Consider equation `((x^(2)+x)^2)+a(x^(2)+x)+4=0` Match the values of `a` in Lits II for the types of roods in Lits I.

Options

A. `{:(,a,b,c,d),((1),p,q,r,s):}`
B. `{:(,a,b,c,d),((2),q,r,r,p):}`
C. `{:(,a,b,c,d),((3),r,p,s,q):}`
D. `{:(,a,b,c,d),((4),q,s,p,r):}`

Answer

1

Explanation

`(x^(2)+x)^(2)+a(x^(2)+x)+4=0`
Let `t=x^(2)+x=(x+1//2)^(2)-1//4`.
`impliestin[-(1)/(4),oo]`
Now, `f(t)=t^(2)+at+4=0" "....(1)`
(a) All four real and distinct roots.
So, equation (1) has both roots greater then `-1//4`.

Following conditions are required:
`(i) Dgt0impliesa^(2)-16gt0implies|a|gt4`
(ii) `f(-1//4)=(1)/(16)-(a)/(4)+4gt0impliesalt65//4`
(iii) `-(B)/(2A)=(a)/(2)gt-(1)/(4)impliesalt(1)/(2)`
`impliesain(-oo,-4)`
(b) Two real roots which are distinct.

`impliesf(-1//4)lt0`
`impliesagt65//4`
`impliesain(65//4,oo)`
(c) All four roots are imaginary.

(i) `Dge0implies|a|ge4`
(ii) `f(-1//4)gt0impliesalt(65)/(4)`
(iii) `-(B)/(2A)lt-(1)/(4)impliesagt(1)/(2)impliesain[4,(65)/(4)]`
Case II:
`Dlt0`

`impliesain(-4,4)" "....(2)`
From case I and case II,
`ain(-4,(65)/(4))`
(d) Four real roots in which two are equal.

(i) `Dgt0implies|a|gt4`
(ii) `f(-1//4)=0impliesa=65//4`
(iii) `-(B)/(2A)gt-(1)/(4)impliesalt(1)/(2)`
No common solution.
`:.ainphi`.