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Zorluk: Çok zorQuadratic Equations and Polynomial Factoring

Let SS be the set of all real solutions to the polynomial equation (x2−5x+6)(x2−7x+12)=(x−3)2(x−2)(x^2 - 5x + 6)(x^2 - 7x + 12) = (x - 3)^2(x - 2). What is the sum of all elements in set SS?

  1. A
    5
  2. B
    7
  3. C
    8
  4. 10Cevap
  5. E
    13

Cevap

The sum of all distinct real elements in set SS is 10.
The correct answer is obtained by completely factoring both quadratic expressions on the left side into (x−2)(x−3)2(x−4)(x - 2)(x - 3)^2(x - 4). Moving all terms to one side gives (x−2)(x−3)2(x−4)−(x−3)2(x−2)=0(x - 2)(x - 3)^2(x - 4) - (x - 3)^2(x - 2) = 0. Factoring out (x−2)(x−3)2(x - 2)(x - 3)^2 yields (x−2)(x−3)2(x−5)=0(x - 2)(x - 3)^2(x - 5) = 0. The distinct solutions are 2, 3, and 5. Their sum is 10.

Adım Adım Çözüm

1
Factor the quadratic expressions on the left side of the equation.
x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3) and x2−7x+12=(x−3)(x−4)x^2 - 7x + 12 = (x - 3)(x - 4). Substituting these into the left side yields (x−2)(x−3)2(x−4)(x - 2)(x - 3)^2(x - 4).
Factoring simplifies higher-degree polynomial expressions into linear factors.
2
Set up the equation with all terms on one side to avoid dividing by variable expressions.
(x−2)(x−3)2(x−4)−(x−3)2(x−2)=0(x - 2)(x - 3)^2(x - 4) - (x - 3)^2(x - 2) = 0.
Dividing by variable expressions can cause the loss of valid roots.
3
Factor out the common polynomial factor (x−2)(x−3)2(x - 2)(x - 3)^2.
(x−2)(x−3)2[(x−4)−1]=0(x - 2)(x - 3)^2 [(x - 4) - 1] = 0, which simplifies to (x−2)(x−3)2(x−5)=0(x - 2)(x - 3)^2(x - 5) = 0.
Factoring allows finding all roots by setting each factor equal to zero.
4
Identify the distinct real solutions forming set SS and compute their sum.
The roots are x=2x = 2, x=3x = 3, and x=5x = 5. Thus, S={2,3,5}S = \{2, 3, 5\}. The sum of elements in SS is 2+3+5=102 + 3 + 5 = 10.
A set contains only unique elements, so repeated roots are counted only once.

Anahtar Kavram

Factoring polynomial equations completely without dividing by variable expressions
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