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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.

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.

sqrt4 =x is simply x^2 = 4 we break that down and x = plus or minus 2 MGMAT is wrong.

-sqrt4 is -2 sqrt(-4) is an imaginary # _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

\(\sqrt{4} = x\) ==> x=2 is a general mathematical rule

You confuse with this one:

\(x^2=4\)

\(x=\pm\sqrt{4}=\pm2\)

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?

\(x^2=4\)

\(x=\pm\sqrt{4}=\pm2\)

sonibubu wrote:

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).

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 But we seems to talk about the same. _________________