saxenashobhit wrote:
I had same question today
\(\sqrt{4}\) = + or - 2. So answer should be + or - x. I don't get how can OA be -x
SOME NOTES:1. GMAT is dealing only with
Real Numbers: Integers, Fractions and Irrational Numbers.
2. Any nonnegative real number has a
unique non-negative square root called
the principal square root and unless otherwise specified,
the square root is generally taken to mean
the principal square root.
When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the
only accepted answer is the positive root.
That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.
Even roots have only non-negative value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
3. \(\sqrt{x^2}=|x|\).
The point here is that as
square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).
So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?
Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).
So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).
What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
BACK TO THE ORIGINAL QUESTION:1. If \(x<0\), then \(\sqrt{-x*|x|}\) equals:A. \(-x\)
B. \(-1\)
C. \(1\)
D. \(x\)
E. \(\sqrt{x}\)
\(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\). Note that as \(x<0\) then \(-x=-negative=positive\), so \(\sqrt{-x*|x|}=-x=positive\) as it should be.
Or just substitute the some negative \(x\), let \(x=-5<0\) --> \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x\).
Answer: A.
Hope it's clear.
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