Examkin

Question

Difficulty: HardQuadratic Equations and Polynomial Factoring

If xx is a real number satisfying the equation (x2−2x)2−(x2−2x)−12=0(x^2 - 2x)^2 - (x^2 - 2x) - 12 = 0, what is the product of all distinct real roots of the equation?

  1. A
    −12-12
  2. −4-4Answer
  3. C
    −3-3
  4. D
    22
  5. E
    1212

Answer

The product of all distinct real roots of the equation is −4-4.
Substituting u=x2−2xu = x^2 - 2x simplifies the given equation to u2−u−12=0u^2 - u - 12 = 0, which factors as (u−4)(u+3)=0(u - 4)(u + 3) = 0. Setting x2−2x=4x^2 - 2x = 4 yields x2−2x−4=0x^2 - 2x - 4 = 0, which has two real roots (since the discriminant is 20>020 > 0) with product −4-4. Setting x2−2x=−3x^2 - 2x = -3 yields x2−2x+3=0x^2 - 2x + 3 = 0, which has a negative discriminant (−8-8) and therefore no real solutions. Thus, the product of all distinct real roots is −4-4.

Step-by-Step Solution

1
Use algebraic substitution to simplify the disguised quadratic equation.
Let u=x2−2xu = x^2 - 2x. Substituting uu into (x2−2x)2−(x2−2x)−12=0(x^2 - 2x)^2 - (x^2 - 2x) - 12 = 0 gives u2−u−12=0u^2 - u - 12 = 0.
Recognizing the repeated expression x2−2xx^2 - 2x reduces a 4th-degree polynomial into a standard 2nd-degree quadratic equation in terms of uu.
2
Solve for uu by factoring.
u2−u−12=(u−4)(u+3)=0u^2 - u - 12 = (u - 4)(u + 3) = 0, so u=4u = 4 or u=−3u = -3.
Factoring provides the possible numerical values for the expression x2−2xx^2 - 2x.
3
Analyze each case for xx and check for real roots using the discriminant.
Case 1: x2−2x=4  ⟹  x2−2x−4=0x^2 - 2x = 4 \implies x^2 - 2x - 4 = 0. Discriminant D1=(−2)2−4(1)(−4)=20>0D_1 = (-2)^2 - 4(1)(-4) = 20 > 0. Since D1>0D_1 > 0, this quadratic has 2 distinct real roots. By Vieta's formula, the product of these two real roots is ca=−41=−4\frac{c}{a} = \frac{-4}{1} = -4.
Case 2: x2−2x=−3  ⟹  x2−2x+3=0x^2 - 2x = -3 \implies x^2 - 2x + 3 = 0. Discriminant D2=(−2)2−4(1)(3)=−8<0D_2 = (-2)^2 - 4(1)(3) = -8 < 0. Since D2<0D_2 < 0, this quadratic has no real roots.
Only roots with a non-negative discriminant belong to the real number domain and contribute to the product of real roots.
4
Determine the product of all distinct real roots.
The total product of all distinct real roots is −4-4.
Since Case 1 produces the only real roots of the original equation, their product −4-4 is the complete product of all real solutions.

Key Concept

Disguised Quadratic Equations and Discriminant Real Root Filtering
Rate this question

Topics

Quadratic Equations and Polynomial Factoring

More questions from this topic