Question

Let a, b, c be real numbers and a ≠ 0. If α is a root of a²x² ₊ bx + c = 0, β is a root a²x² – bx – c = 0, and 0 < α < β then the equation a²x² + 2bx + 2c = 0 has a root γ that always satisfies –

• A. γ = $\frac{1}{2}$ (α + β)
• A. (b) γ = α + $\frac{β}{2}$
• C. γ = α
• D. α < γ < β

Answer 1

Answer: (D)

α < γ < β

Answer 2

Let ƒ(x) = a²x² + 2bx + 2c. From the question,

a²α² + bα + c = 0 and a²β² – bβ – c = 0

Now, ƒ(α) = a²α² + 2bα + 2c = bα + c = –a²α²

ƒ(β) = a²β² + 2bβ + 2c = 3bβ + 3c = 3(bβ + c) = 3a²β²

Clearly, 0 < α < β ⇒ α, β are real

So ƒ(α) < 0, ƒ(β) > 0

So, ƒ(γ) = 0 where α < γ < β.