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Inequalities

Expert replies
Source: — Problem Solving |

Re: Inequalities

by sudhir3127 » Mon Aug 25, 2008 12:23 am
gmattester wrote:Can someone plz. explain this problem
I go with B..

statement 1.X#3 doesnt give any picture ..suppose x = 4

rt(4-3)^2 # 3-4
1 is not equal to -1...................hence in sufficient

statement 2.

-X(ABS (x)) >0 we know from this that X<0 only then the equation will be >0

now for any negative value of X the equation
rt ( X-3)^2 = 3-X

hence B

hope that helps
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by carefreeamit » Mon Aug 25, 2008 12:23 am
statement 1 is not sufficient
statement 2 -
-x|x|> 0
since |x| is positive -x should also be positive i.e. x is negative.
take any negative value of x and substitute.

Ans B
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by LSB » Mon Aug 25, 2008 2:19 am
rt ( X-3)^2 = 3-X (square and multiply)
X^2-6x+9 = 9-6X+X^2
2X^2 = 0
This essentially asks if X=0

1) X <>3 INSUFF ... X could be anything but 3
2) From this it follows X>0.
If X>0, then 2X^2 cannot be zero
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by 4meonly » Mon Aug 25, 2008 8:38 am
LSB wrote: X^2-6x+9 = 9-6X+X^2
2X^2 = 0
X^2-6x+9 = 9-6X+X^2
is the same X^2-6x+9 = X^2-6X+9

how i've got?

i missed something?

2X^2 = 0
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by chisbri » Mon Aug 25, 2008 8:59 am
I tried to solve this by squaring both sides:

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

In which case, the statement would be valid for all values of X. What have I done wrong here?[/img]
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by LSB » Mon Aug 25, 2008 12:11 pm
You're right. Typo on my end (it was late at night). Funny how I still got it right :-) .

It's a tough one because if we do square like I suggested then we get rid of the positive vs negative contraint. I guess the only way I can see is to plug numbers into the original equation then.
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