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Difficulty: HardQuadratic Equations

In the quadratic equation 3x2+kx+12=03x^2 + kx + 12 = 0, where kk is a constant, the difference between the two real solutions is 44. What is the value of k2k^2?

  1. A
    160
  2. B
    192
  3. 288Answer
  4. D
    0

Answer

The correct value of k2k^2 is 288.
The correct value is 288. The difference between the two solutions r1r_1 and r2r_2 of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is given by ∣r1−r2∣=b2−4ac∣a∣|r_1 - r_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}. Substituting a=3a = 3, b=kb = k, and c=12c = 12, we get ∣r1−r2∣=k2−4(3)(12)3=k2−1443|r_1 - r_2| = \frac{\sqrt{k^2 - 4(3)(12)}}{3} = \frac{\sqrt{k^2 - 144}}{3}. Setting this equal to the given difference of 44 yields k2−1443=4\frac{\sqrt{k^2 - 144}}{3} = 4. Multiplying by 3 gives k2−144=12\sqrt{k^2 - 144} = 12, and squaring both sides gives k2−144=144k^2 - 144 = 144, which simplifies to k2=288k^2 = 288. Alternatively, using Viete's formulas, r1+r2=−k3r_1 + r_2 = -\frac{k}{3} and r1r2=4r_1 r_2 = 4. Using the identity (r1−r2)2=(r1+r2)2−4r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2, we have 42=(−k3)2−4(4)4^2 = (-\frac{k}{3})^2 - 4(4), which simplifies to 16=k29−1616 = \frac{k^2}{9} - 16, leading to k29=32\frac{k^2}{9} = 32, or k2=288k^2 = 288.

Step-by-Step Solution

1
Identify the coefficients of the quadratic equation 3x2+kx+12=03x^2 + kx + 12 = 0.
The coefficients are a=3a = 3, b=kb = k, and c=12c = 12.
These values are needed to apply the difference of roots formula or Viete's relations.
2
State the formula for the difference between the two solutions r1r_1 and r2r_2 of a quadratic equation.
∣r1−r2∣=b2−4ac∣a∣|r_1 - r_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}. Substituting the coefficients gives ∣r1−r2∣=k2−4(3)(12)3=k2−1443|r_1 - r_2| = \frac{\sqrt{k^2 - 4(3)(12)}}{3} = \frac{\sqrt{k^2 - 144}}{3}.
This formula relates the difference of the roots directly to the coefficients of the equation.
3
Set the expression for the difference of the roots equal to the given difference of 44 and solve for k2k^2.
k2−1443=4⇒k2−144=12⇒k2−144=144⇒k2=288\frac{\sqrt{k^2 - 144}}{3} = 4 \Rightarrow \sqrt{k^2 - 144} = 12 \Rightarrow k^2 - 144 = 144 \Rightarrow k^2 = 288.
Squaring both sides and isolating k2k^2 yields its value.

Key Concept

Difference of roots and discriminant of a quadratic equation

Alternative Method

Use Viete's relations: The sum of the roots is r1+r2=−k3r_1 + r_2 = -\frac{k}{3} and the product of the roots is r1r2=123=4r_1 r_2 = \frac{12}{3} = 4. The difference between the roots is given as ∣r1−r2∣=4|r_1 - r_2| = 4. Square this relation to get (r1−r2)2=16(r_1 - r_2)^2 = 16. Expand and rewrite the identity as (r1+r2)2−4r1r2=16(r_1 + r_2)^2 - 4r_1 r_2 = 16. Substituting the sum and product, we get (−k3)2−4(4)=16⇒k29−16=16⇒k29=32⇒k2=288(-\frac{k}{3})^2 - 4(4) = 16 \Rightarrow \frac{k^2}{9} - 16 = 16 \Rightarrow \frac{k^2}{9} = 32 \Rightarrow k^2 = 288.
Estimated Time:2m 0s
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Topics

Quadratic Equations

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