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Difficulty: HardQuadratic Equations and Factoring
What is the sum of all real solutions to the equation (x2−3x)2−8(x2−3x)+12=0(x^2 - 3x)^2 - 8(x^2 - 3x) + 12 = 0?
  1. A
    33
  2. 66Answer
  3. C
    88
  4. D
    1212
  5. E
    −6-6

Answer

The sum of all real solutions to the given equation is 6.
By substituting u=x2−3xu = x^2 - 3x, the quartic equation reduces to the quadratic equation u2−8u+12=0u^2 - 8u + 12 = 0, which factors as (u−2)(u−6)=0(u - 2)(u - 6) = 0. Setting x2−3x=2x^2 - 3x = 2 and x2−3x=6x^2 - 3x = 6 yields two distinct quadratic equations: x2−3x−2=0x^2 - 3x - 2 = 0 and x2−3x−6=0x^2 - 3x - 6 = 0. Since both discriminants (1717 and 3333) are strictly positive, each equation has two distinct real solutions. By Vieta's formulas, the sum of the roots for a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is −b/a-b/a. Thus, the sum of the roots for x2−3x−2=0x^2 - 3x - 2 = 0 is 33, and the sum of the roots for x2−3x−6=0x^2 - 3x - 6 = 0 is 33. Adding these together gives a total sum of 3+3=63 + 3 = 6.

Step-by-Step Solution

1
Apply algebraic substitution to reduce the equation to a standard quadratic form.
Let u=x2−3xu = x^2 - 3x. The original equation becomes u2−8u+12=0u^2 - 8u + 12 = 0.
Recognizing repeated quadratic expressions allows for simplification into a single quadratic in terms of uu.
2
Factor the quadratic equation in uu to find its roots.
(u−2)(u−6)=0(u - 2)(u - 6) = 0, which gives u=2u = 2 and u=6u = 6.
The factors of 1212 that sum to −8-8 are −2-2 and −6-6.
3
Substitute x2−3xx^2 - 3x back for uu and check the discriminant of each resulting quadratic equation.
For u=2u = 2: x2−3x−2=0x^2 - 3x - 2 = 0 has discriminant Δ1=(−3)2−4(1)(−2)=17>0\Delta_1 = (-3)^2 - 4(1)(-2) = 17 > 0 (2 distinct real roots).
For u=6u = 6: x2−3x−6=0x^2 - 3x - 6 = 0 has discriminant Δ2=(−3)2−4(1)(−6)=33>0\Delta_2 = (-3)^2 - 4(1)(-6) = 33 > 0 (2 distinct real roots).
Verifying that the discriminant is positive ensures that all four roots are real numbers.
4
Calculate the sum of the roots for each quadratic equation using Vieta's formulas.
For x2−3x−2=0x^2 - 3x - 2 = 0, the sum of roots is −−31=3-\frac{-3}{1} = 3.
For x2−3x−6=0x^2 - 3x - 6 = 0, the sum of roots is −−31=3-\frac{-3}{1} = 3.
According to Vieta's formulas, for a quadratic ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is −b/a-b/a.
5
Add the sums of the roots from both equations to find the total sum of all real solutions.
Total sum =3+3=6= 3 + 3 = 6.
Since all four roots are real and distinct, the overall sum is the sum of the roots of the two constituent quadratics.

Key Concept

Quadratic Substitution and Vieta's Formulas for Root Sums
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Topics

Quadratic Equations and Factoring

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