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19 Feb 2012, 13:22
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Hi All,
MGMAT guide says that sq rt (x^2) = abs val (x)
e.g x^2 =25 -> x = +/- 5
abs (5) = +/- 5
Now when we solve this eqn
sq rt (x) = x-2 we get two solns x=4 and x=1
The way it shows to confirm is by putting these back in the eqn, so
sq rt (4) = 4-2
2 =2 , so correct
sq rt (1) = 1-2
1 = -1 so incorrect and so only 4 is the solution
My question
why sq rt (4) not equal to sq rt (2^2) and as from above +/-2 is the solution
and sq rt (1) not equal, to sq rt (1^1) and as from above +/-1 is the solution
In the above cases I mention sq rt (1^1) = -1 satisfies the eqn.
I am confused on what is the process and logic here to solve GMAT even roots problems.
MGMAT guide says that sq rt (x^2) = abs val (x)
e.g x^2 =25 -> x = +/- 5
abs (5) = +/- 5
Now when we solve this eqn
sq rt (x) = x-2 we get two solns x=4 and x=1
The way it shows to confirm is by putting these back in the eqn, so
sq rt (4) = 4-2
2 =2 , so correct
sq rt (1) = 1-2
1 = -1 so incorrect and so only 4 is the solution
My question
why sq rt (4) not equal to sq rt (2^2) and as from above +/-2 is the solution
and sq rt (1) not equal, to sq rt (1^1) and as from above +/-1 is the solution
In the above cases I mention sq rt (1^1) = -1 satisfies the eqn.
I am confused on what is the process and logic here to solve GMAT even roots problems.
19 Feb 2012, 13:26
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raviram80
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, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\). 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 \(\sqrt[3]{-64} =-4\).
As for \(\sqrt{x}\) and \(x^{\frac{1}{2}}\): they are the same.
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 \(\sqrt{x^2}=|x|\).
Hope it's clear.
General Discussion
19 Feb 2012, 15:55
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I think the difference is in variable and constant
[square_root][25] will always be 5
but sq rt [x2] = abs [x] can be +/- x depending on whether x is positive or negative, x being a variable.
Do you think the above reasoning is correct?
[square_root][25] will always be 5
but sq rt [x2] = abs [x] can be +/- x depending on whether x is positive or negative, x being a variable.
Do you think the above reasoning is correct?
