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.`

Explanation

Since `alpha` is a root of `a^(2)x^(2)+bx+c=0`.
Then `a^(2) alpha^(2)+b alpha +c=0`………i
and `beta` is a root of `a^(2)x^(2)-bx-c=0`. ltbr then `a^(2) beta^(2)-b beta-c=0`….ii
Let `f(x)=a^(2)x^(2)+2bx+2c`
`:.f(alpha)=a^(2) alpha^(2)+2b alpha+2c=a^(2)-2a^(2) alpha^(2)` [from Eq. (i)]
`=-a^(2)alpha^(2)`
`impliesf(alpha)lt0` and `f(beta)=alpha^(2)beta^(2)+2b beta+2c`
`=a^(2) beta^(2)+2a^(2) beta^(2)` [from Eq. (ii)]
`=3a^(2)beta^(2)`
`impliesf(beta)gt0`
Since `f (alpha)` and `f(beta)` are of opposite signs, then it is clear that a root `gamma` of the equation `f(x)=0` lies between `alpha` and `beta`.
Hence `alpha lt gamma lt beta [ :' alpha lt beta]`