|x + 1|

Question

|x + 1| < |x^2 + 2x + 2|

A: (-infinity, +infinity)
B: (-1, +infinity)
C: (-infinity, -1)
D: no solutions
E: (-1, 1)

gurpreetsingh · Accepted Answer

|x + 1| < |x^2 + 2x + 2| since x^2 + 2x + 2 = (x+1)^2 +1 which is always > 0 => |x + 1| < (x+1)^2 +1 take |x+1| = y => y y^2 +1 - y>0 (y-\frac{1}{2})^2 +\frac{3}{4}>0 This is true for all values of y, hence it is true for all values of x. Hence A

shaselai · Answer

Lets try plugin and POE as alternative to the other solution- this is pretty quick as the math isnt hard at all and the answer choices given are super easy to pick easy numbers 1/2
POE:
A. lets skip this for now since if any number doesn't work this will fail. 
B. try -2 since it is supposed to be "invalid". |-2+1| < |(-2)^2 + 2*-2 +2| = 1 <2 which is VALID so NO B
C. try 1 since it is out of scopre - |1+1| < |1^2 + 2*1 +2| = 2 < 5 which is VALID so no C
D. obviously no from above examples (and in general on test "no solutions" is a trap
E. from trying -2 from B we know this is false

So we derived A with POE.

shrouded1 · Answer

|x+1| < |x^2+2x+1+1| |x+1| < |(x+1)^2 + 1| Note that (x+1)^2+1 is always positive, so, |x+1| < x^2+2x+2 If x+1>=0 OR x>=-1, Then x^2+x+1>0 b^2-4ac = -3 ... No real roots and a>0 .. So this function is always positive But assumption is x>-1 So x>=-1 If x+1<0 OR x<-1, Then x^2+3x+3>0 b^2-4ac = -3 ... No real roots and a>0 .. So this function is always positive But assumption is x<-1 So x<-1 Combining, all values of x satisfy this relation Answer is (a)

shrouded1 · Answer

Ok, reading the solutions above ... I officially believe I just used a sledge hammer to kill an ant

Bunuel · Answer

Good approach.

Though you could stop already at the stage  |x + 1|<|(x+1)^2 +1| . As it's clear that this inequality holds true for all x-es: if  |x + 1|\geq{1}  then clearly  |x + 1|<(x+1)^2  and if  |x + 1|<{1}  then  LHS<1  BUT  RHS=|(x+1)^2 +1|>1 .

shrouded1 · Answer

probably beyond the scope of the GMAT but here goes :

No real roots (discriminant < 0) implies the parabola does not intersect the X-axis

And coeff of x^2 (or 'a') > 0 implies it is upward facing

If you plot it out, you will see this is only possible if the parabola only exists in the upper two quadrants ==> it can never take a negative value

mainhoon · Answer

Interestingly if visualized graphically, y=|x+1| is a V shaped plot, with y= 0 at x= -1 place where it turns around...

Y=|(x+1)^2+1| is a parabola and is clearly above the previous V shape. Check lowest it would fall is 1 at x= -1...

Posted from my mobile device

sujoykrdatta · Answer

If you are comfortable with graphs - sketching rough graphs based on the expression, it is very easy to get the solution to the above: 1.JPG Since the graph of |x2 + 2x + 2| is always above the graph of |x+1|, we have: |x + 1| < |x^2 + 2x + 2| for all x Answer A Alternative approach: |x + 1| < |x^2 + 2x + 2| => |x+1| < |(x+1)^2 + 1| Let |x+1| = p => (x+1)^2 = |(x+1)|^2 = p^2 Note that p is always positive since p = |x+1|; thus p can be either 01 => We have: p < |p^2 + 1| # The RHS is always greater than 1, hence is always true for all fractional 0 < p < 1 # For p > 1: p^2 > p => p^2 + 1 > p > 0 Thus, the above is true for all p Answer A

HoudaSR · Answer

Bunuel   KarishmaB  I need your help with this one please.

When I try to solve this algebraically, I get the following two inequalities, for which there are no solutions: x^2+x+1>0 and x^2+3x+3>0.

My first reflex was to answer this question as D: No solutions, but then when I went and I plugged in numbers I realised that whichever numbers I was picking they were making the inequality true.

