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absolute value

by Viper83 » Mon Feb 14, 2011 8:55 am
what is the value of r?

1) abs(r-3)=abs(r-1)

2) abs(r-4)=2

Can anyone help? I don't know the answer, just saw the question on another forum some time ago.
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by Night reader » Mon Feb 14, 2011 9:18 am
r=?
st(1) |r-3|=|r-1| --> r-3=r-1 OR 3-r=1-r, only one value r=-2 Sufficient
st(2) |r-4|=2 --> r-4=2 OR 4-r=2, two values r {2;6} Not Sufficient

IOM A
Viper83 wrote:what is the value of r?

1) abs(r-3)=abs(r-1)

2) abs(r-4)=2

Can anyone help? I don't know the answer, just saw the question on another forum some time ago.
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by clock60 » Mon Feb 14, 2011 12:11 pm
i also agree that the answer is A, but my calculation slightly differs from that of above poster
if r=-2, then
|r-3|=|-2-3|=|-5|=5. and
|r-1|=|-2-1|=|-3|=3, so -2 can`t be the root of above equation to me r=2, is the only valid root
st 2 gives us two valid values of r, 2, and 6 but we need unique value st 2 insuff

also i am not sure in above reasoning
st(1) |r-3|=|r-1| --> r-3=r-1 OR 3-r=1-r,

to me it is
-(r-3)=-(r-1) if r<1 no roots
-(r-3)=(r-1) if 1<r<3 r=2
(r-3)=(r-1) if r>3 again no roots
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by Night reader » Mon Feb 14, 2011 12:58 pm
@clock, :) thanks for correcting me above.
Yes we have to account for the values when our mode becomes either +ve, -ve or 0. The answer is r=2 and choice A
However, we can't arbitrarily assign sign for r; there are three conditions when the mod |r| becomes r>0, r<0 or r=0. Mod |r-3| becomes zero (0) when r=3, the mod |r-1| becomes zero when r=1 and we have to keep these critical values on the number line too.

@Viper, another way to crack the mods is to square both ends --> |r-3|=|r-1| <> (r-3)^2=(r-1)^2, (r^2-6r+9)=(r^2-2r+1), 8=4r and r=2
clock60 wrote: also i am not sure in above reasoning
st(1) |r-3|=|r-1| --> r-3=r-1 OR 3-r=1-r,

to me it is
-(r-3)=-(r-1) if r<1 no roots
-(r-3)=(r-1) if 1<r<3 r=2
(r-3)=(r-1) if r>3 again no roots
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by clock60 » Mon Feb 14, 2011 1:48 pm
@ night reader
i definitely agree that we can`t arbitrary assing sign for r, nor did i assign it in my solution, i followed strictly the definition of module
|a|=a if a>=0 and |a|=-a if a<0
in our case
|r-3|=(r-3) if r-3>=0 or r>=3. |r-3|=-(r-3) if r<3, the same is true for
|r-1|=(r-1) if r-1>=0. r>=1. and |r-1|=-(r-1) if r-1<0 or r<1
now you must review three different intervals r<1. 1=<r<3, and r>=3 and find the value of r if it exists on each interval
thank you for pointing my mistake about r=0, i missed firstly
as for squaring it is efficient approach ut in cases |f(x)|=|g(x)| but try to use it for example
|x-5|-|2x+8|=-12. and you will be less happy
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by Night reader » Mon Feb 14, 2011 9:59 pm
clock60 wrote:@ night reader
i definitely agree that we can`t arbitrary assing sign for r, nor did i assign it in my solution, i followed strictly the definition of module
|a|=a if a>=0 and |a|=-a if a<0
in our case
|r-3|=(r-3) if r-3>=0 or r>=3. |r-3|=-(r-3) if r<3, the same is true for
|r-1|=(r-1) if r-1>=0. r>=1. and |r-1|=-(r-1) if r-1<0 or r<1
now you must review three different intervals r<1. 1=<r<3, and r>=3 and find the value of r if it exists on each interval
thank you for pointing my mistake about r=0, i missed firstly
as for squaring it is efficient approach ut in cases |f(x)|=|g(x)| but try to use it for example
|x-5|-|2x+8|=-12. and you will be less happy
yea only critical values help here, /x-5=0, x=5/; /2x+8=0, x=-4/
-infinity<x<-4 (Or simply x<-4) ---> 5-x+2x+8=-12, x=-25; -4<x<5 ---> 5-x-2x-8=-12, x=3; 5<x<+infinity (Or simply x>5) ---> x-5-2x-8=-12, x=-1. After plug-in we get only x= {-25; 3}
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