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Difficulty: Very hardQuadratic Equations

In the quadratic equation 3x2−kx+12=03x^2 - kx + 12 = 0, kk is a constant. The equation has two distinct real solutions, and the difference between these two solutions is less than 22. Which of the following describes all possible values of kk?

  1. −65<k<−12-6\sqrt{5} < k < -12 or 12<k<6512 < k < 6\sqrt{5}Answer
  2. B
    −65<k<65-6\sqrt{5} < k < 6\sqrt{5}
  3. C
    −122<k<−12-12\sqrt{2} < k < -12 or 12<k<12212 < k < 12\sqrt{2}
  4. D
    k<−12k < -12 or k>12k > 12

Answer

The possible values of kk are −65<k<−12-6\sqrt{5} < k < -12 or 12<k<6512 < k < 6\sqrt{5}
The correct answer is found by combining two constraints. First, for the quadratic equation to have two distinct real solutions, the discriminant must be strictly positive: (−k)2−4(3)(12)>0⇒k2>144(-k)^2 - 4(3)(12) > 0 \Rightarrow k^2 > 144, which means ∣k∣>12|k| > 12. Second, the difference between the roots of ax2+bx+c=0ax^2 + bx + c = 0 is b2−4ac∣a∣\frac{\sqrt{b^2 - 4ac}}{|a|}. Here, the difference is k2−1443\frac{\sqrt{k^2 - 144}}{3}. Setting this difference to be less than 2 gives k2−1443<2⇒k2−144<6⇒k2−144<36⇒k2<180\frac{\sqrt{k^2 - 144}}{3} < 2 \Rightarrow \sqrt{k^2 - 144} < 6 \Rightarrow k^2 - 144 < 36 \Rightarrow k^2 < 180, which means ∣k∣<65|k| < 6\sqrt{5}. Combining these inequalities yields 12<∣k∣<6512 < |k| < 6\sqrt{5}, which translates to −65<k<−12-6\sqrt{5} < k < -12 or 12<k<6512 < k < 6\sqrt{5}.

Step-by-Step Solution

1
Find the condition for the quadratic equation to have two distinct real solutions.
k2>144k^2 > 144, which means k<−12k < -12 or k>12k > 12.
For the equation 3x2−kx+12=03x^2 - kx + 12 = 0 to have two distinct real solutions, its discriminant Δ=b2−4ac\Delta = b^2 - 4ac must be strictly positive. Here, Δ=(−k)2−4(3)(12)=k2−144>0\Delta = (-k)^2 - 4(3)(12) = k^2 - 144 > 0.
2
Express the difference between the two solutions using the quadratic formula.
∣x1−x2∣=k2−1443|x_1 - x_2| = \frac{\sqrt{k^2 - 144}}{3}.
The solutions to the quadratic equation are given by x=k±k2−1446x = \frac{k \pm \sqrt{k^2 - 144}}{6}. The difference between these solutions is x1−x2=(k+k2−144)−(k−k2−144)6=2k2−1446=k2−1443x_1 - x_2 = \frac{(k + \sqrt{k^2 - 144}) - (k - \sqrt{k^2 - 144})}{6} = \frac{2\sqrt{k^2 - 144}}{6} = \frac{\sqrt{k^2 - 144}}{3}.
3
Apply the condition that the difference between the solutions is less than 2.
k2<180k^2 < 180, which means −65<k<65-6\sqrt{5} < k < 6\sqrt{5}.
We set the difference expression to be less than 2: k2−1443<2⇒k2−144<6\frac{\sqrt{k^2 - 144}}{3} < 2 \Rightarrow \sqrt{k^2 - 144} < 6. Squaring both sides gives k2−144<36⇒k2<180k^2 - 144 < 36 \Rightarrow k^2 < 180. Since 180=65\sqrt{180} = 6\sqrt{5}, this yields −65<k<65-6\sqrt{5} < k < 6\sqrt{5}.
4
Combine the inequalities from Step 1 and Step 3 to find the final overlapping range for kk.
−65<k<−12-6\sqrt{5} < k < -12 or 12<k<6512 < k < 6\sqrt{5}.
We must satisfy both k2>144k^2 > 144 (for real solutions) and k2<180k^2 < 180 (for the difference to be less than 2). This gives 144<k2<180144 < k^2 < 180, which corresponds to the union of intervals −65<k<−12-6\sqrt{5} < k < -12 and 12<k<6512 < k < 6\sqrt{5}.

Key Concept

Using the discriminant and quadratic formula to analyze properties of roots under inequality constraints.
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Topics

Quadratic Equations

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