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Difficulty: HardQuadratic Equations and Factoring

For a real constant kk, the quadratic equation x2−2kx+(k2−k−6)=0x^2 - 2kx + (k^2 - k - 6) = 0 has two distinct real roots rr and ss such that r<0<sr < 0 < s and ∣r∣<∣s∣|r| < |s|. Which of the following inequalities expresses all possible values of kk?

  1. 0<k<30 < k < 3Answer
  2. B
    −2<k<3-2 < k < 3
  3. C
    −6<k<3-6 < k < 3
  4. D
    k<−2k < -2 or k>3k > 3
  5. E
    −2<k<0-2 < k < 0

Answer

The inequality expressing all possible values of kk is 0<k<30 < k < 3.
The condition that one root is negative and one root is positive (r<0<sr < 0 < s) requires the product of the roots rs=k2−k−6rs = k^2 - k - 6 to be negative, which resolves to −2<k<3-2 < k < 3. Furthermore, since the positive root ss has a larger absolute magnitude than the negative root rr (∣r∣<∣s∣|r| < |s|), the sum of the roots r+s=2kr + s = 2k must be positive, requiring k>0k > 0. Taking the intersection of −2<k<3-2 < k < 3 and k>0k > 0 yields 0<k<30 < k < 3.

Step-by-Step Solution

1
Apply Vieta's formulas to express the sum and product of the roots in terms of kk.
For x2−2kx+(k2−k−6)=0x^2 - 2kx + (k^2 - k - 6) = 0, the sum of roots is r+s=2kr + s = 2k and the product of roots is rs=k2−k−6rs = k^2 - k - 6.
Vieta's formulas directly relate the coefficients of a quadratic equation to the sum and product of its roots.
2
Analyze the condition r<0<sr < 0 < s.
Since one root is negative and the other is positive, their product must be negative: rs=k2−k−6<0rs = k^2 - k - 6 < 0. Factoring gives (k−3)(k+2)<0(k - 3)(k + 2) < 0, which yields −2<k<3-2 < k < 3.
A positive number multiplied by a negative number produces a negative product.
3
Analyze the condition ∣r∣<∣s∣|r| < |s|.
Since r<0r < 0, ∣r∣=−r|r| = -r. Since s>0s > 0, ∣s∣=s|s| = s. The inequality ∣r∣<∣s∣|r| < |s| becomes −r<s-r < s, which simplifies to r+s>0r + s > 0. Substituting r+s=2kr + s = 2k gives 2k>02k > 0, or k>0k > 0.
The positive root having greater magnitude than the absolute value of the negative root means the sum of the roots must be positive.
4
Combine the conditions to find the valid range for kk.
Combining −2<k<3-2 < k < 3 and k>0k > 0 yields the intersection 0<k<30 < k < 3. (The discriminant condition Δ=4(k+6)>0  ⟹  k>−6\Delta = 4(k + 6) > 0 \implies k > -6 is satisfied for all k∈(0,3)k \in (0, 3)).
The parameter kk must satisfy both root sign constraints simultaneously.

Key Concept

Using Vieta's formulas and root magnitude conditions to solve quadratic parameter inequality problems.
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Topics

Quadratic Equations and Factoring

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