For what values of `m` the equation `(1+m)x^(2)-2(1+3m)x+(1+8m)=0` has `(m in R)`
(i) both roots are imaginary?
(ii) both roots are equal?
(iii) both roots are real and distinct?
(iv) both roots are positive?
(v) both roots are negative?
(vi) roots are opposite in sign?
(vii)roots are equal in magnitude but opposite in sign?
(viii) atleast one root is positive?
(ix) atleast one root is negative?
(x) roots are in the ratio `2:3`?

Answer

(i) `m epsilon (0,3)` (ii) `m=0,3` (iii) `m epsilon (-oo,0)u(3,oo)` (iv) `m epsilon (-oo,-1)uu[3,oo)`
(v) `m epsilon phi`
(vi) `m epsilon (-1,-1//8)`
(vii) `m=-1//3` (viii) `m epsilon(-oo,-1)uu(-1,-1//8)uu[3,oo)` (ix) `m epsilon(-1,-1//8)` (x) `m=(81+-sqrt(6625))/32`

Explanation

`:'(1+m)x^(2)-2(1+3m)x+(1+8m)=0`
`impliesD=4(1+3m)^(2)-4(1+m)(1+8m)=4m(m-3)`
(i) Both roots are imaginary.
`:.Dlt0`
`implies4m(m-3)lt0`
`implies0ltmlt3`
or ` m epsilon (0,3)`
(ii) Both roots are equal
`:.D=0`
`implies4m(m-3)=0`
`impliesm=0,3`
(iii) Both roots are real and distinct
`:.Dgt0`
`implies4x(m-3)gt0`
`impliesmlt0` or `mgt3`
`:.m epsilon(-oo,0)uu(3,oo)`
(iv) Both roots are positive.
Case I Lsum of the roots `gt0`
`implies(2(1+3m))/((1+m))gt0`
`impliesm epsilon (-oo,-1)uu(-1/3,oo)`
Case II Produc of the roots `gt0`
`implies((1+8m))/((1+m))gt0`
`m epsilon (-oo,-1)uu(-1/8,oo)`
Case III` D ge0`
`implies 4m(m-3)ge0`
`m epsilon (-oo,0]uu[3,oo)`
Combining all Cases, we get
`m epsilon (-oo,-1)uu[3,oo)`
(v) Both roots are negative.
Consider the following cases:
Case I Sum of the roots `lt 0implies(2(1+3m))/((1+m))lt0`
`impliesm epsilon (-1,-1/3)`
Case II Product of the roots `gt0implies((1+8m))/((1+m))gt0`
`impliesm epsilon (-oo,1)uu(-1/8,oo)`
Case III `Dge0`
`4m(m-3)ge0impliesm epsilon (-oo,0]uu[3,oo)`
Combining all cases we get
`m epsilon phi`
(vi) Roots are opposite in sign, then
Case I Consider the following cases:
Product of the roots `lt0`
`implies((1+8m))/((1+m))lt0`
` m epsilon (-1,-1/8)`
Case II `D gt0implies4m(m-3)gt0`
`implies m epsilon (-oo,0)uu(3,oo)`
Combining all cases we get
`m epsilon (-1,-1/8)`
(vii) Roots are equal in magnitude but opposite in sign, then Consider the following cases:
Case I Sum of the roots `=0`
`implies(2(1+3m))/((1+m))=0`
`impliesm=-1/3, m !=1`
Case `Dgt0implies4m(m-3)gt0`
`impliesm epsilon (-oo,0)uu(3,oo)`
Combining all cases we get
`m=-1/3`
(viii) Atleast one root is positive, then either one root si positive or bothh roots are positive
i.e. (d) `uu`(f)`
or `m epsilon (-oo,-1)uu(-1.-1/8)uu[3,oo)`
(ix) Atleast one root is negative, then either one root is negative or both roots are negative.
i.e. (e) `duu` (f) or `m epsilon (-1,-1/8)`
(x) Let roots are `2 alpha` are `3 alpha`. Then
Consider the following cases:
Case I Sum of the roots `=2alpha+3 alpha=(2(1+3m))/((1+m))`
`implies alpha=(2(1+3m))/(5(1+m))`
Case II Product of the roots `=2 alpha.3 alpha=((1+8m))/((1+m))`
`implies6 alpha^(2)=((1+8m))/((1+m))`
From Eqs i and ii we get
`6{(2(1+3m))/(5(1+m))}^(2)=((1+8))/((1+m))`
`implies24(1+3m)^(2)=25(1+8m)(1+m)`
`implies24(9m^(2)+6m+1)=25(8m^(2)+9m+1)`
`=16m^(2)-81m-1=0`
or `m=(81+-sqrt((-81)^(2)+64))/32`
`impliesm=(81+-sqrt(6625))/32`