NikhilDev
Hey bunuel,
I got x=-1,2 from the first statement.
But in the second statement, i took positive and negative possibilities and came up with 4 equations.
the 4 equations are
1.(x^2)-x=2
2.(x^2)+x=2
3. -(x^2)+x=2
4. -(x^2)-x=2
Out of these four equations, according to me, only the 1st and the second have real solutions.
So from the 1st equation above i got x=-1,2(Same as the result from our First Statement )
and from the second equation i got x=1,-2.
But everywhere people have written that they got x=-2,2 from the second statement. Can you please explain it to me as to what the issue is.
You are making a mistake while interpreting different scenarios after you open up the absolute sign.
Per statement 2, |x^2-|x||=2
Consider 2 cases,
Case 1: when \(x \geq 0\), \(|x| =x\) --->\(|x^2-|x||=2\) becomes \(|x^2-x|=2\) ---> \(x^2-x=\pm 2\) , giving you 2 quadratic equations
\(x^2-x-2=0\) and \(x^2-x+2=0\) (no real solutions, eliminate). From \(x^2-x-2=0\) , you get x=2 and x=-1 but x=-1 is NOT allowed as you are assuming that for this case, \(x \geq 0\).
Thus x=2 is the only possible solution from this scenario.
Case 2: when \(x<0\)---> \(|x| = -x\) ---> \(|x^2-|x||=2\) becomes \(|x^2+x|=2\)---> \(x^2+x=\pm 2\) , giving you 2 quadratic equations again,
\(x^2+x-2=0\) and\(x^2+x+2=0\) (no real solutions, eliminate). From \(x^2+x-2=0\), you get x=-2 and x=1 but x=1 is NOT allowed as you are assuming that for this case, \(x < 0\).
Thus x=-2 is the only possible solution from this scenario.
Finally, when you combine both the scenarios, you see that the allowed values are \(x=\pm 2\). Thus this statement is NOT sufficient to answer the question asked.
Hope this help. You have to be very careful about your assumptions in absolute value questions.