Bunuel wrote:
CristianJuarez wrote:
stan3544 wrote:
And here we go again. Sometimes i feel as if GMAT has it's own version of math that goes against universal one. I understand that on verbal there can be a whole lot of inerprteations of rules, simply because it is VERBAL, as in not a precise science which is basically a human imagination. But math has to be precise. And now, according to
e-GMAT -b^m if m is even --> positive number and you are stating that −2^4=−16 and i made a whole lot of mistakes when practicing quant from both
OG and
scholaranium that involved sqr.root of a number, because i didn't include the negative value as well. Can comeone give a link or a screenshot of an
OG question where when we take sqr root we don't consider the negative value?
The gmat doesn't have its own version of math. The square root is a function, that's why it is always positive. As a function, it can't take a positive and a negative value (if it did, it wouldn't be a function, you can look at the graph to check that out).
When you solve x^2 = 4, applying the square root function and assuming two results for x is incorrect, because you would only get x = 2. The formal and correct method is:
x^2 = 4
x^2 - 4 =0
(x + 2)(x - 2)=0 and from here get both results for x.
That's correct. + 1.
Another way of solving x^2 = 4 would be:
\(x = \sqrt{4}=2\) or \(x = -\sqrt{4}=-2\).
Yes, actually I'd like to correct myself, I hope it isn't more confusing.
Bunuel's solution applying square root function comes from the fact that the function f(x)=x^2 allows x to take positive or negative values (or 0):
I don't have 5 posts so I can't include the image
, please search for it on google.
Meanwhile, the function \(f(x)= \sqrt{x}\) doesn't (this is why square root is always positive):
(check image on google)
So, if people want to solve this excercise by applying square root function, you must consider that this is incorrect:
\(x^2=4\) Apply \(\sqrt{}\)
\(\sqrt{x^2}=\sqrt{4}\)
\(x = \sqrt{4}\)
\(x=2\) or \(x= -2\)
It is incorrect because \(\sqrt{4}\) is always 2 and you lost a solution in the process.
So, you have to consider that mathemathically, \(\sqrt{x^2} = |x|\). This is only because \(f(x)=x^2\) can take a positive or a negative value for x (or 0 of course). Indeed, this doesn't mean the square root takes a positive or a negative solution, its always positive (or zero)!! (remember, \(|x|\) is always \(>= 0\) for any value of \(x\)).
You can check it out like this:
\(\sqrt{4} = \sqrt{2^2}= |2|=2, not -2!!\) or:
\(\sqrt{4} = \sqrt{(-2)^2}= |-2|=2, not -2!!\) It's impossible to get to \(-2\) like that
So the correct process by applying square root function is:
\(x^2=4\) Apply \(\sqrt{}\)
\(\sqrt{x^2}=\sqrt{4}\)
\(|x| = 2\)
\(x=2\) or \(x= -2\)