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In the above example, x = 2, since (2)(2) = 4. While it is true that (-2)(-2) = 4, the GMAT follows the standard convention that a radical (root) sign denotes only the non-negative root of a number. Thus, 2 is the only solution for x.
Does this seem valid to anybody else? Seems sort of untrue to me, x should be +/- 2 in this case, not just +2.
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In the above example, x = 2, since (2)(2) = 4. While it is true that (-2)(-2) = 4, the GMAT follows the standard convention that a radical (root) sign denotes only the non-negative root of a number. Thus, 2 is the only solution for x.
Does this seem valid to anybody else? Seems sort of untrue to me, x should be +/- 2 in this case, not just +2.
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sqrt4 =x is simply x^2 = 4 we break that down and x = plus or minus 2 MGMAT is wrong.
\(\sqrt{4} = x\) ==> x=2 is a general mathematical rule
You confuse with this one:
\(x^2=4\)
\(x=\pm\sqrt{4}=\pm2\)
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Huh? There are always 2 square roots of any positive number, so sq(4) = +/- 2. If the GMAT doesn't acknowledge this, then fine, but mathematically that's not correct.
Doesn't (-2)(-2) = 4?
From wikipedia:
Every positive number x has two square roots. One of them is sq(x), , which is positive, and the other is -sq(x), which is negative. Together these two square roots are denoted +/- sq(x).
Huh? There are always 2 square roots of any positive number, so sq(4) = +/- 2. If the GMAT doesn't acknowledge this, then fine, but mathematically that's not correct.
Doesn't (-2)(-2) = 4?
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\(x^2=4\)
\(x=\pm\sqrt{4}=\pm2\)
sonibubu
From wikipedia:
Every positive number x has two square roots. One of them is sq(x), , which is positive, and the other is -sq(x), which is negative. Together these two square roots are denoted +/- sq(x).
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From wikipedia:
In mathematics, a square root of a number x is a number r such that r2 = x, or in words, a number r whose square (the result of multiplying the number by itself) is x. Every non-negative real number x has a unique non-negative square root, called the principal square root and denoted with a radical symbol as \(\sqrt{x}\)
In mathematics, an nth root of a number a is a number b such that bn=a. When referring to the nth root of a real number a it is assumed that what is desired is the principal nth root of the number, which is denoted \(\sqrt[n]{a}\) using the radical symbol (\(\sqrt{a}\)). The principal nth root of a real number a is the unique real number b which is an nth root of a and is of the same sign as a. Note that if n is even, negative numbers will not have a principal nth root.
\(\sqrt(x)\) is a principal root and \(\sqrt(4)=2\)
Sorry, maybe I misunderstand your question :dunnow But we seems to talk about the same.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.