If x^2 - |x + 2 | + x 0, then which of the following range(s) gives

Question

Competition Mode Question If x^2 - |x + 2 | + x > 0 , then which of the following range(s) gives all possible values of x? A. x < -\sqrt{2} B. -\sqrt{2} < x < \sqrt{2} C. x > \sqrt{2} D. 0 < x < \sqrt{2} E. x < -\sqrt{2} or x > \sqrt{2} Are You Up For the Challenge: 700 Level Questions

exc4libur · Accepted Answer

x^2-|x+2|+x>0 |x+2|<0, when x<-2: x^2-(-x-2)+x>0, x^2+2x+2>0, test: x=0: 2>0, x=-1: 1>0 for any x, the inequality is true. |x+2|>0, when x>-2 x^2-(x+2)+x>0, x^2-2>0, x>√2, x<-√2 ans (E)

GMATinsight · Answer

Let's substitute x = 1 to check whether it's an acceptable value or not

at x = 1, x^2−|x+2|+x = 1-3+1 which is NOT greater than 0 hence all option with vlalue 1 are OUT
B and D are out

at x = -2, x^2−|x+2|+x = 4-0-2 > 0 which is greater than 0 hence all option with value x=-2 are OUT
C is out

at x = +2, x^2−|x+2|+x = 4-4+2 > 0 which is greater than 0 hence 
A is out

Answer: Option E

freedom128 · Answer

Surely, we can solve this analytically, but the smartest way would be picking a good number to eliminate wrong choices quickly.

x^2 + x > |x + 2|

x=1 --> 1 + 1 > |1+2| (No!)...  eliminate (B), (D) 
x=-2 --> 4 - 2 > |-2+2| (Yes)...  eliminate (C)

Only (A) and (E) remain. Pick a number (say, x=2) that isn't shared by both choices. 
x=2 --> 4 + 2 > |2 + 2| (Yes)...  eliminate (A)

FINAL ANSWER IS (E)
x < -√2 or x > √2

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gurmukh · Answer

X=1 does not satisfy so option B and D are out.
X=-1.5 and 1.5 satisfy the equation so option E is the answer.

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LeenaPG · Answer

Solution:

x2−|x+2|+x>0

assume x+2 is positive.

x^2 - x-2 +x>0

x^2> 2
x > √2 or x< -√2

now assign values to the original equation and only x> √2 appear to satisfy all values.

So Answer = C

MayankSingh · Answer

IMO E x^2−|x+2|+x>0 x< -2 x^2+2x+2>0 (x+1)^2+1>0 X = {-∞ , +∞} x>-2 x^2-2>0 x>√2 & x < -√2 range x = -2< x < √2 & x> √2 Only E satisfies this.

tyson1990 · Answer

Options given helps to find range in this problem.
\sqrt{2} = 1.4 approx

Lets take x = -3
F(x) = 9 -|-3+2|-3 = 5 holds thru.

Hence, range must contain x=-3. B, C & D options are eliminated.

Similarly, take x=3
F(x) = 9-|3+2|+3 = 7 holds true. Option E is correct.

For cross-checking, F(x) can be determined taking value of 'x' as +1 & -1.

verydisco · Answer

IMO, (E).

Here's how I solved it. We have two possible equations:

x^2- +x > 0 , and  x^2-(x+2)+x > 0

Through the latter equation, we get the solution that x >  \sqrt{2}  or x <  -\sqrt{2} .

But are either of these cases valid for the former equation? Plug in any values within these ranges to find your answer. For simplicity, we can take the values +2 and -2, both of which are accepted values.

On plugging in the values, we get that the expression  x^2- +x > 0  is positive, i.e., greater than 0. Thus, values of x in this range hold true, and hence the answer is (E).

debjit1990 · Answer

Ans: E
x2−∣x+2∣+x>0

If x≤−2

x2+x+2+x

=x2+2x+2
x2+2x+2>0 (always)

If x>−2

x2−x−2+x>0

x2−2>0
x<-root 2 or x>root 2

NitishJain · Answer

E ?

It can be:
Option 1: x^2- (x+2) +x >0
==> x^2 -2 >0
==> (x+ √2)(x-√2)>0
x>√2 or x< -√2

Option 2: x^2 + x + 2 +x >0
x^2 +2x +2 >0
(x+1)^2 + 1 >0 ==> Always true

Hence E.

sambitspm · Answer

See the attachment.

aviddd · Answer

I substituted values and narrowed down the options.

Is there a more elegant way,  Bunuel ? Can I square the equation and arrive at possible roots?

Kinshook · Answer

If x^2 - |x + 2 | + x > 0 , then which of the following range(s) gives all possible values of x? Case 1: x < - 2; x+2 < 0 x^2 - (-x-2) + x > 0 x^2 + 2x + 2 > 0 (x+1)^2 + 1 > 0 True for all values of x. Case 2: x >= -2; x+2 >=0 x^2 - (x+2) + x > 0 x^2 - 2> 0 (x-\sqrt{2})(x+\sqrt{2}) > 0 x > \sqrt{2} or x < -\sqrt{2} Since x < -\sqrt{2} caters to all values of x < -2 IMO E

If x^2 - |x + 2 | + x > 0, then which of the following range(s) gives

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