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Difficulty: HardQuadratic Equations

The quadratic equation 3x2−12x+c=03x^2 - 12x + c = 0, where cc is a constant, has two real solutions, rr and ss. If 1r+1s=32\frac{1}{r} + \frac{1}{s} = \frac{3}{2}, what is the value of cc?

  1. A
    -8
  2. 8Answer
  3. C
    18
  4. D
    24

Answer

8
The correct answer is 8. By Vieta's formulas, the sum of the solutions is r+s=−−123=4r + s = -\frac{-12}{3} = 4 and their product is rs=c3rs = \frac{c}{3}. The given equation 1r+1s=32\frac{1}{r} + \frac{1}{s} = \frac{3}{2} can be rewritten as r+srs=32\frac{r+s}{rs} = \frac{3}{2}. Substituting the expressions for the sum and product gives 4c/3=12c=32\frac{4}{c/3} = \frac{12}{c} = \frac{3}{2}. Solving this equation yields c=8c = 8. Checking the discriminant of the resulting quadratic equation 3x2−12x+8=03x^2 - 12x + 8 = 0 gives Δ=(−12)2−4(3)(8)=48\Delta = (-12)^2 - 4(3)(8) = 48, which is positive and confirms the existence of two real solutions.

Step-by-Step Solution

1
Use Vieta's formulas to express the sum and product of the solutions in terms of the coefficients of the quadratic equation.
The sum of the solutions is r+s=−ba=−−123=4r + s = -\frac{b}{a} = -\frac{-12}{3} = 4, and the product of the solutions is rs=ca=c3rs = \frac{c}{a} = \frac{c}{3}.
Vieta's formulas relate the sum and product of the roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 directly to its coefficients.
2
Express the given algebraic relation 1r+1s\frac{1}{r} + \frac{1}{s} in terms of r+sr+s and rsrs by finding a common denominator.
1r+1s=s+rrs=r+srs\frac{1}{r} + \frac{1}{s} = \frac{s + r}{rs} = \frac{r + s}{rs}.
Finding a common denominator allows us to rewrite the sum of reciprocals using the sum and product of the solutions.
3
Substitute the sum and product from Step 1 into the rewritten equation and solve for cc.
Substituting the values gives 4c3=32\frac{4}{\frac{c}{3}} = \frac{3}{2}, which simplifies to 12c=32\frac{12}{c} = \frac{3}{2}. Solving for cc yields 3c=243c = 24, or c=8c = 8.
This substitution turns the given relation into a single linear equation in terms of the variable cc.
4
Verify that c=8c = 8 yields two real solutions by checking the discriminant of the quadratic equation.
The discriminant is Δ=b2−4ac=(−12)2−4(3)(8)=144−96=48\Delta = b^2 - 4ac = (-12)^2 - 4(3)(8) = 144 - 96 = 48. Since Δ>0\Delta > 0, there are indeed two real solutions.
The problem states that the equation has two real solutions, so we must confirm that c=8c = 8 satisfies this condition.

Key Concept

Applying Vieta's formulas and algebraic manipulation to solve for constants in a quadratic equation.
Estimated Time:2m 0s
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Topics

Quadratic Equations

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