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If x and y are numbers such that (x+9)(y-9)=0, what is the

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sandra123
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If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink] New post   Updated on: 15 Sep 2013, 12:37
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If x and y are numbers such that (x+9)(y-9)=0, what is the smallest possible value of x^2 + y^2

A. 0
B. 9
C. 18
D. 81
E. 162

I understand its a basic question, but I fail to understand the concept. I simply put x+9=0, so x=-9. in the same way, y-9=0, so y=9
so x^2 + y^2 = 81+81=162
what am I doing wrong?
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D

Originally posted by sandra123 on 15 Sep 2013, 12:35.
Last edited by Bunuel on 15 Sep 2013, 12:37, edited 1 time in total.
Edited the question.
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Re: If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink] New post   15 Sep 2013, 12:41
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sandra123 wrote:
If x and y are numbers such that (x+9)(y-9)=0, what is the smallest possible value of x^2 + y^2

A. 0
B. 9
C. 18
D. 81
E. 162

I understand its a basic question, but I fail to understand the concept. I simply put x+9=0, so x=-9. in the same way, y-9=0, so y=9
so x^2 + y^2 = 81+81=162
what am I doing wrong?


From (x+9)(y-9)=0 it follows that either x=-9 or y=9. Thus either x^2=81 or y^2=81.

Now, if x^2=81, then the least value of y^2 is 0, so the least value of x^2 + y^2 = 81 + 0 = 81.
Similarly if y^2=81, then the least value of x^2 is 0, so the least value of x^2 + y^2 = 0 + 81 = 81.

Answer: D.

Hope it's clear.
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tchang
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Re: If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink] New post   15 Sep 2013, 17:49
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In the given equation, if x = -9 then y could = ANYTHING

If y = 9, then x could = ANYTHING

They DON'T BOTH have to be true at the same time.

9^2 + 0^2 = 81

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D


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Re: If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink] New post   22 Sep 2013, 05:26
sandra123 wrote:
If x and y are numbers such that (x+9)(y-9)=0, what is the smallest possible value of x^2 + y^2

A. 0
B. 9
C. 18
D. 81
E. 162

I understand its a basic question, but I fail to understand the concept. I simply put x+9=0, so x=-9. in the same way, y-9=0, so y=9
so x^2 + y^2 = 81+81=162
what am I doing wrong?


This question could me solved in this way.

(x+9)(x-9) = 0
=> xy - 81 = 9(x-y)
=> (x-y) = (xy/9) - 9
=> Squaring on both sides and cancelling equal terms
=> x^2 + y^2 = (xy)^2/81 + 81

Notice that 81 is added to (xy)^2/81, so the absolute minimum value of the above expression will be equal to 81 when x=9 and y=9 => x^2 + y^2 = 162.
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Re: If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink] New post   04 Sep 2017, 08:54
Since (x+9)(y-9) = 0 => x= -9 OR y=9

If x= -9 => X^2 = 81
The smallest value y can now take is y =0
Hence x^2+y^2 = 81

Same is the case with Y=9
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Re: If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink] New post   18 Mar 2021, 06:58
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Re: If x and y are numbers such that (x+9)(y-9)=0, what is the [#permalink]
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