• Oct 12th 2006, 11:06 AM
fima2001
for what values of M

i added the question in a picture
Attachment 1058
• Oct 12th 2006, 12:55 PM
topsquark
Quote:

Originally Posted by fima2001
for what values of M

i added the question in a picture
Attachment 1058

I assume you mean to solve for when y = 0 and x1, x2 are the two roots?

0 = (m-2)x^2 - 2(m-2)x + (4-m)

So
a = m-2
b = -2(m-2) = -2m + 4
c = 4 - m

x = [-b (+/-) sqrt(b^2 - 4*a*c)]/(2a)

x = [-(-2m+4) (+/-) sqrt({-2m+4}^2 - 4*(m-2)*(4-m))]/(2*(m-2))

x = [2m-4 (+/-) sqrt(4m^2 - 16m + 16 - 4*(-m^2 +6m - 8))]/(2m-4)

x = [(2m-4) (+/-) sqrt(8m^2 - 40m + 48)]/(2m+4)

x = [(2m-4) (+/-) 2*sqrt(2m^2 - 10m + 12)]/(2m+4)

x = 1 (+/-) [2/(2m+4)]*sqrt(2m^2 - 10m + 12)

x = 1 (+/-) sqrt(2m^2 - 10m + 12)/(m+2)

So the larger of these, x2, must use the "+" and x1 must use the "-".

Consider the x2 condition:
2<x2<6

Thus
2 < 1 + sqrt(2m^2 - 10m + 12)/(m+2) < 6

or

1 < sqrt(2m^2 - 10m + 12)/(m+2) < 5
This has critical points where the numerator and denominator are zero. Thus it has critical points at m = -2, m = 2, and m = 3. So split the number line into 4 intervals and check the inequality:
(-infinity, -2): No.
(-2, 2): Yes
(2, 3): No.
(3, infinity): No.

So we know that -2 < m < 2.

Now let's look at x1:
1 < x1 < 2

1 < 1 - sqrt(2m^2 - 10m + 12)/(m+2) < 2

0 < - sqrt(2m^2 - 10m + 12)/(m+2) < 1

Again we find the critical points at m = -2, 2, and 3. We already know that -2 < m < 2, so all we need to do is check the inequality on this interval. Also note that m must be negative in order for this to hold. So I'm only going to check the inequality on (-2, 0).

(-2, 0): No.

Thus it is my conclusion that there is no such m giving roots x1 and x2 the desired properties. (I have verified this for several representative values of m.)

-Dan
• Oct 12th 2006, 05:11 PM
ThePerfectHacker
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