Consider the equation `x^(4) + 2ax^(3) + x^(2)+2ax + 1 =0`, where `a in R`.
If exactly two roots are positive and two roots are negative, then the number of intergal values of a is

Options

A. `(-oo,-1//4)`
B. `(5//4, oo)`
C. `(-oo,-3//4)`
D. none of these

Answer

3

Explanation

Given equations is
`x^(4) + 2ax^(3) + x^(2) + 2ax + 1 =0`
or `(x + (1)/(x^(2))) + 2a(x+(1)/(x))+ 1 =0`
or `(x +(1)/(x))^(2) + 2a(x+(1)/(x)) - 1=0`
`rArr t^(2) + 2at -1=0`
Where `t = x+ (1//x)`. Now,
`(x+(1)/(x))^(2) ge 2 or (x+(1)/(x)) le-2`
`therefore tge 2 or t le-2`
Now, Eq.(1) will have at least tow positive roots, when at least one roots of Eq.(2) will be greater than 2. From Eq. (2),
`D= 4a^(2) - 4(-1) = 4(1+a^(2)) gt 0, AA a in R`
Let the roots of Eq. (2) be `alpha, bet`. If `alpha, beta le 2`, then
`rArr f(2) ge and(-B)/(2A) lt 2`
`rArr 4+4a -1 ge 0 and -(2a)/(2) lt2`
`rArr a ge-(3)/(4) and a gt - 2`
`rArr a gt -(3)/(4)`
Therefore, at least one root will be gerter than 2. Then
`a lt-(3)/(4)`
Combining (3) and (4) , we get .
`a lt -(3)/(4)`
Hence, at least one root will positive if a `in(-oo,-3//4)`