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Zorluk: OrtaLinear Equations in One and Two Variables

If xx is a real number that satisfies the equation 2x1=3x+11|2x - 1| = 3x + 11, what is the value of x2+2xx^2 + 2x?

  1. 00Cevap
  2. B
    33
  3. C
    88
  4. D
    3535
  5. E
    120120

Cevap

The value of x2+2xx^2 + 2x is 00.
Solving the equation 2x1=3x+11|2x - 1| = 3x + 11 produces two algebraic candidates: x=12x = -12 and x=2x = -2. Substituting x=12x = -12 into the right side yields 3(12)+11=253(-12) + 11 = -25. Since absolute value expressions cannot be negative, x=12x = -12 is an extraneous solution. Substituting x=2x = -2 yields 5=5|-5| = 5, which is valid. Evaluating x2+2xx^2 + 2x at x=2x = -2 gives (2)2+2(2)=0(-2)^2 + 2(-2) = 0.

Adım Adım Çözüm

1
Set up the two linear cases for the absolute value equation 2x1=3x+11|2x - 1| = 3x + 11.
Case 1: 2x1=3x+112x - 1 = 3x + 11; Case 2: 2x1=(3x+11)2x - 1 = -(3x + 11).
By definition, u=c|u| = c implies u=cu = c or u=cu = -c (provided c0c \geq 0).
2
Solve each linear equation for candidate values of xx.
From Case 1: x=12x = -12. From Case 2: 2x1=3x11    5x=10    x=22x - 1 = -3x - 11 \implies 5x = -10 \implies x = -2.
Isolate xx algebraically in both equations.
3
Check candidate solutions in the original equation to eliminate extraneous roots.
For x=12x = -12: 2(12)1=25=25|2(-12) - 1| = |-25| = 25, but 3(12)+11=25253(-12) + 11 = -25 \neq 25 (extraneous). For x=2x = -2: 2(2)1=5=5|2(-2) - 1| = |-5| = 5, and 3(2)+11=53(-2) + 11 = 5 (valid solution).
An absolute value cannot equal a negative number; substituting back is mandatory.
4
Evaluate the target expression x2+2xx^2 + 2x using the valid root x=2x = -2.
(2)2+2(2)=44=0(-2)^2 + 2(-2) = 4 - 4 = 0.
Substitute the verified real solution into the given expression.

Anahtar Kavram

Solving Absolute Value Linear Equations and Checking for Extraneous Solutions
Tahmini Süre:1m 30s
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