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. If x is a positive number and 1/2 the square root of x is the cube

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. If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 25 Aug 2016, 01:25
1
1
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (01:33) correct 27% (01:28) wrong based on 165 sessions

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Re: . If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 25 Aug 2016, 05:37
1
Bunuel wrote:
If x is a positive number and 1/2 the square root of x is the cube root of x, then x =

A. 64
B. 32
C. 16
D. 4
E. 1



1/2 the square root of x is cube root of x.

if x = 64..

then 1/2 the square root of x = 4 and cube of x is 64.

Option A.
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Re: . If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 03 Mar 2019, 17:14
Hello,

Could someone please explain how to solve this algebraically?

Thanks in advance.
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Re: . If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 03 Mar 2019, 18:59
freedomfox2121 wrote:
Hello,

Could someone please explain how to solve this algebraically?

Thanks in advance.



(x)^1/2 / 2 = (x) ^ 3

(x) ^ 1/2 / (x)^3 = 2

(x) ^ 1/6 = 2

x = 64
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. If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 05 Mar 2019, 20:27
jake092 wrote:
freedomfox2121 wrote:
Hello,

Could someone please explain how to solve this algebraically?

Thanks in advance.



(x)^1/2 / 2 = (x) ^ 3

(x) ^ 1/2 / (x)^3 = 2

(x) ^ 1/6 = 2

x = 64


I set up the problem as (x)^1/2 / 2 = (x)^1/3
Is X^1/3 the correct way to write a cube root? My understanding is that x^3 is x cubed.
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Re: . If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 06 Mar 2019, 20:45
freedomfox2121 wrote:
jake092 wrote:
freedomfox2121 wrote:
Hello,

Could someone please explain how to solve this algebraically?

Thanks in advance.



(x)^1/2 / 2 = (x) ^ 3

(x) ^ 1/2 / (x)^3 = 2

(x) ^ 1/6 = 2

x = 64


I set up the problem as (x)^1/2 / 2 = (x)^1/3
Is X^1/3 the correct way to write a cube root? My understanding is that x^3 is x cubed.


I set up the formula as (x)^1/2 / 2 = (x)^1/3
Then simplified to (x)^1/2 = 2(x)^1/3 -> [(x)^1/2] / [(x)^1/3] = 2
Which then turns into (x)^1/6 = 2
6√(x) = 2
Finally, [6√(x)]^6 = (2)^6
x = 64
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Re: . If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 29 May 2019, 08:23
Bunuel wrote:
If x is a positive number and 1/2 the square root of x is the cube root of x, then x =

A. 64
B. 32
C. 16
D. 4
E. 1


x>0
1/2 (x^1/2) = x^1/3
square and cube both side. Or take power of 6 on both side.
1/2^6 *x^3 = x^2
x^3 = 2^6 * x^2
x^2(x-2^6) = 0
since x =/=0
Hence x = 2^6 = 64

Answer A
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Re: . If x is a positive number and 1/2 the square root of x is the cube  [#permalink]

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New post 01 Jun 2019, 06:18
Bunuel wrote:
If x is a positive number and 1/2 the square root of x is the cube root of x, then x =

A. 64
B. 32
C. 16
D. 4
E. 1


We can create the equation:

(1/2)√x = ^3√x

(½)x^(½) = x^(⅓)

We can raise each side to the 6th power, obtaining:

[(½)^6](x^3) = x^2

Dividing each side by x^2 (which is possible because x is positive), we have:

(1/64)x = 1

x = 64

Answer: A
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Re: . If x is a positive number and 1/2 the square root of x is the cube   [#permalink] 01 Jun 2019, 06:18
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