Is x

Question

Is x < 0?

(1) x^3 + x^2 + x + 2 = 0

(2) x^2 - x - 2 < 0

idontknowwhy94 · Answer

Guys IHO ,

the answer is  A  as the only way for the equation to be 0 is when x is -ve

for st 2 ; we have 2 answers

rohit8865 · Answer

1) if x>=0 then the expression will never be 0. thus x<0.........suff 2) X^2-x-2<0 (x+1)(x-2)<0 X satisfies for range -10 or <0 ..........insuff.... Ans A

mohit_w · Answer

You can solve the equation algebraically as well.

1) x^3+x^2+x+2=0

on solving you get the following values of x :- -2/-3,-2.In all cases x<0. Therefore x must be <0. Statement is Sufficient.

2) on Solving statement number 2, we get 2 values of x, x<2 OR x<-1. Two different solns. i.e X can be +ve and X can be -ve. Therefore,statement is not sufficient.

Ans: A

KarishmaB · Answer

Responding to a pm:

In stmnt 1, note that the point is that x cannot be positive or 0. So it has to be negative (since it has to be real).
If x is positive, all terms are positive and their sum cannot be 0.
If x is 0, you get 2 = 0 which is not valid.
So x has to be negative. No need to look for the actual value. It will not be an integer.

x^{3} + x^{2}+ x + 2 = 0  
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0
So somewhere in between -1 to -2, x will take a value such that you will get 0 = 0

elainetianfong · Answer

1) x^{3} + x^{2} + x + 2 = 0 => x^2(x+1) + (x + 1) + 1 =0 => (x+1)(x^2+1) + 1 = 0 => (x+1)(x^2+1) = -1 RHS= -1 < 0 => LHS <0. As x^2+1 > 0, then x +1 <0 => x < -1 => x <0 (suff) 2)2) x^{2} -x-2 < 0 => -1 < x < 2 => Insuff

lavverma · Answer

Can you provide the solution for x^3+x^2+x+2=0
Thanks in advance!

KarishmaB · Answer

This equation is not conducive to being solved easily. You cannot get the first root by easy hit n trial here. You only need to observe here that all terms are positive. So the root must be negative as explained in my comment above: https://gmatclub.com/forum/is-x-0-1-x-3 ... l#p1747591

Archit3110 · Answer

#1
x^3 + x^2 + x + 2 = 0
x has to be in between -1 to -2 so sufficient
#2
solving expression we get x=-1 and x=2
insufficient
IMO A

chetan2u · Answer

A different way that turns this question into a very simple question. Let us use the number properties to get to our answer. 1) x^3+x^2+x+2=0 Now x^2+2>0 , so x^3+x<0........x(x^2+1)<0 But x^2+1>0 , so x<0 Sufficient 2) x^2-x-2<0 x^2-2x+x-2<0......(x+1)(x-2)<0 Two cases 1) If x+1<0 or x<-1, then x-2>0 or x>2.....NO values exist 2) If x+1>0 or x>-1, then x-2<0 or x<2....Possible values -1

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Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0

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