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√-x |x| = -x?

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by Frankenstein » Fri Jul 08, 2011 11:34 am
Hi,
|x| = x, if x>0
|x| = -x, if x<0
So, √-x |x| = √-x*(-x) = √x^2 = |x|
As x<0, |x| = -x
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by winniethepooh » Fri Jul 08, 2011 11:53 am
(Square)Root of a negative number is an imaginary number digit , which is certainly not equal to any number to the left of 0 on the number line!
Also, as mentioned in the OG, Gmat Quant will deal only with real numbers. So, you need not bother!
Last edited by winniethepooh on Fri Jul 08, 2011 12:06 pm, edited 1 time in total.
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by MBA.Aspirant » Fri Jul 08, 2011 11:59 am
Frankenstein wrote:Hi,
|x| = x, if x>0
|x| = -x, if x<0
So, √-x |x| = √-x*(-x) = √x^2 = |x|
As x<0, |x| = -x
just to confirm can you illustrate with an example? the result of the sqrt here is -ve or is it -(-ve)?
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by Frankenstein » Fri Jul 08, 2011 12:04 pm
Hi,
consider x =-2
|x| = 2 (which is actually -x right?)
√-x |x| = √-(-2)*2 = √4 = 2 and not -2 because square root of a number is always positive.
this 2 is actually -x right?
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by MBA.Aspirant » Fri Jul 08, 2011 12:52 pm
so in both cases the result is +ve? whether x> or < 0?

when does taking a sqrt render a -ve root?

@winnie we're not dealing with sqrt of a -ve here at all.
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by Frankenstein » Fri Jul 08, 2011 8:37 pm
MBA.Aspirant wrote:so in both cases the result is +ve? whether x> or < 0?

when does taking a sqrt render a -ve root?

@winnie we're not dealing with sqrt of a -ve here at all.
Hi,
|x| is always non-negative.
sqrt is always non-negative. Don not confuse this with the following
x^2 = a^2. Find x?
In this case were finding the roots of x, which means we are finding the values of x which satisfy the equation.
For x^ = a^2, x = a or x = -a
But sqrt(a^2) is always |a| i.e. positive when a not equal to zero.
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