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Difficulty: Very hardQuadratic Equations and Polynomial Factoring

The quadratic equation x2−6x+a=0x^2 - 6x + a = 0 has two distinct real roots α\alpha and γ\gamma, and the quadratic equation y2−by+21=0y^2 - by + 21 = 0 has two distinct real roots β\beta and δ\delta. If α<β<γ<δ\alpha < \beta < \gamma < \delta and the four roots form an arithmetic progression in that order, what is the value of a+ba + b?

Answer: 15

Answer

The value of a+ba + b is 15.
By representing the four ordered roots as r,r+d,r+2d,r+3dr, r+d, r+2d, r+3d, Vieta's formula for the sum of roots of the first equation yields r+(r+2d)=2(r+d)=6r + (r+2d) = 2(r+d) = 6, which implies r+d=3r+d = 3. Thus, the second root is β=3\beta = 3. Using the product of roots for the second equation, 3δ=213\delta = 21 gives δ=7\delta = 7. The common difference is d=(7−3)/2=2d = (7-3)/2 = 2, which gives the roots 1,3,5,71, 3, 5, 7. Finally, a=1×5=5a = 1 \times 5 = 5 and b=3+7=10b = 3 + 7 = 10, so a+b=15a + b = 15.

Step-by-Step Solution

1
Apply Vieta's formulas to both quadratic equations.
α+γ=6\alpha + \gamma = 6, αγ=a\alpha\gamma = a, β+δ=b\beta + \delta = b, and βδ=21\beta\delta = 21.
Vieta's relations connect the coefficients of a quadratic polynomial to the sum and product of its roots.
2
Set up the arithmetic progression representation for the roots.
α=r\alpha = r, β=r+d\beta = r + d, γ=r+2d\gamma = r + 2d, and δ=r+3d\delta = r + 3d, where d>0d > 0.
The roots form an increasing arithmetic progression in the order α,β,γ,δ\alpha, \beta, \gamma, \delta.
3
Substitute the expressions for α\alpha and γ\gamma into α+γ=6\alpha + \gamma = 6.
r+(r+2d)=2r+2d=2(r+d)=6  ⟹  r+d=3r + (r + 2d) = 2r + 2d = 2(r + d) = 6 \implies r + d = 3.
Combining terms simplifies the sum of the first and third terms of the arithmetic progression.
4
Identify the value of β\beta and solve for δ\delta.
β=r+d=3\beta = r + d = 3, so βδ=21  ⟹  3δ=21  ⟹  δ=7\beta\delta = 21 \implies 3\delta = 21 \implies \delta = 7.
Since β=r+d\beta = r + d, its value is directly determined as 3, allowing δ\delta to be solved from the product relation.
5
Calculate the common difference dd and the first term rr.
δ−β=2d=7−3=4  ⟹  d=2\delta - \beta = 2d = 7 - 3 = 4 \implies d = 2, and r=3−2=1r = 3 - 2 = 1.
The difference between the fourth and second terms of an AP is equal to 2d2d.
6
Find aa, bb, and their sum a+ba + b.
a=αγ=1×5=5a = \alpha\gamma = 1 \times 5 = 5, b=β+δ=3+7=10b = \beta + \delta = 3 + 7 = 10, so a+b=5+10=15a + b = 5 + 10 = 15.
With all four roots determined (1,3,5,71, 3, 5, 7), the missing coefficients are calculated using Vieta's formulas.

Key Concept

Combining Vieta's Formulas with Arithmetic Progressions to Solve Quadratic Systems
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Topics

Quadratic Equations and Polynomial Factoring

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