I am a bit confused about how to approach these type of absolute value questions that involve quadratics that turn out to have no solutions.

Thank you in advance.

KarishmaB · Answer

How did you get these inequalities: x^2+x+1>0 and x^2+3x+3>0 ? By removing the absolute value signs? Ok.

But I am not sure what you mean by "no solutions." 
They don't have any roots (the discriminant is negative) because their parabolas lie above the X axis. So if you put x^2+x+1 = 0 you won't get any values of x. The value of x^2+x+1 is always positive, for all values of x and that is why all values of x satisfy x^2+x+1>0.

KarishmaB · Answer

If you are looking for the exact set of values that satisfy this condition, best method here would be to try some values. 
Put x = 0. It satisfies. It means the answer is not (C), (D) or (E).

Put x = -2. It satisfies. It means answer is not (B).

By elimination, answer (A)

TheBipedalHorse · Answer

|x + 1| < |x^2 + 2x + 2|

|(x+1)^2 + 1| > |x+1|

|B^2 + 1| > |B| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . where B = x+1

The above condition is always true no matter what value of B you take (0, -1, -10, +10) because - Squaring a number and adding a positive integer to it will always give a value that is greater than the original value

alternatively

the parabola y = x^2 + 1 has no roots (real) and is always above the x axis where as y = |x| actually touches the x axis at (0,0).................................(condition 1)
also
x^2 + 1 = x has no real roots, therefore y= x^2 + 1 and y = x never meet each other at any point.......................................(condition 2)

Combining both conditions, it is fair to say that no matter that, x^2 + 1 will always be greater than x

HoudaSR · Answer

Thank you KarishmaB for your response. Yes you are right, it isn't that x^2+x+1>0 has no solution, it is x^2+x+1=0 that has no solution. Usually, I would solve this kind of question algebraically as following: Abv (x+1) < Abv (x^2+2x+2) Sqrt (x+1)^2 < sqrt (x^2+2x+2)^2 (x+1)^2 < (x^2+2x+2)^2 (x+1)^2 - (x^2+2x+2)^2 < 0 (x^2+3x+3) (x^2+x+1) >0 At this stage, I would usually set the roots on the number line and figure out the solution range for the inequality, but here both x^2+3x+3 and x^2+x+1 have no roots! How should I proceed further? How can I algebraically figure out that this inequality is always positive for any value of x?

KarishmaB · Answer

HoudaSR 
It's a conceptual thing. A quadratic will have no roots in two cases - either when it is completely above X axis or when it is completely below X axis. It doesn't cut the X axis at all. 
So when you see that a quadratic has no roots, just put x = 0. 
Say put x = 0 in (x^2+3x+3), we get y = 3. This means that the value of y will always be positive only and hence (x^2+3x+3) is always positive.

HoudaSR · Answer

Thank you so much  KarishmaB  for your insightful response! I have learnt a lot.

Are there any similar questions you can share with me for practice?

Thanks again!

KarishmaB · Answer

I suggest you to check out my YouTube video on this topic for clarity: https://youtu.be/QOSVZ7JLuH0 A heads up though, it's work in progress and we will be updating it but at least it will give you an idea.

Kinshook · Answer

gmatbull
What is the question ? Do you want to know domain/ range of x ?
If yes, then the solution is provided below.

|x + 1| < |x^2 + 2x + 2|
|x+1| < |(x+1)ˆ2+1| = (x+1)ˆ2 + 1 = xˆ2 + 2x + 2

Case 1: x<-1
-1-x < xˆ2 + 2x + 2
xˆ2 + 3x + 3 > 0
(x+3/2)ˆ2 + (3-9/4) > 0
(x+3/2)ˆ2 + 3/4 > 0
Valid for all values of x.

Case 2: x>=-1
x+1 < xˆ2 + 2x + 2
xˆ2 + x + 1 > 0
(x+.5)ˆ2 + (1-.25) > 0
Valid for all values of x.

x = (-infinity, + infinity)

IMO A

gmatbull

|x + 1| < |x^2 + 2x + 2|

A: (-infinity, +infinity)
B: (-1, +infinity)
C: (-infinity, -1)
D: no solutions
E: (-1, 1)

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

|x + 1| < |x^2 + 2x + 2|

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

|x + 1| < |x^2 + 2x + 2|

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards