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# I read the following in a Manhattan GMAT book: If sq(4) = x,

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I read the following in a Manhattan GMAT book: If sq(4) = x, [#permalink]  25 Jan 2008, 11:33
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I read the following in a Manhattan GMAT book:

If sq(4) = x, what is x?

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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Re: square roots on the GMAT [#permalink]  25 Jan 2008, 11:38
sonibubu wrote:
I read the following in a Manhattan GMAT book:

If sq(4) = x, what is x?

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 #
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Re: square roots on the GMAT [#permalink]  25 Jan 2008, 11:44
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Manhattan GMAT is correct -

$\sqrt{x}$ on the GMAT can always be assumed to be the positive root of $x$.
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Re: square roots on the GMAT [#permalink]  25 Jan 2008, 11:55
solaris1 wrote:
Manhattan GMAT is correct -

$\sqrt{x}$ on the GMAT can always be assumed to be the positive root of $x$.

you're right. i misinterpreted the question. we cannot have a negative under a radical since GMAT does not allow imaginary numbers.
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Re: square roots on the GMAT [#permalink]  25 Jan 2008, 13:37
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Expert's post
$\sqrt{4} = x$ ==> x=2 is a general mathematical rule

You confuse with this one:

$x^2=4$

$x=\pm\sqrt{4}=\pm2$
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Re: square roots on the GMAT [#permalink]  28 Jan 2008, 09:37
walker wrote:
$\sqrt{4} = x$ ==> 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).
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Re: square roots on the GMAT [#permalink]  28 Jan 2008, 11:17
Expert's post
sonibubu wrote:
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.
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Re: square roots on the GMAT   [#permalink] 28 Jan 2008, 11:17
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