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Re: If (x^2)^(1/2)/(x^3)^(1/3) = -1and x 0, what is the value of [#permalink]
gmatophobia wrote:
Bunuel wrote:
­If \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\) and \(x \neq 0\), what is the value of \(\frac{2|x| + 6}{3 - x}\) ?

A. -2
B. 1
C. 2
D. 3
E. Cannot be determined from the given information­
­

\(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}\)

= \(\frac{|x|}{x}\)

\(\frac{|x|}{x} = -1\)

Hence, we can infer that that \(x\) is negative. 
­
Let's assume \(x = -1\)

= \(\frac{2|x| + 6}{3 - x}\)

= \(\frac{2|-1| + 6}{3 - (-1)}\)

= \(\frac{8}{4}\)

= \(2\)

­Option C

­I have a doubt. The numerator can be either postive or negative for every case of denominator can only be positive or negative.
In that case, won't the equation be always either 1 or -1. Instead of only -1. This leads to an undefined behaviour right?

Bunuel Can you please help?
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Re: If (x^2)^(1/2)/(x^3)^(1/3) = -1and x 0, what is the value of [#permalink]
 
OokandGluk wrote:
­I have a doubt. The numerator can be either postive or negative for every case of denominator can only be positive or negative.
In that case, won't the equation be always either 1 or -1. Instead of only -1. This leads to an undefined behaviour right?

Bunuel Can you please help?

­The value of the numerator can only be positive.

For example: 

\(\sqrt{4} = 2\) and not \(\pm2\)­
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Re: If (x^2)^(1/2)/(x^3)^(1/3) = -1and x 0, what is the value of [#permalink]
Expert Reply
OokandGluk wrote:
gmatophobia wrote:
Bunuel wrote:
­If \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\) and \(x \neq 0\), what is the value of \(\frac{2|x| + 6}{3 - x}\) ?

A. -2
B. 1
C. 2
D. 3
E. Cannot be determined from the given information­
­

\(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}\)

= \(\frac{|x|}{x}\)

\(\frac{|x|}{x} = -1\)

Hence, we can infer that that \(x\) is negative. 
­
Let's assume \(x = -1\)

= \(\frac{2|x| + 6}{3 - x}\)

= \(\frac{2|-1| + 6}{3 - (-1)}\)

= \(\frac{8}{4}\)

= \(2\)

­Option C

­I have a doubt. The numerator can be either postive or negative for every case of denominator can only be positive or negative.
In that case, won't the equation be always either 1 or -1. Instead of only -1. This leads to an undefined behaviour right?

Bunuel Can you please help?

\(­\sqrt{x^2} = |x|\) and |x| cannot be negative. Generally, the square root sign always indicates non-negative square root.

Mathematically, \(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

√9 = 3, NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3. Because x^2 = 9 means that x =-√9 =-3 or x = √9 = 3.

Hope it helps.
 ­
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Re: If (x^2)^(1/2)/(x^3)^(1/3) = -1and x 0, what is the value of [#permalink]
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Kudos
Bunuel wrote:
­If \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\) and \(x \neq 0\), what is the value of \(\frac{2|x| + 6}{3 - x}\) ?

A. -2
B. 1
C. 2
D. 3
E. Cannot be determined from the given information­


This is a PS Butler Question

­

­
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Re: If (x^2)^(1/2)/(x^3)^(1/3) = -1and x 0, what is the value of [#permalink]
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