If x^2 + 6|x| - 40 = 0, how many real values does x have?

Question

A. 2
B. 4
C. 1
D. 6
E. 8

sujoykrdatta · Answer

If x≥0, |x| = x => x^2 + 6x - 40 = 0 x = -10 or 4 But x ≥ 0 => x = 4 If x<0, |x| = -x => x^2 - 6x - 40 = 0 x = 10 or -4 But x < 0 => x = -4 Ans A Alt: x^2 = |x|^2 Replacing: |x|^2 + 6|x| - 40 = 0 => (|x|+10)(|x|-4)=0 => |x| = 4 or -10 Since |x| cannot be negative, we have |x| = 4 => x = ±4 Ans A

BrushMyQuant · Answer

We need to find how many real values does x have if  x^2 + 6|x| - 40 = 0

We have two cases as we have |x|  
 
   -   Case 1:  x ≥ 0
 
=> |x| = x
 =>  x^2 + 6x - 40 = 0 
=>  x^2 + 10x - 4x - 40 = 0  
=> x * (x + 10) - 4 * (x + 10) = 0
=> (x - 4) * (x + 10) = 0
=> x = -10, 4

But the condition was x ≥ 0
  => x = 4 is a  SOLUTION
    
  -   Case 2:  x ≤ 0
 
=> |x| = -x
 =>  x^2 - 6x - 40 = 0 
=>  x^2 - 10x + 4x - 40 = 0  
=> x * (x - 10) + 4 * (x - 10) = 0
=> (x - 10) * (x + 4) = 0
=> x = 10, -4
 
But the condition was x ≤ 0
  => x = -4 is a  SOLUTION

So, there are two solutions for x = -4, 4

So,  Answer will be A 
Hope it helps!

Watch the following video to MASTER Absolute Values

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

empirekid1301 · Answer

I might be completely wrong here but cant we use the discriminant method , for case 1 we take x as +ve then B^2 -4ac = 36 - (4*1*-40) or 36 + 120 which we know is +ve when D is +ve 2 real roots ,

for case 2 we take x is -ve so B^2 - 4ac becomes 36 (we are squaring so no chnage in sign) - 4*-1(cause we assume x is -ve) *-40 or 36 -(120) which we know is -ve and when D is -ve there are no real roots

So X is +ve: 2 and X is -ve : 0 , therefore total 2 roots

Some one Please confirm if this is viable method as it is much faster than the others mentioned here but i dont wanna use a wrong method for some other question and mess up

If x^2 + 6|x| - 40 = 0, how many real values does x have?

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We have two cases as we have \|x\|
-Case 1: x ≥ 0 => \|x\| = x => \(x^2 + 6x - 40 = 0\) => \(x^2 + 10x - 4x - 40 = 0\) => x * (x + 10) - 4 * (x + 10) = 0 => (x - 4) * (x + 10) = 0 => x = -10, 4 But the condition was x ≥ 0 => x = 4 is a SOLUTION	-Case 2: x ≤ 0 => \|x\| = -x => \(x^2 - 6x - 40 = 0\) => \(x^2 - 10x + 4x - 40 = 0\) => x * (x - 10) + 4 * (x - 10) = 0 => (x - 10) * (x + 4) = 0 => x = 10, -4 But the condition was x ≤ 0 => x = -4 is a SOLUTION

GMAT Problem Solving (PS) Questions

If x^2 + 6|x| - 40 = 0, how many real values does x have?

Prep Toolkit

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More than Quant/Verbal Abilities: Speed vs. Accuracy

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