solve `x^2+2x+4 = 0`

Options

A. `-3` and 1
B. `3/2` and 2
C. `-1` and 1
D. 0 and 2

Answer

B

Explanation

We have `1/3lt(x^(2)-2x+4)/(x^(2)+2x+4)lt3,AA xepsilonR`
`1/2lt(x^(2)+2x+4)/(x^(2)-2x+4)lt3,AAxepsilonR`
Let `y=(9.3^(2x)+6.3^(x)+4)/(9.3^(2x)-6.3^(x)+4)=((3^(x+1))^(2)+2.3^(x+1)+4)/((e^(x+1))^(2)-2.3^(x+1)+4)`
`(t^(2)+2t+4)/(t^(2)-2t+4)` where `t=3^(x+1)`
`(y-1)t^(2)-2(y+1)t+4(y-1)=0`
By the given condition for every `t epsilonR`
`1/2ltylt3`.....i
But`t=3^(x+1)gt0`
We have product of the roots `=4gt0` which is true.
And sum of the roots `=(2(y+1))/((y-1)ge0`
`implies(y+1)/(y-1)gt0`
`:.yepsilon (-oo,-1)uu(1,o)`...ii
From Eqs (i) and (ii) we get
`1ltylt3`