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PareshGmat
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 10 Apr 2014, 21:29
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B
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65% (hard)

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40% (01:36) correct 60% (02:24) wrong based on 5 sessions

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If \(\frac{\sqrt{x-y}}{3} = \sqrt{\frac{17}{x+y}}\),

then \(x^2 + y^2 =\)


A: 170

B: 185

C: 298

D: 153

E: 241



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B
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ajithkumar
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 10 Apr 2014, 22:07
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I don't know if it is a fool proof method.. but this is how I would do it.

Square on both the sides..
(x-y)/9 = 17/(x+y)
(x-y)(x+y) = 17 * 9

(17 is a prime number)

so x+y = 17
x-y = 9

by solving you will get x=13 and y = 4
so x^2+y^2=185
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PareshGmat
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 10 Apr 2014, 23:44
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\(\frac{\sqrt{x-y}}{3} = \sqrt{\frac{17}{x+y}}\)

\(\sqrt{x-y} . \sqrt{x+y} = \sqrt{17} * 3\)

Squaring both sides

(x-y) (x+y) = 17 * 9

\(x^2 - y^2 = 169 - 16\)

\(x^2 - y^2 = 13^2 - 4^2\)

x = 13; y = 4

\(x^2 + y^2 = 169 + 16 = 185\)

Answer = B
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 11 Apr 2014, 03:02
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PareshGmat
If \(\frac{\sqrt{x-y}}{3} = \sqrt{\frac{17}{x+y}}\),

then \(x^2 + y^2 =\)


A: 170

B: 185

C: 298

D: 153

E: 241

\(\frac{\sqrt{x-y}}{3} = \sqrt{\frac{17}{x+y}}\)

\(\sqrt{x-y} . \sqrt{x+y} = \sqrt{17} * 3\)

Squaring both sides

(x-y) (x+y) = 17 * 9

\(x^2 - y^2 = 169 - 16\)

\(x^2 - y^2 = 13^2 - 4^2\)

x = 13; y = 4

\(x^2 + y^2 = 169 + 16 = 185\)

Answer = B
Show more

That's not correct: \(x^2 - y^2 = 153\) have infinitely many solutions. For example,

\(y=0\) and \(x=\sqrt{153}\);
\(y=1\) and \(x=\sqrt{154}\);
...
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 11 Apr 2014, 03:11
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Bunuel
PareshGmat
If \(\frac{\sqrt{x-y}}{3} = \sqrt{\frac{17}{x+y}}\),

then \(x^2 + y^2 =\)


A: 170

B: 185

C: 298

D: 153

E: 241

\(\frac{\sqrt{x-y}}{3} = \sqrt{\frac{17}{x+y}}\)

\(\sqrt{x-y} . \sqrt{x+y} = \sqrt{17} * 3\)

Squaring both sides

(x-y) (x+y) = 17 * 9

\(x^2 - y^2 = 169 - 16\)

\(x^2 - y^2 = 13^2 - 4^2\)

x = 13; y = 4

\(x^2 + y^2 = 169 + 16 = 185\)

Answer = B

That's not correct: \(x^2 - y^2 = 153\) have infinitely many solutions. For example,

\(y=0\) and \(x=\sqrt{153}\);
\(y=1\) and \(x=\sqrt{154}\);
...
Show more


Agreed.

Just to add, for finding finite number of solutions for two independent variables, we need two different equations.
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 12 Apr 2014, 10:18
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ajithkumar
I don't know if it is a fool proof method.. but this is how I would do it.

Square on both the sides..
(x-y)/9 = 17/(x+y)
(x-y)(x+y) = 17 * 9

(17 is a prime number)

so x+y = 17
x-y = 9

by solving you will get x=13 and y = 4
so x^2+y^2=185
Show more

Could you please elaborate on the 3rd part of the solution?

Why does x+y = 17 (since 17 is a prime number)?
Is there a certain trick behind this?
Or do you simply look at (x-y)(x+y) = 17 * 9 and determine that either (x-y) or (x+y) has to
equal 17 (or 9 for that matter)?

