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# GMAT Diagnostic Test Question 13

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GMAT Diagnostic Test Question 13 [#permalink]

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07 Oct 2013, 00:12
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GMAT Diagnostic Test Question 13
Field: Algebra
Difficulty: 700

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is

A. -2
B. 0
C. 1
D. 3
E. 5
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Kudos [?]: 129488 [0], given: 12201

Math Expert
Joined: 02 Sep 2009
Posts: 41913

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Re: GMAT Diagnostic Test Question 13 [#permalink]

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07 Oct 2013, 00:14
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Explanation:

If $$x=\sqrt[4]{x^3+6x^2}$$, then the sum of all possible solutions for x is:

A. -2
B. 0
C. 1
D. 3
E. 5

Take the given expression to the 4th power: $$x^4=x^3+6x^2$$;

Re-arrange and factor out x^2: $$x^2(x^2-x-6)=0$$;

Factorize: $$x^2(x-3)(x+2)=0$$;

So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

The sum of all possible solutions for x is 0+3=3.

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Re: GMAT Diagnostic Test Question 13 [#permalink]

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07 Oct 2013, 08:20
So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

Why exactly is the even fourth root be negative.

Lets say we have a number 16 whose square root can be +- 4

Fourth roots can be +-2 -----> (-2) * (-2) * (-2) * (-2) = 16

Hence if $$\sqrt{expression}$$ = $$\sqrt{16}$$

Why cant x= -2 ? We are selecting all -2 here

I might be wrong, but best to ask and clear any ambiguity.

Kudos [?]: 22 [0], given: 34

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129488 [2], given: 12201

Re: GMAT Diagnostic Test Question 13 [#permalink]

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07 Oct 2013, 09:48
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irda wrote:
So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

Why exactly is the even fourth root be negative.

Lets say we have a number 16 whose square root can be +- 4

Fourth roots can be +-2 -----> (-2) * (-2) * (-2) * (-2) = 16

Hence if $$\sqrt{expression}$$ = $$\sqrt{16}$$

Why cant x= -2 ? We are selecting all -2 here

I might be wrong, but best to ask and clear any ambiguity.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root. That is, $$\sqrt{25}=5$$, NOT +5 or -5.

In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.
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Re: GMAT Diagnostic Test Question 13 [#permalink]

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08 Jun 2014, 05:33
Bunuel wrote:
irda wrote:
So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

Why exactly is the even fourth root be negative.

Lets say we have a number 16 whose square root can be +- 4

Fourth roots can be +-2 -----> (-2) * (-2) * (-2) * (-2) = 16

Hence if $$\sqrt{expression}$$ = $$\sqrt{16}$$

Why cant x= -2 ? We are selecting all -2 here

I might be wrong, but best to ask and clear any ambiguity.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root. That is, $$\sqrt{25}=5$$, NOT +5 or -5.

In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.

I would like to know if this is officially declared in GMAC's website- that GMAT only accepts +ve root. This is a clear ambiguity in terms of answer logic.

Kudos [?]: [0], given: 25

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129488 [0], given: 12201

Re: GMAT Diagnostic Test Question 13 [#permalink]

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08 Jun 2014, 05:43
lakhyajd wrote:
Bunuel wrote:
irda wrote:
So, the roots are $$x=0$$, $$x=3$$ and $$x=-2$$. But $$x$$ cannot be negative as it equals to the even (4th) root of some expression ($$\sqrt{expression}\geq{0}$$), thus only two solution are valid $$x=0$$ and $$x=3$$.

Why exactly is the even fourth root be negative.

Lets say we have a number 16 whose square root can be +- 4

Fourth roots can be +-2 -----> (-2) * (-2) * (-2) * (-2) = 16

Hence if $$\sqrt{expression}$$ = $$\sqrt{16}$$

Why cant x= -2 ? We are selecting all -2 here

I might be wrong, but best to ask and clear any ambiguity.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root. That is, $$\sqrt{25}=5$$, NOT +5 or -5.

In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.

I would like to know if this is officially declared in GMAC's website- that GMAT only accepts +ve root. This is a clear ambiguity in terms of answer logic.

Official Guide for GMAT® Review, 13th Edition:
$$\sqrt{n}$$ denotes the positive number whose square is n. For example, $$\sqrt{9}$$ denotes 3.

Hope it helps.
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Kudos [?]: 129488 [0], given: 12201

Re: GMAT Diagnostic Test Question 13   [#permalink] 08 Jun 2014, 05:43
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# GMAT Diagnostic Test Question 13

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