A line has eaquation y=mx-1,where m is a constant
A curve has equation y=x^2-5x+3
a) show that the x-coorinate of any point of intersection of the line and the curve satisfies the equation
X^2-(5+m)x+4=0
b)find the values of m for which the equation x^2-(5+m)x+4=0 has equal roots.
c)describe geometrically the situation when m takes either of the values found in part (b)
AS maths
2006-11-24 9:50 am
回答 (2)
2006-11-24 8:38 pm
✔ 最佳答案
a)y = mx-1 -- (1)
y = x^2-5x+3 -- (2)
(2) - (1),
x^2 - (5+m)x + 4 = 0
b)
For equal roots, delta = 0
delta = (5+m)^2-4(1)(4) = 0
(5+m)^2 - 16 = 0
(5+m+4)(5+m-4) = 0
(m+9)(m+1) = 0
m=-9 or m=-1
c)
If m is a value between -9 and -1, the line and the curve will not intersect.
Otherwise, both of them will intersect at 2 points.
參考: my mathematical knowledge
2006-11-24 10:10 am
a)
1) y=mx-1
2) y=x^2-5x+3
Sub. 1) into 2), we have
X^2-(5+m)x+4=0
b) For the equation x^2-(5+m)x+4=0 has equal roots, we have (x+2)^2=0 or (x-2)^2=0, i.e. x = 2 or x = -2
so for x = 2, m = -1, and for x = -2, m = -9
1) y=mx-1
2) y=x^2-5x+3
Sub. 1) into 2), we have
X^2-(5+m)x+4=0
b) For the equation x^2-(5+m)x+4=0 has equal roots, we have (x+2)^2=0 or (x-2)^2=0, i.e. x = 2 or x = -2
so for x = 2, m = -1, and for x = -2, m = -9
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