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Re: Which of the following is always equal to sqrt(9+x^2-6x)? [#permalink]
07 Apr 2015, 09:48
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ankittiss wrote:
Bunuel Can you please help here? According to my understanding there are two cases in the GMAT: Case 1 :
sqrt[(-5)*(-5)] = -5
Case 2 :
sqrt[(5)*(5)] = 5
Is my understanding correct?
Thanks in advance Ankit
No, it's totally wrong.
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:
\(\sqrt{9} = 3\), NOT +3 or -3; \(\sqrt[4]{16} = 2\), NOT +2 or -2;
Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
Again, the point here is that since 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 \(\sqrt{x^2}=|x|\).
I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
So in other words sq rt of (x-3)^2 is +ve or -ve x-3? Sorry, but i don't really get that. I thought when we have a square of a sq rt, the answer is just what's under the sq rt, without any power.
It's not sq rt of 9 to be +-3, it's sq rt of 9^2, which is 9.
Please correct me if i am wrong.
I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
by definition |x| = sqrt(x^2)
since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.
I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
by definition |x| = sqrt(x^2)
since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.
Whew! Finally it makes sense. Thanks KS.
wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.
I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
by definition |x| = sqrt(x^2)
since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.
Whew! Finally it makes sense. Thanks KS.
wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.
But sqrt(x^2) has only one i.e x not -x.
the unknown x in sqrt(x^2) can be negative - the outcome will be positive ---> sqrt(-2^2) = 2 ---> x=-2
I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
by definition |x| = sqrt(x^2)
since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.
Whew! Finally it makes sense. Thanks KS.
wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.
But sqrt(x^2) has only one i.e x not -x.
No... ... In bold, we can only say : sqrt(x^2) = |x|
KillerSquirrel gave an exemple that proves it as well
I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.
by definition |x| = sqrt(x^2)
since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.
Whew! Finally it makes sense. Thanks KS.
wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.
But sqrt(x^2) has only one i.e x not -x.
No... ... In bold, we can only say : sqrt(x^2) = |x|
KillerSquirrel gave an exemple that proves it as well
Fig, I am still not convinced. I am in line with what Honghu and hobbit said here:
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