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If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is

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Bunuel
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If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   06 Dec 2019, 03:05
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If \(xy\neq0\) and \(x\sqrt{12}+y\sqrt{51}=\sqrt{z}(2x+y\sqrt{17})\), what is the value of z?

A. \(\sqrt{3}\)
B. 2
C. 3
D. \(\sqrt{5}\)
E. 7

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C
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freedom128
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Re: If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   06 Dec 2019, 03:36
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xy ≠ 0

x.12^(1/2) + y.51^(1/2)
=3^(1/2) * [2x +y.17^(1/2)]
= z^(1/2) * [2x +y.17^(1/2)]

Z=3

Final answer is (C)

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Re: If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   06 Dec 2019, 04:47
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Quote:
If xy≠0 and x√12 + y√51 = √z(2x +y√17), what is the value of z?

A. √3
B. 2
C. 3
D. √5
E. 7


x√12 + y√51 = √z(2x +y√17)
x√(4*3) + y√(3*17) = √z(2x +y√17)
√3(2x+y√17)=√z(2x +y√17)
√3=√z, z=3

Ans (C)
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Re: If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   06 Dec 2019, 08:41
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We are given than xy≠0 and that x√12+y√51 = √z(2x+y√17). We are to find z.
Simplifying the RHS: x√12+y√51 = 2x√3+y√3*√17 = √3(2x+√17)
Now, equating LHS to RHS: √3(2x+√17)=√z(2x+y√17)
√3=√z
Hence z=3.

The answer is therefore C.
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Re: If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   06 Dec 2019, 18:43
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If x*y ≠0 and x√12 +y√51= √z*(2x+y√17), what is the value of z?

x√12 +y√51 =x√3√4+y√3√17 =2x√3+y√3√17 =√3(2x+y√17)

--> √3(2x+y√17) =√z*(2x+y√17)
x*y ≠0--> we can simplify that --> √z=√3 --> z=3

The answer is C.
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If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   Updated on: 09 Dec 2019, 00:39
we can solve LHS and RHS
x√12+y√51=√z(2x+y√17)
x√3*4+y√3*17 = √z(2x+y√17)
√3(2x+√17) = √z(2x+√17)
√z=√3
z=3
IMO C


If xy≠0xy≠0 and x12‾‾‾√+y51‾‾‾√=z√(2x+y17‾‾‾√)x12+y51=z(2x+y17), what is the value of z?

A. 3‾√3
B. 2
C. 3
D. 5‾√5
E. 7

Originally posted by Archit3110 on 07 Dec 2019, 02:48.
Last edited by Archit3110 on 09 Dec 2019, 00:39, edited 1 time in total.
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Re: If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   07 Dec 2019, 17:58
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√3 can be taken common from above equation

x√12+y√51=x√4.√3+y√3.√17=√3(2x+y√17)


So z=3

OA:C

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Re: If xy ≠ 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   10 Dec 2019, 09:47
If x√12+y√51= 2x√7 + √7*y√17 = x√4z + y√(17z)
Then x√12+y√51=x√4z + y√(17z)
By deduction 4z = 12 or 17z= 51, therefore z=3.
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Re: If xy 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink] New post   18 Mar 2024, 01:55
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Re: If xy 0 and x12^(1/2) + y51^(1/2) = z^(1/2)(2x +y 17^(1/2)), what is [#permalink]
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