If x is a negative integer and x^2 - |x + 3| + x 0, what is the grea

Question

If x is a negative integer and  x^2 - |x + 3| + x > 0 , what is the greatest possible value of x?

A. -1
B. -2
C. -3
D. -4
E. -5

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nick1816 · Accepted Answer

x^2 - |x + 3| + x > 0 Case 1- 0>x>-3 x^2 -x-3+x > 0 x^2 -3>0 x> √3 (Not lie in our range) or x < - √3 Greatest integer that is smaller than -√3 and greater than -3 is -2 B Or Just put given options in the inequality Option A- (-1)^2- |-1+3|-1 = -2<0 ( reject ) Option B- (-2)^2- |-2+3|-2 = 1>0 ( Answer )

sri46 · Answer

x=-ve integer

By substitution:
x=-1 value of equation is -2 <0 
x=-2 value of equation is 1 >0 
Greatest possible value of x=-2
B

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ScottTargetTestPrep · Answer

Instead of solving the inequality algebraically, we can just try out the given answer choices.

A. -1

If x = -1, x^2 - |x + 3| + x = 1 - 2 - 1 = -2, which is not greater than 0.

B. -2

If x = -2, x^2 - |x + 3| + x = 4 - 1 - 2 = 1, which IS greater than 0.

Answer: B

Kinshook · Answer

Asked: If x is a negative integer and x^2 - |x + 3| + x > 0 , what is the greatest possible value of x? Case 1: x<-3 x^2 - (-x-3) + x > 0 x^2 + 2x + 3 > 0 (x+1)(x+2) > 0 x>2 or x<-1; x>2 is not feasible x<-3 Case 2: x>=-3 x^2 - (x+3) + x >0 x^2 -3 > 0 x<-\sqrt{3} or x>\sqrt{3} Since x is a negative integer x < - 1.71 x_{max} = - 2 IMO B

prathameshrathi · Answer

x2+2x+3>0 
(x+1)(x+2)>0 ..... wrong factorization!?

BrushMyQuant · Answer

x is a negative integer and  x^2 - |x + 3| + x > 0

As we have |x + 3| in the equation so we will have two cases 
 
   -   Case 1:  x + 3 ≥ 0 => x ≥ - 3
 
=> |x + 3| = x + 3
=>  x^2 - (x + 3) + x > 0 
=>  x^2 - x - 3 + x > 0 
=>  x^2 > 3  
=> x >  \sqrt{3}  or x < - \sqrt{3}  
 
But we know that x ≥ - 3 and x is a negative integer 
 => Solution will be -3 ≤ x ≤ - \sqrt{3}

=> x = -3 , -2  
  -   Case 2:  x + 3 ≤ 0 => x ≤ -3
 
 => |x + 3| = -(x + 3)
 =>  x^2 - (-(x + 3)) + x > 0 
 =>  x^2 + x + 3 + x > 0 
 =>  x^2 + 2x + 3 > 0 
 => x*(x+2) > -3

Now, x  ≤ -3
  => Both x and x+2 will be negative and their product will be positive and will ALWAYS be > -3

=> Equation is true for all values of x ≤ -3

=> Combined solution is -3 ≤ x ≤ - \sqrt{3}  and x ≤ -3 
 => x ≤ - \sqrt{3}  
 => x ≤ -1.732

Greatest possible value of x = -2

So,  Answer will be B 
Hope it helps!

Watch the following video to MASTER Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

BrushMyQuant · Answer

Yes, that is wrong factorization. We can solve it by doing x*(x+2) > -3. Explained in my response above.

Arya5577 · Answer

Yes, actually the determinant is negative so this inequality holds true for any value of x.

luisdicampo · Answer

Deconstructing the Question

We are given

x^2-|x+3|+x>0

where   x   is a negative integer.

We need the greatest possible value of  x , so test the negative integers starting from the largest:

-1, -2, -3, \dots

Step-by-step

Test  x=-1

(-1)^2-|(-1)+3|+(-1)

=1-|2|-1

=1-2-1=-2

This is not greater than  0 .

So  x=-1  does not work.

Test  x=-2

(-2)^2-|(-2)+3|+(-2)

=4-|1|-2

=4-1-2=1

This is greater than  0 .

So  x=-2  works.

Since  -2  works and  -1  does not, the greatest possible value of  x  is

-2

Answer B: -2

If x is a negative integer and x^2 - |x + 3| + x > 0, what is the grea

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As we have \|x + 3\| in the equation so we will have two cases
-Case 1: x + 3 ≥ 0 => x ≥ - 3 => \|x + 3\| = x + 3 => \(x^2 - (x + 3) + x > 0\) => \(x^2 - x - 3 + x > 0\) => \(x^2 > 3 \) => x > \(\sqrt{3}\) or x < -\(\sqrt{3}\) But we know that x ≥ - 3 and x is a negative integer => Solution will be -3 ≤ x ≤ -\(\sqrt{3}\) => x = -3 , -2	-Case 2: x + 3 ≤ 0 => x ≤ -3 => \|x + 3\| = -(x + 3) => \(x^2 - (-(x + 3)) + x > 0\) => \(x^2 + x + 3 + x > 0\) => \(x^2 + 2x + 3 > 0\) => x*(x+2) > -3 Now, x ≤ -3 => Both x and x+2 will be negative and their product will be positive and will ALWAYS be > -3 => Equation is true for all values of x ≤ -3

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