3 Inequations

[arjmen]

Find the range of x for these inequations:

1) |x^2 - 7x + 12| > x^2 - 7x + 12

2) 2 <= |x+3| < 11

3) |x+8| < |x-3|

[maria1_03]

Originally Posted by arjmen

Find the range of x for these inequations:

1) |x^2 - 7x + 12| > x^2 - 7x + 12

2) 2 <= |x+3| < 11

3) |x+8| < |x-3|

1) x is between (3;4)

2) x is between (-14 ; -5) U [-1; 8)

3) x < - 2.5

[chix475ntu]

1) 3<x<4
2) -14<x<=-5 U -1<=x<8
3)x<-2.5

[4567]

Could someone please write out how to solve the first one for old-timers such as myself

[Synergy79]

I agree with the first two. Could somebody please explain how the third eq was resolved?

[dsaqwert]

Originally Posted by Synergy79

I agree with the first two. Could somebody please explain how the third eq was resolved?

|x+8| < |x-3|
x<=-8
-x-8<-x+3
-8<3
x<=-8

-8<x<3
x+8<-x+3
-8<x<-2.5

x>3
x+8<x-3
8<-3

overall x<-2.5

[arjmen]

Solutions:

1) |x^2 - 7x + 12| > x^2 - 7x + 12

If |y| > y, clearly y<0

Similarly,
x^2 - 7x + 12 < 0
=> (x-4)(x-3) < 0
Therefore, 3 < x < 4

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2) 2 <= |x+3| < 11

Textbook approach:
Break into 2 inequations
i) |x+3| < 11
implies -11 < (x+3) < 11
implies -14 < x < 8

ii) |x+3| >= 2
implies x >= -1 OR x <= -5

Merging the 2 solution sets gives,
-14 < x <= -5 OR -1 <= x < 8

Number line method:
http://mathforum.org/library/drmath/view/53126.html

I find that sketching a quick graph, helps me see the answer immediately

-------------------------------------------------------------------------

3) |x+8| < |x-3|
Clearly, x<0

We have 2 inequations:
i) -(x+8) < -(x-3) --------> where (x+8) <= 0
implies -8 < 3 ---> Insightful!

ii) (x+8) < -(x-3) ---------> where (x+8) > 0
implies x < -5/2

Therefore, x < -5/2

In Solution 3,

Why is -(x-3) the RHS in both inequalities? why not (x-3).

|x+8| < |x+3|

=> (x+8) < (x+3) ; -(x+8) < (x+3)

OR

(x+8) < -(x+3) ; -(x+8) < -(x+3)

Please explain...

[chix475ntu]

|x+8| < |x+3|
From the above, there can be 4 equations possible.
-(x+8)=-(x+3) ==> (x+8)=(x+3)
-(x+8)=(x+3) ==> (x+8)=-(x+3)
(x+8)=-(x+3)
(x+8)=(x+3)

From the above you can see that there are actually 2 equations only. How you want to put - and + on both side of equations depends from each person I think

[Dimas]

Q1:Q2: nothing added to arjmen

Q3: I recommend solving it geometrically. Much easier for linear absolute values.