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04 May 2016, 19:42
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re-wording a question.
say it says solve for x^2
(1) x = Sq root (5x-1)
the answer says to take square both sides and get x^2=5x-1, than set it up as a quadratic equation and solve.
normally, everything under the square root has to be positive, so you'd basically get x^2 = l 5x - 1 l
so technically it "should" be insufficient, however, since x^2 can only be be 0 or positive, than l 5x-1 l can't be negative, so it is sufficient.
say if the question was to solve for x and (1) 4 = sq root (5x - 1), than you'd have two solutions (x=1 and x= -3/5), but you can eliminate the negative as X can only be positive.
is my thinking correct?
say it says solve for x^2
(1) x = Sq root (5x-1)
the answer says to take square both sides and get x^2=5x-1, than set it up as a quadratic equation and solve.
normally, everything under the square root has to be positive, so you'd basically get x^2 = l 5x - 1 l
so technically it "should" be insufficient, however, since x^2 can only be be 0 or positive, than l 5x-1 l can't be negative, so it is sufficient.
say if the question was to solve for x and (1) 4 = sq root (5x - 1), than you'd have two solutions (x=1 and x= -3/5), but you can eliminate the negative as X can only be positive.
is my thinking correct?
04 May 2016, 22:24
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jasonfodor
Here are the key points about squares and square roots:
1.
\(x^2 = 4\)
The equation has 2 roots: x = 2 or -2.
Logical since either value of 2 or -2 will give 4 when squared.
2.
\(\sqrt{4} = 2\) only
Now, this is a bit odd, right? If you square -2, you will get 4 too. Hence \(\sqrt{4}\) should be 2 or -2. But it isn't. And here is the reason: When we write \(\sqrt{x}\), this is the principal square root of the number. x has two roots, the positive square root (or principle square root) written as \(\sqrt{x}\) and negative square root written as \(-\sqrt{x}\). So whenever we write \(\sqrt{x}\), it means we are talking about positive square root only. And that is the reason \(\sqrt{4} = 2\) only.
3.
\((\sqrt{x})^2 = x\)
We don't have to worry about positive negative here since x is under square root. It will be positive only.
\(\sqrt{-4}\) is not defined.
4.
\(\sqrt{x^2} = |x|\)
Again, this has to do with the principal square root concept. Square root of x^2 should be x (which could have been either positive or negative such as
\(\sqrt{(-2)^2}\) which is fine. But when you take \(\sqrt{4}\), you get only 2. Hence you need to put the absolute value sign around x. Doesn't matter whether originally x was positive or negative - after coming out of the square root, it will be positive only.
General Discussion
04 May 2016, 21:43
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Hi jasonfodor,
The result of a square root calculation can ONLY be positive, but there still could be more than one correct answer to the question that is asked.
As a simple example, the square root of (4) is +2
However, if there is a squared term under a square root sign, then there could be additional options. For example:
If the square root of (X^2) = 4, then X could be +4 OR -4.
In this second example, you are mathematically allowed to 'square' both sides; you'd end up with:
X^2 = 16, which gives us X = +4 or -4
GMAT assassins aren't born, they're made,
Rich
The result of a square root calculation can ONLY be positive, but there still could be more than one correct answer to the question that is asked.
As a simple example, the square root of (4) is +2
However, if there is a squared term under a square root sign, then there could be additional options. For example:
If the square root of (X^2) = 4, then X could be +4 OR -4.
In this second example, you are mathematically allowed to 'square' both sides; you'd end up with:
X^2 = 16, which gives us X = +4 or -4
GMAT assassins aren't born, they're made,
Rich
