YanisBoubenider wrote:
\sqrt{}
Bunuel wrote:
goldfinchmonster wrote:
Hi..
According to me X^2 < 2 can be written as
Mod (X) < 2, which in tern yields the range of x as -2 < x < 2... Could you tell me where iam going wrong..
Thanks in advance..
If you take the square root from x^2 < 2 you get \(|x| < \sqrt{2}\), which in turn gives \(-\sqrt{2} < x <\sqrt{2}\).
Hope it's clear.
Hi Bunuel
I'm struggling a little with this one.
Indeed I assumed that when GMAT gives the symbol of a square root, we should only consider the positive number of x. In other words if GMAT asks x^16 therefore X = 4 or x = -4 BUT if GMAT asks √16 therefore we should only consider 4 and NOT -4.
Coming back to the question, I assumed that x^2 < 2 and therefore √x<√2 and therefore x must be strictly positive and be between 0 and √2
Even though I kind of understand your reasoning, where am I doing wrong here?
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, \(\sqrt{16}=4\), NOT +4 or -4.
Even roots have only a positive value on the GMAT.In contrast, the equation \(x^2=16\) has TWO solutions, +4 and -4.
Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
Now, about the problem itself:
MUST KNOW: \(\sqrt{x^2}=|x|\):The point here is that since
square root function cannot 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 \(\sqrt{x^2}=|x|\).
For example if x = -5, then \(\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|\)
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