F.4 A-MTH ( Please be quick !!! ) (Part 1)

Let f(x)=x^2+2x-1 and g(x)=-x^2+2kx-9 ( where k is a constant ).
(a) Suppose the graph of y=f(x) cuts the x-axis at the pointsP and Q, and the graph of y=g(x) cuts the x-axis at the points R and S.
(i) find the lengths of PQ and RS (in term of k ).

let a,b be the x-intercept of f(x), ie roots of f(x)
and c,d be the x-intercept of g(x), ie roots of g(x)

sum of roots of f(x)
a+b=-2
product of roots of f(x)
ab=-1
the lengths of PQ=|b-a|
=√(b-a)^2
=√[(b+a)^2-4ab]
=√[(-2)^2-4(-1)]
=√[4+4]
=√8

sum of roots of g(x)
c+d=2k
product of roots of g(x)
cd=9
the lengths of RS=|d-c|
=√(d-c)^2
=√[(d+c)^2-4cd]
=√[(2k)^2-4(9)]
=√[4k^2-36]


(ii) find, in terms of k, the x-coordinate of the mid-point of RS.

the x-coordinate of the mid-point of RS
=(c+d)/2
=(2k)/2
=k

(iii) if the mid-points of PQ and RS coincide with each other, find the value of k.

the x-coordinate of the mid-point of PQ
=(a+b)/2
=(-2)/2
=-1
x-coordinate of the mid-point of PQ
= the x-coordinate of the mid-point of RS
so k=-1

首先想問f(x)=x^2+2x-1 or x^2-2x+1
如果你的問題真是無錯的話

i) f(x)=0=x^2+2x-1
x= 0.41 or -2.41
lengthof PQ = 0.41-(-2.41)= 2.82

g(x)=0=-x^2+2kx-9
x= -2k+[(2k)^2-36]^1/2 or -2k-[(2k)^2-36]^1/2
lengthof RS =2 [(2k)^2-36]^1/2

ii)x-coordinate of the mid-point of RS= [(2k)^2-36]^1/2

iii)2.82/2=[(2k)^2-36]^1/2
1.41=[(2k)^2-36]^1/2
k=3.08