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Zorluk: OrtaInequalities and Absolute Value Equations

If xx is a real number satisfying the equation ∣2x−3∣=3x+1|2x - 3| = 3x + 1, what is the value of x2+2xx^2 + 2x?

  1. A
    −1625-\frac{16}{25}
  2. B
    825\frac{8}{25}
  3. 2425\frac{24}{25}Cevap
  4. D
    88
  5. E
    5625\frac{56}{25}

Cevap

The value of x2+2xx^2 + 2x is 2425\frac{24}{25}.
The equation ∣2x−3∣=3x+1|2x - 3| = 3x + 1 requires 3x+1≥03x + 1 \ge 0, or x≥−13x \ge -\frac{1}{3}. Setting 2x−3=3x+12x - 3 = 3x + 1 gives x=−4x = -4, which violates x≥−13x \ge -\frac{1}{3} and is extraneous. Setting −(2x−3)=3x+1-(2x - 3) = 3x + 1 yields −2x+3=3x+1-2x + 3 = 3x + 1, which simplifies to 5x=25x = 2, or x=25x = \frac{2}{5}. Since 25≥−13\frac{2}{5} \ge -\frac{1}{3}, this is the unique valid solution. Substituting x=25x = \frac{2}{5} into x2+2xx^2 + 2x gives (25)2+2(25)=425+2025=2425\left(\frac{2}{5}\right)^2 + 2\left(\frac{2}{5}\right) = \frac{4}{25} + \frac{20}{25} = \frac{24}{25}.

Adım Adım Çözüm

1
Set up the conditions for the absolute value equation ∣2x−3∣=3x+1|2x - 3| = 3x + 1.
Since the absolute value ∣2x−3∣|2x - 3| must be non-negative, the right-hand side must satisfy 3x+1≥03x + 1 \ge 0, which means x≥−13x \ge -\frac{1}{3}.
An absolute value expression cannot equal a negative number.
2
Solve Case 1: 2x−3≥02x - 3 \ge 0 (so x≥32x \ge \frac{3}{2}).
2x−3=3x+1  ⟹  x=−42x - 3 = 3x + 1 \implies x = -4.
Remove the absolute value bars directly when the inside expression is non-negative.
3
Check Case 1 candidate solution against constraints.
x=−4x = -4 is extraneous because −4<32-4 < \frac{3}{2} and 3(−4)+1=−11<03(-4) + 1 = -11 < 0. Reject x=−4x = -4.
Plugging x=−4x = -4 into the original equation yields ∣−11∣=−11|-11| = -11, which is false.
4
Solve Case 2: 2x−3<02x - 3 < 0 (so x<32x < \frac{3}{2}).
−(2x−3)=3x+1  ⟹  −2x+3=3x+1  ⟹  5x=2  ⟹  x=25-(2x - 3) = 3x + 1 \implies -2x + 3 = 3x + 1 \implies 5x = 2 \implies x = \frac{2}{5}.
Negate the expression inside the absolute value when it is negative.
5
Check Case 2 candidate solution and evaluate the target expression.
Since x=25≥−13x = \frac{2}{5} \ge -\frac{1}{3}, it is valid. Then x2+2x=(25)2+2(25)=425+45=2425x^2 + 2x = \left(\frac{2}{5}\right)^2 + 2\left(\frac{2}{5}\right) = \frac{4}{25} + \frac{4}{5} = \frac{24}{25}.
Substitute the single valid real solution into x2+2xx^2 + 2x.

Anahtar Kavram

Absolute Value Equations and Extraneous Solution Elimination
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Konular

Inequalities and Absolute Value Equations

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