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Difficulty: MediumReal Numbers, Number Line, and Absolute Value

If xx is a real number such that ∣2x−5∣=∣x+4∣|2x - 5| = |x + 4|, which of the following could be the value of x2x^2? Select all such values.

  1. 19\frac{1}{9}Answer
  2. B
    13\frac{1}{3}
  3. C
    11
  4. D
    99
  5. 8181Answer

Answer

The possible values of x2x^2 are 19\frac{1}{9} and 8181.
To solve ∣2x−5∣=∣x+4∣|2x - 5| = |x + 4|, set 2x−52x - 5 equal to both x+4x + 4 and −(x+4)-(x + 4). Solving 2x−5=x+42x - 5 = x + 4 gives x=9x = 9, which leads to x2=81x^2 = 81. Solving 2x−5=−x−42x - 5 = -x - 4 gives 3x=13x = 1, so x=13x = \frac{1}{3}, which leads to x2=19x^2 = \frac{1}{9}. Therefore, the options representing 19\frac{1}{9} and 8181 are correct.

Step-by-Step Solution

1
Set up equations to remove the absolute value bars
Two linear equations: 2x−5=x+42x - 5 = x + 4 or 2x−5=−(x+4)2x - 5 = -(x + 4).
For real numbers AA and BB, ∣A∣=∣B∣|A| = |B| implies A=BA = B or A=−BA = -B.
2
Solve the first equation 2x−5=x+42x - 5 = x + 4
x=9x = 9.
Subtract xx and add 55 to both sides.
3
Solve the second equation 2x−5=−x−42x - 5 = -x - 4
3x=1  ⟹  x=133x = 1 \implies x = \frac{1}{3}.
Add xx and add 55 to both sides.
4
Compute x2x^2 for each possible value of xx
If x=9x = 9, then x2=81x^2 = 81. If x=13x = \frac{1}{3}, then x2=19x^2 = \frac{1}{9}.
The question asks for the values of x2x^2, not xx.

Key Concept

Solving equations with absolute values on both sides requires considering positive and negative case equivalences.
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Topics

Real Numbers, Number Line, and Absolute Value

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