If x is an integer, how many possible values

Question

If x is an integer, how many possible values of x exist for  x^2+5|x|+6=0 ?

A. 4
B. 2
C. 3
D. 1
E. 0

generis · Accepted Answer

One method: check the signs of the terms.

The squared term is positive (or nonnegative if x=0).

The term whose product is (+5) * (some nonnegative or positive number because absolute value nonnegative or positive), is positive or nonnegative (if x = 0).

The constant is positive.

You cannot sum three positive numbers, or two zeros (if x=0) and a positive, and get zero. No values will work.

Answer E

Another way: If "check the signs method" doesn't occur to you, try factoring the quadratic as if there were no absolute value bars around the x in the second term.

x^2+5x+6=0 
(x + 3)(x + 2)

So x = -3 or -2

Check the values. Plug -3 and -2 into original equation.

Neither works:
-3:  x^2+5|x|+6=0 
-3: 9 + (5*3) + 6 = 
-3: 9 + 15 + 6 does not equal zero.

From the pattern of positive terms that results when plugging in a negative number (squared term is positive, absolute value term is positive, constant is positive), -2 will not work either.

From the  (+)  +  (+)  +  (+)  pattern: you cannot get to 0 with three positive numbers. No values will work.

Answer E

RooIgle · Answer

Humbly I think that it´s much easier to think in the way that  x^2+5|x|+6=0  is gonna have to suffer a check for strenuous roots afterwards; thus, it´s not necessary to solve or to do absolutely anything with the equation, and it´s just enough that your bubble lights up with the strenuous nuance and say to youself:

"okay, the absolute-value term is always positive, and so is the quadratic term, so there is simply no way that a solution exists that can make the left side side equal to the right".

Conclusion: 0 solutions.

-

BeingHan · Answer

E
When x <0
The equation will become x^2 -5x+6=0. The two roots are 3 and 2 going against the range of x .
If we take x>0, x^2 +5x+6=0 and two roots will be -3, and -2. 
This too goes against the range of x. So zero solutions.
Another way of looking at this will be to observe that all the three terms of the equation are each greater than zero. So the equation will never be zero and hence no or zero solution.

Sent from my Moto G (5) Plus using  GMAT Club Forum mobile app

aghosh54 · Answer

By range of x what are you suggesting that? X is an integer.

Sent from my Redmi 4 using  GMAT Club Forum mobile app

DHAR · Answer

genxer123 
I like your "One method: check the signs of the terms." very much.+1 kudos given. So, it will be true for any quadratic equation in the form ax^2+b|x|+c=0 as long a,b and c are positive?

generis · Answer

Yes -- but also nonnegative (x = 0 especially, or a AND b = 0) . . . as long as c is positive. In other words, if your first two terms result in the nonnegative 0, check c. Positive? No solution. You can't add a positive number to zero and get zero. Sometimes you will see posters insist that there is no such thing as the absolute value of 0. Because absolute value is a distance (from point of origin, often 0) there is such a thing: |0| is 0. Zero is 0 distance away from zero. Come to think of it, though the coefficients wouldn't work in the factor method part, I'm going to amend the first part my answer to include nonnegative! Thanks. :-)

cbh · Answer

a tricky one

x^2+5|x|+6=0 can be written as (|x| + 3) (|x| + 2) = 0 as |x|^2 always = x^2 then no solution for 0 as there is no negative value for |x|

hellosanthosh2k2 · Answer

x^2 = |x||x |
given  x^2 + 5|x| + 6 = 0

Method1:
Let a = |x|
 a^2 + 5a + 6 = 0
roots: a = -2, -3
|x| = -2, |x| = -3
modulus can never be negative, so no solution exists

Method 2:
  x^2 + 5|x| + 6 = 0  => to hold this,  x^2 + 5|x| = -6 , but modulus can never be negative as the minimum value is 0, so solution exists

generis · Answer

RooIgle 
Surely, a few travelers will linger, then halt, when a collection of oddly graceful words unfurls. 
Surely, those stock-still few will wonder: of what stuff is this "strenuous nuance" made? Is it balletic, like a thought? Or sensible, like a nod? 
Surely, as they wander away, they will dare to see beyond content (zero does equal zero, after all).
Surely, they will know enough to murmur, "Who gets  that close  to the oxymoronic but does not collide with it?" 
Brave soul. Perilous territory. 
No. Not "surely." But this time, yes. At least one. Nicely done.

spetznaz · Answer

+1 for option E. The equation can be re-written as |x|^2+5|x|+6=0. Solving we get |x|=-3,-2. This clearly not possible. Hence no value of x will yield the required value. Hence option E.

