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Zorluk: OrtaLinear Equations in One and Two Variables
If xx is a real number that satisfies the absolute value equation ∣2x−7∣=3x−11|2x - 7| = 3x - 11 what is the value of x2−3xx^2 - 3x?
  1. A
    2.162.16
  2. 44Cevap
  3. C
    1616
  4. D
    2828
  5. E
    270270

Cevap

The value of x2−3xx^2 - 3x is 44.
Solving the absolute value equation ∣2x−7∣=3x−11|2x - 7| = 3x - 11 yields two potential roots: x=4x = 4 and x=3.6x = 3.6. Checking both in the original equation shows that x=3.6x = 3.6 makes the right-hand side negative (−0.2-0.2), making it an extraneous solution. The only valid solution is x=4x = 4. Substituting x=4x = 4 into x2−3xx^2 - 3x yields 42−3(4)=16−12=44^2 - 3(4) = 16 - 12 = 4.

Adım Adım Çözüm

1
Set up equations based on the definition of absolute value
Case 1: 2x−7=3x−112x - 7 = 3x - 11; Case 2: 2x−7=−(3x−11)2x - 7 = -(3x - 11)
An absolute value equation ∣A∣=B|A| = B splits into A=BA = B or A=−BA = -B, with the requirement that B≥0B \ge 0.
2
Solve Case 1: 2x−7=3x−112x - 7 = 3x - 11
3x−2x=11−7  ⟹  x=43x - 2x = 11 - 7 \implies x = 4
Isolate the variable xx by algebraic rearrangement.
3
Solve Case 2: 2x−7=−3x+112x - 7 = -3x + 11
5x=18  ⟹  x=185=3.65x = 18 \implies x = \frac{18}{5} = 3.6
Isolate xx for the negative case.
4
Check both potential solutions in the original equation ∣2x−7∣=3x−11|2x - 7| = 3x - 11
For x=4x = 4: ∣2(4)−7∣=∣1∣=1|2(4) - 7| = |1| = 1 and 3(4)−11=13(4) - 11 = 1 (Valid).
For x=3.6x = 3.6: ∣2(3.6)−7∣=∣0.2∣=0.2|2(3.6) - 7| = |0.2| = 0.2 but 3(3.6)−11=−0.23(3.6) - 11 = -0.2 (Extraneous).
The right side 3x−113x - 11 must be non-negative. Since 3(3.6)−11=−0.2<03(3.6) - 11 = -0.2 < 0, x=3.6x = 3.6 is extraneous.
5
Evaluate the target expression x2−3xx^2 - 3x using the valid root x=4x = 4
42−3(4)=16−12=44^2 - 3(4) = 16 - 12 = 4
Substitute the single valid root into the requested expression.

Anahtar Kavram

Absolute Value Linear Equations and Extraneous Solution Checking
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