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Inequalities and Roots

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Re: Inequalities and Roots  [#permalink]

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New post 25 Jul 2013, 21:26
1
gmatter0913 wrote:
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?


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.
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Re: Inequalities and Roots  [#permalink]

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New post 26 Jul 2013, 02:05
Hi Karishma,

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)
Veritas Prep GMAT Instructor
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Joined: 16 Oct 2010
Posts: 8288
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Re: Inequalities and Roots  [#permalink]

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New post 26 Jul 2013, 04:23
1
gmatter0913 wrote:
Hi Karishma,

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)


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)
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Re: Inequalities and Roots  [#permalink]

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New post 04 Jul 2018, 07:53
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Re: Inequalities and Roots &nbs [#permalink] 04 Jul 2018, 07:53

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