`:'(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`
