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by Night reader » Thu Jan 06, 2011 12:43 am
If Sqrt[x^2 - 6x +9] + |x+y+2| = 0, find y

A.-5
B.-3
C.-2
D. 2
E. 3

[spoiler]my solution (see you've not decided on this problem): If Sqrt[(x-3)^2] + |x+y+2| = 0 => (1) x-3+|x+y+2|, x<0 & (2) x-3+|x+y+2|, x>=0 => (1) y=-5 & (2) y=1-x^2 => plug-in/solve for x => -5=1-x^2, x^2=6 OR x=|Sqrt(6)|, finally -5=1-2|Sqrt(6)| => pick A[/spoiler]
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by Anurag@Gurome » Thu Jan 06, 2011 1:26 am
Night reader wrote:If Sqrt[x^2 - 6x +9] + |x+y+2| = 0, find y

A.-5
B.-3
C.-2
D. 2
E. 3
.... √(x² - 6x + 9) + |x + y + 2| = 0
=> √(x - 3)² + |x + y + 2| = 0
=> |x - 3| + |x + y + 2| = 0

Now |x - 3| and |x + y + 2| both are non-negative number. Thus these two can sum up to result a zero only if both of them are individually equal to zero.

Hence, |x - 3| = 0 and |x + y + 2| = 0
=> x = 3 and (x + y + 2) = 0
=> (y + 5) = 0
=> y = -5

The correct answer is A.
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by mike34 » Thu Jan 06, 2011 1:32 am
The key is:

Don't overcalculate or you'll spend a lot of time !!!

Answer: A

My solution:

sqrt(X² - 6X + 9) + X = -Y - 2

If X = 0, then sqrt(X² - 6X + 9) + X must be equal to 3

So Y = -5

You can check in pluging an other value of X.
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by Night reader » Thu Jan 06, 2011 1:46 am
mike34 wrote:The key is:

Don't overcalculate or you'll spend a lot of time !!!

Answer: A

My solution:

sqrt(X² - 6X + 9) + X = -Y - 2

If X = 0, then sqrt(X² - 6X + 9) + X must be equal to 3

So Y = -5

You can check in pluging an other value of X.
@mike, what do you mean by don't overcalculate? sqrt(X² - 6X + 9) + X = -Y - 2 is not the only solution, what about sqrt(X² - 6X + 9) - X = Y + 2? And why do we need to set x=0? I don't feel safe with your approach :(
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