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Statement 1: \((x-y)(x+y)=0\), so either \(x-y=0\) or \(x+y=0\), in either case we have two variables and one equation. hence insufficient
Statement 2: Solve it to get \(\frac{(x^2+y^2)}{xy}=0\)
or \(x^2+y^2 = 0\). Sum of two positive numbers is 0 only when both the numbers are 0
Hence \(x=0\). Sufficient
Option B
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In order to make (2) works, both x, y have to be different from 0. So, how can x = y = 0 ? I don't think this is a valid solution. Plz show the source of the question, thanks!
Statement 1: \((x-y)(x+y)=0\), so either \(x-y=0\) or \(x+y=0\), in either case we have two variables and one equation. hence insufficient
Statement 2: Solve it to get \(\frac{(x^2+y^2)}{xy}=0\)
or \(x^2+y^2 = 0\). Sum of two positive numbers is 0 only when both the numbers are 0
Hence \(x=0\). Sufficient
Option B
In order to make (2) works, both x, y have to be different from 0. So, how can x = y = 0 ? I don't think this is a valid solution. Plz show the source of the question, thanks!
For the source of the question, kindly consult owner of the question Bunuel
As far as the equation in statement 2 is concerned, in an equality if RHS is \(0\), then LHS has to be \(0\). Here the LHS is an algebraic expression that is not presented in its simplest form. We need to find the roots of this quadratic equation. Hence on simplification it yields \(x^2+y^2\) which will yield \(x=y=0\). In graphical terms it will be a DOT in the x-y plane at the origin.
Hi Bunuel - can you provide more technical clarity on this.
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