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Zorluk: OrtaQuadratic Equations and Polynomial Factoring

What is the sum of all distinct real solutions to the equation (x2−3x)2−2(x2−3x)−8=0(x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0?

  1. A
    2
  2. B
    3
  3. 6Cevap
  4. D
    0
  5. E
    8

Cevap

The sum of all distinct real solutions is 6.
By defining u=x2−3xu = x^2 - 3x, the original equation reduces to u2−2u−8=0u^2 - 2u - 8 = 0, which factors as (u−4)(u+2)=0(u - 4)(u + 2) = 0, giving u=4u = 4 or u=−2u = -2. Substituting back for xx gives two quadratic equations: x2−3x−4=0x^2 - 3x - 4 = 0 (yielding roots x=4x = 4 and x=−1x = -1) and x2−3x+2=0x^2 - 3x + 2 = 0 (yielding roots x=1x = 1 and x=2x = 2). The four distinct real roots are −1,1,2,-1, 1, 2, and 44, and their sum is −1+1+2+4=6-1 + 1 + 2 + 4 = 6.

Adım Adım Çözüm

1
Use algebraic substitution to simplify the disguised quadratic equation.
Let u=x2−3xu = x^2 - 3x. The equation becomes u2−2u−8=0u^2 - 2u - 8 = 0.
Recognizing the repeating quadratic block x2−3xx^2 - 3x transforms the 4th-degree polynomial into a standard 2nd-degree quadratic in terms of uu.
2
Factor the quadratic equation in terms of uu to find its roots.
(u−4)(u+2)=0  ⟹  u=4 or u=−2(u - 4)(u + 2) = 0 \implies u = 4 \text{ or } u = -2.
Finding the values of uu establishes the two quadratic equations in terms of xx that need to be solved.
3
Substitute x2−3xx^2 - 3x back for uu and solve for xx in Case 1 (u=4u = 4).
x2−3x=4  ⟹  x2−3x−4=0  ⟹  (x−4)(x+1)=0  ⟹  x=4 or x=−1x^2 - 3x = 4 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1.
The discriminant b2−4ac=(−3)2−4(1)(−4)=25>0b^2 - 4ac = (-3)^2 - 4(1)(-4) = 25 > 0, so there are two distinct real solutions: x=4x = 4 and x=−1x = -1.
4
Substitute x2−3xx^2 - 3x back for uu and solve for xx in Case 2 (u=−2u = -2).
x2−3x=−2  ⟹  x2−3x+2=0  ⟹  (x−1)(x−2)=0  ⟹  x=1 or x=2x^2 - 3x = -2 \implies x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2.
The discriminant b2−4ac=(−3)2−4(1)(2)=1>0b^2 - 4ac = (-3)^2 - 4(1)(2) = 1 > 0, so there are two distinct real solutions: x=1x = 1 and x=2x = 2.
5
Sum all distinct real solutions.
−1+1+2+4=6-1 + 1 + 2 + 4 = 6.
All four solutions {−1,1,2,4}\{-1, 1, 2, 4\} are distinct and real.

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