Quadratic EquationHard
Question
Let a, b, c be real numbers, a ≠ 0. If α is a root of a2x2 + bx + c = 0, β is the root of a2x2 - bx - c = 0 and 0 < α < β, then the equation a2x2 + 2bx + 2c = 0 has a root γ that always satisfies
Options
A.
B.γ = α + 
C.γ = α
D.α < γ < β
Solution
Since, α is a root of a2x2 + bx + c = 0
⇒ a2α2 + bα + c = 0 .......(i)
and β is a root of a2x2 - bx - c = 0
⇒ a2β2 - bβ - c = 0 .......(ii)
Let f(x) = a2x2 + 2bx + 2c
∴ f(α) = a2α2 + 2bα + 2c
= a2α2 + 2b2α2 = - a2α2 [from Eq. (i)]
and f(β) = a2β2 + 2bβ + 2c
= a2β2 + 2b2β2 = 3a2β2 [from Eq. (ii)]
⇒ f(α)f(β) < 0
∴ f(x) must have a root lying in the open interval (α, β)
∴ α < γ < β
⇒ a2α2 + bα + c = 0 .......(i)
and β is a root of a2x2 - bx - c = 0
⇒ a2β2 - bβ - c = 0 .......(ii)
Let f(x) = a2x2 + 2bx + 2c
∴ f(α) = a2α2 + 2bα + 2c
= a2α2 + 2b2α2 = - a2α2 [from Eq. (i)]
and f(β) = a2β2 + 2bβ + 2c
= a2β2 + 2b2β2 = 3a2β2 [from Eq. (ii)]
⇒ f(α)f(β) < 0
∴ f(x) must have a root lying in the open interval (α, β)
∴ α < γ < β
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