Additionally, I understand that you can basically search for the solution of x^2-y^2=153 by force. However,
your approach seems more elegant.
Thanks in advance.
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ind23
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 12 Apr 2014, 11:12
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NicoGER
ajithkumar
I don't know if it is a fool proof method.. but this is how I would do it.

Square on both the sides..
(x-y)/9 = 17/(x+y)
(x-y)(x+y) = 17 * 9

(17 is a prime number)

so x+y = 17
x-y = 9

by solving you will get x=13 and y = 4
so x^2+y^2=185

Could you please elaborate on the 3rd part of the solution?

Why does x+y = 17 (since 17 is a prime number)?
Is there a certain trick behind this?
Or do you simply look at (x-y)(x+y) = 17 * 9 and determine that either (x-y) or (x+y) has to
equal 17 (or 9 for that matter)?

Additionally, I understand that you can basically search for the solution of x^2-y^2=153 by force. However,
your approach seems more elegant.
Thanks in advance.
Show more

(x-y)(x+y) = 17*9 = 153
Given, 17 is a prime number it cannot be broken further into factors but 9 can be written as 3*3

so, (x-y)(x+y) = 17*9 = 17*3*3

Now for +ve x & y, x+y > x-y, hence x+y can be either 17 or 17*3
and x-y can be either 9 or 3
for x+y = 17*3, x-y = 3, we will get very high values of x and y and x^2+y^2 would be much higher than any of the options.

hence x+y = 17 and x-y = 9
=> x= 13, y = 4


-----------------------
Kudos, if the answer helped
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NicoGER
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 12 Apr 2014, 11:48
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ind23
NicoGER
ajithkumar
I don't know if it is a fool proof method.. but this is how I would do it.

Square on both the sides..
(x-y)/9 = 17/(x+y)
(x-y)(x+y) = 17 * 9

(17 is a prime number)

so x+y = 17
x-y = 9

by solving you will get x=13 and y = 4
so x^2+y^2=185

Could you please elaborate on the 3rd part of the solution?

Why does x+y = 17 (since 17 is a prime number)?
Is there a certain trick behind this?
Or do you simply look at (x-y)(x+y) = 17 * 9 and determine that either (x-y) or (x+y) has to
equal 17 (or 9 for that matter)?

Additionally, I understand that you can basically search for the solution of x^2-y^2=153 by force. However,
your approach seems more elegant.
Thanks in advance.

(x-y)(x+y) = 17*9 = 153
Given, 17 is a prime number it cannot be broken further into factors but 9 can be written as 3*3

so, (x-y)(x+y) = 17*9 = 17*3*3

Now for +ve x & y, x+y > x-y, hence x+y can be either 17 or 17*3
and x-y can be either 9 or 3
for x+y = 17*3, x-y = 3, we will get very high values of x and y and x^2+y^2 would be much higher than any of the options.

hence x+y = 17 and x-y = 9
=> x= 13, y = 4


-----------------------
Kudos, if the answer helped
Show more

I think I understand. Thanks for replying!
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Bunuel
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If [m][fraction]\sqrt{x-y}/3[/fraction] = \sqrt{[fraction]17 13 Apr 2014, 06:32
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NicoGER
ajithkumar
I don't know if it is a fool proof method.. but this is how I would do it.

Square on both the sides..
(x-y)/9 = 17/(x+y)
(x-y)(x+y) = 17 * 9

(17 is a prime number)

so x+y = 17
x-y = 9

by solving you will get x=13 and y = 4
so x^2+y^2=185

Could you please elaborate on the 3rd part of the solution?

Why does x+y = 17 (since 17 is a prime number)?
Is there a certain trick behind this?
Or do you simply look at (x-y)(x+y) = 17 * 9 and determine that either (x-y) or (x+y) has to
equal 17 (or 9 for that matter)?

Additionally, I understand that you can basically search for the solution of x^2-y^2=153 by force. However,
your approach seems more elegant.
Thanks in advance.
Show more

Please notice that this solution is not correct. Refer to my post above.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Bunuel
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109739 posts