Question

Difficulty: HardQuadratic Equations and the Quadratic Formula

For a constant kk, the quadratic equation 13x2k6x+(k10)=0\frac{1}{3}x^2 - \frac{k}{6}x + (k - 10) = 0 has two real roots, r1r_1 and r2r_2. If the sum of the reciprocals of the roots, 1r1+1r2\frac{1}{r_1} + \frac{1}{r_2}, is equal to 12-\frac{1}{2}, what is the value of kk?

Answer: 7.5

Answer

The value of kk is 7.57.5.
By writing the sum of the reciprocals in terms of the sum and product of the roots, we find that 1r1+1r2=r1+r2r1r2\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}. Substituting the values from Vieta's formulas (r1+r2=k2r_1 + r_2 = \frac{k}{2} and r1r2=3(k10)r_1 r_2 = 3(k - 10)) yields k6(k10)=12\frac{k}{6(k - 10)} = -\frac{1}{2}. Solving this equation for kk results in k=7.5k = 7.5.

Step-by-Step Solution

1
Identify the quadratic coefficients in terms of the constant kk.
a=13a = \frac{1}{3}, b=k6b = -\frac{k}{6}, and c=k10c = k - 10
Applying Vieta's formulas requires the coefficients from the standard form ax2+bx+c=0ax^2 + bx + c = 0.
2
Determine the sum and the product of the roots using Vieta's formulas.
r1+r2=k2r_1 + r_2 = \frac{k}{2} and r1r2=3(k10)r_1 r_2 = 3(k - 10)
Vieta's formulas state that the sum of the roots is ba-\frac{b}{a} and the product of the roots is ca\frac{c}{a}.
3
Express the sum of the reciprocals of the roots in terms of kk.
1r1+1r2=r1+r2r1r2=k6(k10)\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{k}{6(k - 10)}
Finding a common denominator allows the sum of the reciprocals to be written as the ratio of the sum of the roots to the product of the roots.
4
Equate the expression to 12-\frac{1}{2} and solve the resulting equation for kk.
k=7.5k = 7.5
Solving k6(k10)=12\frac{k}{6(k - 10)} = -\frac{1}{2} gives 2k=6(k10)    8k=60    k=7.52k = -6(k - 10) \implies 8k = 60 \implies k = 7.5.

Key Concept

Vieta's Formulas and Algebraic Relationships of Roots
Estimated Time:2m 30s
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