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Let a, b, c be real numbers, a != 0. If ...

Let `a, b, c` be real numbers, `a != 0.` If `alpha` is a zero of `a^2 x^2+bx+c=0, beta` is the zero of `a^2x^2-bx-c=0 and 0 , alpha < beta` then prove that the equation `a^2x^2+2bx+2c=0` has a root `gamma` that always satisfies `alpha < gamma < beta.`

A

`gamma = (!)/(2) (alpha + beta)`

B

`gamma = alpha + (1)/(2) beta`

C

`gamma = alpha + beta`

D

`alpha lt gamma lt beta`

Text Solution

Verified by Experts

The correct Answer is:
D
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