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705-805 Level|   Algebra|      
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achloes
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If x and z are positive constants, for how many values of y is 04 Jun 2023, 10:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

51% (01:56) correct 49% (01:39) wrong based on 78 sessions

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If x and z are positive constants, for how many values of y is x(y)^2 = z(y)^4?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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C
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gmatophobia
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If x and z are positive constants, for how many values of y is 04 Jun 2023, 22:57
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achloes
If x and z are positive constants, for how many values of y is x(y)^2 = z(y)^4?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Show more

Given

  • x and z are positive constants

\(xy^2 = zy^4\)

\(y^2(x - zy^2)=0\)

Case 1: y = 0

Case 2: \(x - zy^2 = 0\)

\(x = zy^2\)

\(y^2 = \frac{x}{z}\)

\(|y| = \sqrt{\frac{x}{z}}\)

y can either take a positive or negative value.

Hence, three values of y are possible.

Option C
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If x and z are positive constants, for how many values of y is 12 Nov 2024, 04:13
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gmatophobia Have a little doubt. how root x/z is negative ? underroots are always positive. In order to have both positive and negative value, x and z must be perfect square ?

gmatophobia
achloes
If x and z are positive constants, for how many values of y is x(y)^2 = z(y)^4?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Given

  • x and z are positive constants

\(xy^2 = zy^4\)

\(y^2(x - zy^2)=0\)

Case 1: y = 0

Case 2: \(x - zy^2 = 0\)

\(x = zy^2\)

\(y^2 = \frac{x}{z}\)

\(|y| = \sqrt{\frac{x}{z}}\)

y can either take a positive or negative value.

Hence, three values of y are possible.

Option C
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Bunuel
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If x and z are positive constants, for how many values of y is 12 Nov 2024, 06:51
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Can,Will
gmatophobia Have a little doubt. how root x/z is negative ? underroots are always positive. In order to have both positive and negative value, x and z must be perfect square ?

gmatophobia
achloes
If x and z are positive constants, for how many values of y is x(y)^2 = z(y)^4?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Given

  • x and z are positive constants

\(xy^2 = zy^4\)

\(y^2(x - zy^2)=0\)

Case 1: y = 0

Case 2: \(x - zy^2 = 0\)

\(x = zy^2\)

\(y^2 = \frac{x}{z}\)

\(|y| = \sqrt{\frac{x}{z}}\)

y can either take a positive or negative value.

Hence, three values of y are possible.

Option C
Show more

You are right; the square root of a number cannot be negative, but we have a different case here.



When the GMAT provides the square root symbol for an even root, like a square root, fourth root, etc., only the non-negative root is accepted as the answer. For example: That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).­

So here we have \(y^2 = \frac{x}{z}\), which gives \(y = \sqrt{\frac{x}{z}}\) or \(y = - \sqrt{\frac{x}{z}}\).

Hope it's clear.
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