BrentGMATPrepNow · Answer

Take: x² + 5|x| + 6 = 0
Subtract 6 from both sides to get:  x²  +  5|x|  = -6

KEY CONCEPT: x² ≥ 0 and |x| ≥ 0 for  all values of x 
In other words,  x² will always be  greater than or equal to 0  
And |x| will always be  greater than or equal to 0 , which means  5|x| will always be  greater than or equal to 0

So, we can take our equation,  x²  +  5|x|  = -6, and rewrite it as follows: 
( some number that's greater than or equal to zero ) + ( some number that's greater than or equal to zero ) = -6
As we can see, it's impossible for the left side of the equation to equal a NEGATIVE value. 
As such, there can be no solution.

Answer: E

Cheers,
Brent

fskilnik · Answer

?\,\,\,:\,\,\,\# \,\,\,{\mathop{\rm int}} \,\,\,{\rm{roots}}\,\,\,{\rm{for}}\,\,\,\,{x^2} + 5\left| x \right| + 6 = 0

\left. \matrix{
 {x^2} \ge 0 \hfill \cr 
 \left| x \right|\,\, \ge 0\,\, \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,\,:\,\,\,{x^2} + 5\left| x \right|\, + 6\,\,\, \ge 6\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

SchruteDwight · Answer

squaring both sides produces

0=x^4-13x^2+36 \implies 0=(x^2-4)(x^2-9)

However,  x_1=3, x_2=-3, x_3=2, x_4=-2  are all extraneous solutions

Fdambro294 · Answer

Trap question.......trying to trick us GMAT.

Taking each expression at a time:

(X)^2 will always be greater than or equal to 0. Thus, the minimum Value we can have for (X)^2 is = 0

+5 *

For this expression, the output of the Modulus will always be greater than or equal to zero. Since we are multiplying a +Pos Value with a NON Negative Value, the Minimum Value we can get from this expression is 0

MIN occurs when X = 0

0 + 0 + 6 can NEVER equal 0

-E-

There are no solutions to the equation

Another way to see this more easily is to move 6 to the right side of the equation my subtracting each side by 6

(X)^2 + 5  = -6

The left hand side of the equation can NEVER equal a (-)Negative Value.

Posted from my mobile device

Chejain · Answer

Can this question be also solved using b^2-4ac rules to find number of roots of a quadratic equation? pls help me understand how.

ueh55406 · Answer

Aah Nice one, I almost went with A, but felt weird why is there 0 as an option.

A red flag for me was, how can addition of all +ve's = 0 ?

x^2+5|x|+6=0 ?  break this down

x^2  = no matter what x is +ve or -ve, x^2 will always be +ve.

+ 5|x| , No matter what x is (+ ve or -Ve) |x| will always be +ve and |x| * 5 will be positive as well.

+6 >> it's already +ve.

there's no value of x , +ve, -ve, 0 that could give the result as 0.

CrackverbalGMAT · Answer

Solution:

A mod function will always return a non negative value.

So 5|x| will always be a non negative value and x^2 is always a non negative value irrespective of the sign of x.

Thus x2+5|x|+6 should be a non negative value too.

However it is given x2+5|x|+6 = 0 = >x2+5|x| = -6 and this is NOT possible. Thus there are no solutions. (option e)

BrushMyQuant · Answer

We need to find how many possible values of x exist for  x^2+5|x|+6=0 ?

As we have |x| in the equation so we will have two cases 
 
   -   Case 1:  x ≥ 0
 
=> |x| = x
=>  x^2 + 5x + 6=0 
 =>  x^2 + 2x + 3x + 6=0  
 => x * (x + 2) + 3*(x + 2) = 0
=> (x + 2) * (x + 3) = 0
=> x = -2, -3

But our condition was x ≥ 0 and -2, -3 are NOT ≥ 0  
 =>  NO SOLUTION   
  -   Case 2:  x ≤ 0
 
 => |x| = -x
=>  x^2 - 5x + 6=0 
 =>  x^2 - 2x - 3x + 6=0  
 => x * (x - 2) - 3*(x - 2) = 0
=> (x - 2) * (x - 3) = 0
=> x = 2, 3

But our condition was x ≤ 0 and 2, 3 are NOT ≤ 0 
 =>  NO SOLUTION

So,  Answer will be E 
Hope it helps!

To learn how to solve absolute value problems, watch the following video

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

If x is an integer, how many possible values

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

As we have \|x\| in the equation so we will have two cases
-Case 1: x ≥ 0 => \|x\| = x => \(x^2 + 5x + 6=0\) => \(x^2 + 2x + 3x + 6=0\) => x * (x + 2) + 3(x + 2) = 0 => (x + 2) (x + 3) = 0 => x = -2, -3 But our condition was x ≥ 0 and -2, -3 are NOT ≥ 0 => NO SOLUTION	-Case 2: x ≤ 0 => \|x\| = -x => \(x^2 - 5x + 6=0\) => \(x^2 - 2x - 3x + 6=0\) => x * (x - 2) - 3(x - 2) = 0 => (x - 2) (x - 3) = 0 => x = 2, 3 But our condition was x ≤ 0 and 2, 3 are NOT ≤ 0 => NO SOLUTION

GMAT Problem Solving (PS) Questions

If x is an integer, how many possible values

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