Find value of x if `x^2+5|x|+6=0`

Answer

`a epsilon (-oo,-1)uu((sqrt(3))/2,oo)`

Explanation

We have `4^(x^(2))+2(2a+1)2^(x^(2))+4a^(2)-3gt0`….i
Putting `t=2^(x^(2))` in the Eq. (i) we get
`t^(2)+2(2a+1)t+4a^(2)-3gt0`
Let `f(t)=t^(2)+2(2a+1)t+4a^(2)-3[ :' tgt0, :.2^(x^(2))gt0]`
`:' f(t)gt0`

Case Sum of the roots `gt0`
`-2((2a+1))/1gt0`
`:. a epsilon (-oo,-1/2)`
Case II Product of the roots `gt0`
`implies(4a^(2)-3)/1gt0`
or `a^(2)gt3/4`
or `a epsilon (-oo,-(sqrt(3))/2)uu((sqrt(3))/2,oo)`
Case III `Dlt0`
`implies4(2a+1)^(2)-4.1.(4a^(2)-3)lt0`
`implies 4a+4lt0`
`:.alt-1`
or `a epsilon (-oo,-1)`
Combining all cases we get
`a epsilon (-oo,-1)uu((sqrt(3))/2,oo)`