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25 Jul 2013, 21:26
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gmatter0913
Hi Karishma,
I tried the problem x-1 < sqrt (7-x) as below:
As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)
x-1 can be -ve or +ve
When x-1<=0; x<=1 --------> (2)
When x-1>=0; x>=1 --------> (3)
As both sides are +ve, we can square both the sides
(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0
-2<x<3 ------------>(4)
The answer to this problem is (x<3). I am not sure how to arrive at that from hereon. Could you please help me?
I tried the problem x-1 < sqrt (7-x) as below:
As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)
x-1 can be -ve or +ve
When x-1<=0; x<=1 --------> (2)
When x-1>=0; x>=1 --------> (3)
As both sides are +ve, we can square both the sides
(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0
-2<x<3 ------------>(4)
The answer to this problem is (x<3). I am not sure how to arrive at that from hereon. Could you please help me?
You have done the process correctly. Now you need to understand what this implies.
You got x <= 7
Case 1: x-1< 0
When x-1< 0; x < 1
Note that when x-1 is negative, it will always be less than \(\sqrt{(7-x)}\)
So whenever x<1, the inequality will always hold.
Case 2: x-1 >= 0
If x-1 is non negative, we can square the inequality.
From this, you get -2<x<3.
The inequality holds in this range.
From the two cases, we see that the inequality holds for the range x < 3.
26 Jul 2013, 02:05
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Hi Karishma,
I have one more doubt on my solution posted earlier.
I have one more doubt on my solution posted earlier.
Quote:
I tried the problem x-1 < sqrt (7-x) as below:
As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)
x-1 can be -ve or +ve
When x-1<=0; x<=1 --------> (2)
When x-1>=0; x>=1 --------> (3)
As both sides are +ve, we can square both the sides
(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0
-2<x<3 ------------>(4) Shouldn't this be 1<=x<3 (as x>=1 is the pre-supposed condition to square them)
As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)
x-1 can be -ve or +ve
When x-1<=0; x<=1 --------> (2)
When x-1>=0; x>=1 --------> (3)
As both sides are +ve, we can square both the sides
(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0
-2<x<3 ------------>(4) Shouldn't this be 1<=x<3 (as x>=1 is the pre-supposed condition to square them)
26 Jul 2013, 04:23
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gmatter0913
Hi Karishma,
I have one more doubt on my solution posted earlier.
I have one more doubt on my solution posted earlier.
Quote:
I tried the problem x-1 < sqrt (7-x) as below:
As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)
x-1 can be -ve or +ve
When x-1<=0; x<=1 --------> (2)
When x-1>=0; x>=1 --------> (3)
As both sides are +ve, we can square both the sides
(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0
-2<x<3 ------------>(4) Shouldn't this be 1<=x<3 (as x>=1 is the pre-supposed condition to square them)
As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)
x-1 can be -ve or +ve
When x-1<=0; x<=1 --------> (2)
When x-1>=0; x>=1 --------> (3)
As both sides are +ve, we can square both the sides
(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0
-2<x<3 ------------>(4) Shouldn't this be 1<=x<3 (as x>=1 is the pre-supposed condition to square them)
Most certainly. The only reason I don't care about the values from -2 to 1 is that these values are already covered in the first case. We know they already hold for the inequality. We only get 1 to 3 extra values and that's what we care about.
In a stand alone question, the presupposed condition must be satisfied (x <= 7 AND x >= 1)
