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# Is sqrt = 3-x ? 1) x is not equal to 3 2) -x*lxl > 0 why

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Is sqrt = 3-x ? 1) x is not equal to 3 2) -x*lxl > 0 why [#permalink]

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06 May 2009, 19:29
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Is sqrt [(x-3)^2] = 3-x ?

1) x is not equal to 3
2) -x*lxl > 0

why am I not allowed to take the Square of both the left and the right side of the equation?

Then it would equal :

(x-3)^2 = (3-x)^2

If I do this, then A is true. I'm just not sure why you can't. (this is how I learned it back in HS).

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Intern
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Location: Austin

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06 May 2009, 19:44
Because:

sqrt [(x-3)^2] = (x-3) if (x>3)
and
sqrt [(x-3)^2] = -(x-3) if (x<3)

A:does not say if x is ">" than or "<" than 3
B:-x|x| > 0 implies x<0; hence B is sufficient

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Manager
Joined: 08 Feb 2009
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07 May 2009, 14:21
Is sqrt [(x-3)^2] = 3-x ?

1) x is not equal to 3
2) -x*lxl > 0

why am I not allowed to take the Square of both the left and the right side of the equation?

Then it would equal :

(x-3)^2 = (3-x)^2

If I do this, then A is true. I'm just not sure why you can't. (this is how I learned it back in HS).

So,

Can you explain how you solved this question ?

Kudos [?]: 58 [0], given: 3

Retired Moderator
Joined: 18 Jul 2008
Posts: 964

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08 May 2009, 06:47
The way I did it was wrong, unfortunately.

I squared both sides right away (I somehow remembered learning this method a LONG time ago *sigh*), the equation is true for any value of X. So for me, A was sufficient. It's going to be hard to unlearn it.

the OA is B.

goldeneagle94 wrote:
Is sqrt [(x-3)^2] = 3-x ?

1) x is not equal to 3
2) -x*lxl > 0

why am I not allowed to take the Square of both the left and the right side of the equation?

Then it would equal :

(x-3)^2 = (3-x)^2

If I do this, then A is true. I'm just not sure why you can't. (this is how I learned it back in HS).

So,

Can you explain how you solved this question ?

Kudos [?]: 285 [0], given: 5

Manager
Joined: 08 Feb 2009
Posts: 145

Kudos [?]: 58 [0], given: 3

Schools: Anderson

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08 May 2009, 08:38
The way I did it was wrong, unfortunately.

I squared both sides right away (I somehow remembered learning this method a LONG time ago *sigh*), the equation is true for any value of X. So for me, A was sufficient. It's going to be hard to unlearn it.

the OA is B.

goldeneagle94 wrote:
Is sqrt [(x-3)^2] = 3-x ?

1) x is not equal to 3
2) -x*lxl > 0

why am I not allowed to take the Square of both the left and the right side of the equation?

Then it would equal :

(x-3)^2 = (3-x)^2

If I do this, then A is true. I'm just not sure why you can't. (this is how I learned it back in HS).

So,

Can you explain how you solved this question ?

Old habits don't go away so easy.
I have the same issue as yours.
So, to overcome that, I have started taking the inequalities as is and first using the Substitution method.
For e.g.

In this problem, I would keep the question as is:
$$sqrt [(x-3)^2] = (3-x)$$

1) Substitute different values (e.g. x = -4, -1, 0, 1, 4) and check for its equality.

2) Subsitute different values in the statement itself to realize that x < 0. Then check the question for its equality with x = -4, -1,

This approach of NOT taking variables from LHS (left hand side of eqn) to RHS or vice-versa has been helping me.
If substitution doesn't work, then I go for shifting the variables from LHS or RHS to the other side.

Hope it helps you.

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Retired Moderator
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08 May 2009, 11:12
This is a good tip. Thank you.

*NOTE TO SELF* plug in numbers before manipulating equation!

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Intern
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17 May 2009, 14:50
Hi Guys... correct me if i am wrong, but... when x =1 then

lhs = 2 and rhs = 2, the two sides are equal... same thing goes for x = 1, x = 0 and x = -1 and so on

basically the lhs is always positive ... for any x, while the RHS will be positive for x < 3... not 0

i think that the answer is that neither of the statements are sufficient...

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Intern
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17 May 2009, 14:54
but then if x < 0 then it is less then 3 too so the answer is B... go it...

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Re: Inequality rule check   [#permalink] 17 May 2009, 14:54
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