0, Is there any intersection points between them? Or mistakes in the question?
Please check!
2007-08-29 17:22:10 補充：
y=(x+1)(x-5)=x^2-4x-5---(1),put (2)into(1):x^2-6x-1=0, x=3+10^1/2 or 3-10^1/2, put them into (2), intersection points are (3+10^1/2, 2+2(10)^1/2) and (3-10^1/2, 2-2(10)^1/2)
收錄日期: 2021-04-13 13:09:17
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