Question

Difficulty: MediumQuadratic Equations and Polynomial Factoring

What is the sum of all distinct real values of xx that satisfy the equation x(x25x+6)=2(x25x+6)x(x^2 - 5x + 6) = 2(x^2 - 5x + 6)?

  1. A
    2
  2. B
    3
  3. 5Answer
  4. D
    7
  5. E
    9

Answer

The sum of all distinct real solutions is 5.
To solve the equation x(x25x+6)=2(x25x+6)x(x^2 - 5x + 6) = 2(x^2 - 5x + 6), move all terms to the left side to get x(x25x+6)2(x25x+6)=0x(x^2 - 5x + 6) - 2(x^2 - 5x + 6) = 0. Factoring out (x25x+6)(x^2 - 5x + 6) gives (x2)(x25x+6)=0(x - 2)(x^2 - 5x + 6) = 0. Factoring the quadratic part yields (x2)(x2)(x3)=0(x - 2)(x - 2)(x - 3) = 0, or (x2)2(x3)=0(x - 2)^2(x - 3) = 0. The real roots are x=2x = 2 and x=3x = 3. Taking the sum of these distinct real roots gives 2+3=52 + 3 = 5.

Step-by-Step Solution

1
Move all terms to one side of the equation to avoid dividing by a variable expression.
x(x25x+6)2(x25x+6)=0x(x^2 - 5x + 6) - 2(x^2 - 5x + 6) = 0
Dividing by (x25x+6)(x^2 - 5x + 6) directly would lose valid roots where x25x+6=0x^2 - 5x + 6 = 0.
2
Factor out the common quadratic expression (x25x+6)(x^2 - 5x + 6).
(x2)(x25x+6)=0(x - 2)(x^2 - 5x + 6) = 0
Applying the distributive property isolates the linear factor (x2)(x - 2).
3
Factor the quadratic term x25x+6x^2 - 5x + 6 into linear factors.
(x2)(x2)(x3)=0    (x2)2(x3)=0(x - 2)(x - 2)(x - 3) = 0 \implies (x - 2)^2(x - 3) = 0
The quadratic factors into (x2)(x3)(x - 2)(x - 3) because (2)+(3)=5(-2) + (-3) = -5 and (2)×(3)=6(-2) \times (-3) = 6.
4
Identify all distinct real solutions and compute their sum.
Real solutions are x=2x = 2 and x=3x = 3. Distinct sum: 2+3=52 + 3 = 5.
The root x=2x = 2 has multiplicity 2, but only distinct values must be summed.

Key Concept

Polynomial Factoring and Variable Cancellation Rules
Estimated Time:1m 30s
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