Which of the following has exactly one solution ?

Question

A.  x|x|=2^x

B.  x+|x|=2^x

C.  2|x|=2^x

D.  2|x|=2x-1

E.  |x+2|=-x

GMATinsight · Answer

A.  x|x|=2^x  True for x = 2 and 4 ELIMINATED

B.  x+|x|=2^x  True for x = 1 and 2

C.  2|x|=2^x  True for x = 2 and 1 Eliminated

D.  2|x|=2x-1  NO SOLUTION (This can NOT be true for any positive value of x and negative value in turn gives us positive value 1/4 NOT Acceptable)

E.  |x+2|=-x  x=-1

Answer: Option E

skhemani · Answer

How is x=1 not a solution for C?

Bunuel · Answer

It is. Both x = 1 and x = 2 satisfy 2|x| = 2^x.

kris19 · Answer

C. 2|x| = 2^x  
case (i) x is -ve, then
  |x| = x --> 2*x = \frac{1}{2^{x}}  , which is not possible for any x
case (ii) x is +ve, then
  |x| = x --> 2*x = 2^x  
above equation holds good for x = 1 (2*1 = 2^1) and x = 2 (2*2 = 2^2), so option C has two solutions for x, so not correct

E. |x+2| = -x
case (i) x is -ve, then take 
  x = -1 --> |-1+2| = -(-1)  
  1 = 1  , so x takes -1 ( one solution)
  x= -2 --> |-2+2| = -(-2)  
  0 = 2  , so x can't take -2
  x = -3 --> |-3+2| = -(-3)   so x can't take -3, and so for other negative numbers

case (ii) x is +ve, then
  |x+2| = -x   becomes x+2 = -x --> x = -1
so, x can take -1, from case (i) & (ii), x has only one solution (-1)

so option E is correct

ScottTargetTestPrep · Answer

Solution:

We can solve the equations starting with D since D is the easiest equation of the ones given.

D. 2|x| = 2x - 1

If x is positive, we have 2x = 2x - 1, which yields no solution.

If x is negative, we have -2x = 2x - 1. Solving it, we get:

-4x = -1

x = -1/-4 = 1/4

Since we assumed x was negative, the positive value x = 1/4 may not satisfy the equation. Indeed, if we substitute x = 1/4, we get:

2|1/4| ≟ 2(1/4) - 1

2 * 1/4 ≟ 1/2 - 1

1/2 ≟ -1/2

Since 1/2 is not equal to -1/2, x = 1/4 is not a solution for 2|x| = 2x - 1. Thus, this equation has no solutions.

Next, let’s solve the equation in answer choice E since it is similarly easy to solve.

E. |x + 2| = -x

If (x + 2) is positive, then:

x + 2 = -x

2x = -2

x = -1

We can verify that x = -1 is indeed a solution for |x + 2| = -x.

If (x + 2) is negative, then:

-x - 2 = -x

x + 2 = x

No value of x will satisfy the equation above. Hence, the equation |x + 2| = -x has only one solution.

Answer: E

Sevil92 · Answer

Is this an official question? The rule is that absolute value cannot give us a negative result but here we take it as the correct answer. How can |1| be equal to -1?

Bunuel · Answer

|x + 2| = −x has one solution x = -1:

|-1 + 2| = −(-1);

|1| = 1;

1 = 1.

abvoluptatum · Answer

we do have to check for the possible solutions right , |x+2|=-x has 1 solution

egmat · Answer

Hi abvoluptatum,

Yes, you've got it exactly right, and this is a habit worth locking in permanently:  whenever you break an absolute-value equation into cases, every candidate you find must be checked against the condition of the case it came from.

Here's why. When you drop the bars, you assume something about the sign inside:

-  Case 1 (inside ≥ 0):  x + 2 = -x → 2x = -2 →  x = -1 . This case assumes x + 2 ≥ 0, i.e. x ≥ -2. Since -1 ≥ -2, the candidate  survives . ✓
-  Case 2 (inside < 0):  -(x + 2) = -x → -2 = 0, which is impossible. No candidate at all.

So |x + 2| = -x gives  exactly one  valid solution,  x = -1 . That's why E is the answer.

Why the check is not optional

Look at choice  D  in the thread - that's the clearest proof of your point. Solving the negative case gave  x = 1/4 , but that case assumed x < 0. Since  1/4  is positive, it  violates its own condition , so it gets thrown out and D ends up with zero solutions. If you had skipped the check, you'd have wrongly counted a solution there.

The rule: a candidate is real only if it satisfies  both  the equation  and  the sign assumption of the case that produced it.

Quick one to cement the habit

Try solving |x| = x - 3:

- Case x ≥ 0: x = x - 3 → 0 = -3, impossible.
- Case x < 0: -x = x - 3 →  x = 3/2 . But  3/2  is  not  negative, so discard it.

Zero solutions - you only know that because you checked. Same discipline every time.

Answer: E

abvoluptatum · Answer

|x + 2| = −x has one solution x = -1:

|-1 + 2| = −(-1);

|1| = 1;

1 = 1.
easy solution

Which of the following has exactly one solution ?

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