Soru

Zorluk: ZorExponents, Radicals, and Algebraic Expressions

If xx is a real number satisfying the equation 15−2x=2x−3\sqrt{15 - 2x} = 2x - 3, what is the sum of all real values of xx that satisfy this equation?

  1. 33Cevap
  2. B
    52\frac{5}{2}
  3. C
    −12-\frac{1}{2}
  4. D
    72\frac{7}{2}
  5. E
    11

Cevap

The sum of all real values of xx satisfying the equation is 33.
Squaring both sides of 15−2x=2x−3\sqrt{15 - 2x} = 2x - 3 gives 15−2x=4x2−12x+915 - 2x = 4x^2 - 12x + 9, which simplifies to 2x2−5x−3=02x^2 - 5x - 3 = 0. Factoring yields candidate roots x=3x = 3 and x=−1/2x = -1/2. Substituting x=3x = 3 into the original equation gives 9=3\sqrt{9} = 3, which is true. Substituting x=−1/2x = -1/2 yields 16=−4\sqrt{16} = -4, which is false because principal radicals cannot evaluate to negative values. Therefore, x=3x = 3 is the sole valid solution, making the sum equal to 33.

Adım Adım Çözüm

1
Isolate the radical and state domain constraints.
The principal square root 15−2x\sqrt{15 - 2x} must be non-negative, requiring 15−2x≥0  ⟹  x≤7.515 - 2x \ge 0 \implies x \le 7.5, and 2x−3≥0  ⟹  x≥1.52x - 3 \ge 0 \implies x \ge 1.5.
Radical expressions produce non-negative principal square roots.
2
Square both sides of the equation.
15−2x=(2x−3)2  ⟹  15−2x=4x2−12x+915 - 2x = (2x - 3)^2 \implies 15 - 2x = 4x^2 - 12x + 9.
Eliminate the radical to form a polynomial equation.
3
Rearrange into standard quadratic form and solve for xx.
4x2−10x−6=0  ⟹  2x2−5x−3=0  ⟹  (2x+1)(x−3)=04x^2 - 10x - 6 = 0 \implies 2x^2 - 5x - 3 = 0 \implies (2x + 1)(x - 3) = 0, giving candidate solutions x=3x = 3 and x=−1/2x = -1/2.
Solve the quadratic equation using factoring.
4
Test candidate solutions in the original equation to eliminate extraneous roots.
For x=3x = 3: 15−2(3)=9=3\sqrt{15 - 2(3)} = \sqrt{9} = 3 and 2(3)−3=32(3) - 3 = 3 (Valid). For x=−1/2x = -1/2: 15−2(−1/2)=16=4\sqrt{15 - 2(-1/2)} = \sqrt{16} = 4, but 2(−1/2)−3=−42(-1/2) - 3 = -4 (Extraneous).
Squaring an equation can introduce extraneous roots.

Anahtar Kavram

Solving Radical Equations and Identifying Extraneous Solutions
Bu soruyu puanla