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(x + y)(x − y) = 0 xy ≠ 0

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sandy
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(x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   Updated on: 31 Oct 2018, 06:21
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\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(6\sqrt{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
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C

Originally posted by sandy on 28 Oct 2018, 16:54.
Last edited by chetan2u on 31 Oct 2018, 06:21, edited 1 time in total.
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ruposh6240
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Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   30 Oct 2018, 03:28
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B is answer
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chetan2u
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Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   31 Oct 2018, 03:56
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sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(\sqrt[6]{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C
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Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   31 Oct 2018, 06:19
I have confusion about the answer C
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Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   10 Nov 2018, 10:50
chetan2u wrote:
sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(\sqrt[6]{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C


May I know how you derived the 36 that was included into the square root?
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ShefaliSahu
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Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   11 Nov 2018, 21:11
The answer is C.
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Salsanousi
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Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   19 Nov 2018, 17:03
Mercychee wrote:
chetan2u wrote:
sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(\sqrt[6]{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C


May I know how you derived the 36 that was included into the square root?


I think what chetan tried to say is that there is probably a typo because it should be 6 * square root of (19/2x^2)

Square root of 36 is 6.

Because it was outside the root so he brought it inside as 6^2.
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Re: (x + y)(x y) = 0 xy 0 [#permalink] New post   15 Aug 2022, 07:39
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One simple way to solve this is the following:
(x+y)(x−y)=0 --> x^2 - y^2 = 0 (difference of squares pattern)
So x^2 = y^2. We can take the square root of both sides --> x = y

Now, we can plug 1 for both and square both square roots.
We will be left with
A: 18(19/2) = 342
B: 342

Forgive me if I made any mistake.
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Re: (x + y)(x y) = 0 xy 0 [#permalink]
15 Aug 2022, 07:39
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