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

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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?

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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?

­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]
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?

$$­\sqrt{x^2} = |x|$$ 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]
1
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

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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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