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Difficulty: HardQuadratic Equations and Factoring

The quadratic equation x2−2mx+(m2−4m+12)=0x^2 - 2mx + (m^2 - 4m + 12) = 0, where mm is a real constant, has two distinct real roots x1x_1 and x2x_2. If the distance between the two roots on the real number line is 434\sqrt{3}, what is the value of x12+x22x_1^2 + x_2^2?

  1. A
    48
  2. 96Answer
  3. C
    120
  4. D
    144
  5. E
    192

Answer

96
By Vieta's formulas, the sum of the roots is x1+x2=2mx_1 + x_2 = 2m and the product of the roots is x1x2=m2−4m+12x_1 x_2 = m^2 - 4m + 12. The distance between the roots is given by ∣x1−x2∣=(x1+x2)2−4x1x2|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2}. Squaring both sides gives (43)2=48=(2m)2−4(m2−4m+12)=16m−48(4\sqrt{3})^2 = 48 = (2m)^2 - 4(m^2 - 4m + 12) = 16m - 48. Solving 16m−48=4816m - 48 = 48 yields m=6m = 6. Substituting m=6m = 6 into the expressions for the sum and product of the roots gives x1+x2=12x_1 + x_2 = 12 and x1x2=24x_1 x_2 = 24. Finally, applying the identity x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 yields 122−2(24)=144−48=9612^2 - 2(24) = 144 - 48 = 96.

Step-by-Step Solution

1
Apply Vieta's formulas to determine the sum and product of the roots in terms of mm.
For x2−2mx+(m2−4m+12)=0x^2 - 2mx + (m^2 - 4m + 12) = 0, the sum of roots is x1+x2=2mx_1 + x_2 = 2m and the product of roots is x1x2=m2−4m+12x_1 x_2 = m^2 - 4m + 12.
Vieta's relations relate quadratic coefficients directly to root sums and products.
2
Express the distance between the roots ∣x1−x2∣|x_1 - x_2| in terms of mm and solve for mm.
∣x1−x2∣2=(x1+x2)2−4x1x2=(2m)2−4(m2−4m+12)=16m−48|x_1 - x_2|^2 = (x_1 + x_2)^2 - 4x_1 x_2 = (2m)^2 - 4(m^2 - 4m + 12) = 16m - 48. Given ∣x1−x2∣=43|x_1 - x_2| = 4\sqrt{3}, we have 16m−48=(43)2=48  ⟹  16m=96  ⟹  m=616m - 48 = (4\sqrt{3})^2 = 48 \implies 16m = 96 \implies m = 6.
The difference between two roots of a quadratic is linked to the discriminant via ∣x1−x2∣=b2−4ac∣a∣|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}.
3
Calculate the numerical values of the root sum x1+x2x_1 + x_2 and root product x1x2x_1 x_2 using m=6m = 6.
x1+x2=2(6)=12x_1 + x_2 = 2(6) = 12 and x1x2=62−4(6)+12=24x_1 x_2 = 6^2 - 4(6) + 12 = 24.
Substituting m=6m = 6 gives the exact values needed for algebraic evaluation.
4
Compute x12+x22x_1^2 + x_2^2 using the identity x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2.
x12+x22=122−2(24)=144−48=96x_1^2 + x_2^2 = 12^2 - 2(24) = 144 - 48 = 96.
The sum of squares identity isolates x12+x22x_1^2 + x_2^2 without requiring explicit calculation of individual root values.

Key Concept

Quadratic Root Relationships and Vieta's Formulas
Estimated Time:2m 30s
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Topics

Quadratic Equations and Factoring

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