If y=|x+1|/x and x!=0, is xy>0?

@Zarrolou,
Regret for not responding within time.


If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

A)\(x^2+2x+1>0\)

B)\(y\neq{0}\)


First lets simplify the inequality

\(\frac{|x+1|}{x}\) –y = 0 ----------> \(\frac{|x+1|-xy}{x}\) = 0 -----> we know \(x\neq{}0\), then |x+1|-xy must be zero. Hence |x+1| - xy = 0 --------> |x+1| = xy


We are asked whether xy>0 --------> Whether |x+1| > 0 ? --------> We know the expression within modules can either be zero or greater than zero. For xy to be greater than zero |x+1| has to be greater than zero.
|x+1| will be zero only when x=-1 and for any other value of x, |x+1| will always be greater than zero
So the question can be rephrased as whether \(x\neq{-1}\)


Statement 1) \(x^2\) + 2x + 1 > 0

This is an quadratic inequality.
Rule :- For any quadratic inequation \(ax^2\) + bx + c > 0, if \(b^2\) – 4ac = 0 and a > 0 then the inequality holds true outside the interval of roots
In our case \(b^2\) – 4ac = 4 – 4 = 0 and a > 0 so \(x^2\) + 2X + 1 > 0 will hold true for all values beyond the Root(s) of equation (Towards any direction - Positive or Negative)
\(x^2\) + 2x + 1 = 0 --------> x(x+1) +1(x + 1) = 0 ---------> (x+1)(x+1) = 0 ----------> x=Root = -1
So \(x^2\) + 2x + 1 > 0 will hold true for any of x except for -1
This reveals that \(x\neq{-1}\) and xy>0 ----------------> Sufficient


Statement 2) \(y\neq{0}\)

From the question stem we know \(x\neq{}0\)
As per Statement 2, \(y\neq{0}\)-------------->That means both X and Y are nonzero.

|x+1| = xy
xy can be either Positive or Negative
|x+1| can be Zero or Positive
Combining both these inferences we can conclude that XY must be Positive. Sufficient

Answer = D

Regards,

Narenn