If the equations `x^(2)+2lambdax+lambda^(2)+1=0`, `lambda in R` and `ax^(2)+bx+c=0` , where `a`, `b`, `c` are lengths of sides of triangle have a common root, then the possible range of values of `lambda` is

Options

A. `(0,2)`
B. `(sqrt(3),3)`
C. `(2sqrt(2),3sqrt(2))`
D. `(0,oo)`

Answer

A

Explanation

`(a)` `(x+lambda)^(2)+1=0` has clearly imaginery roots
So, both roots of the equations are common
`:. (a)/(1)=(b)/(2lambda)=(c )/(lambda^(2)+1)=k`(Say)
Then `a=k`, `b=2lambdak`, `c=(lambda^(2)+1)k`
As `a`, `b`, `c` are sides of triangle
`a+b gt implies 2lambda+1 gt lambda^(2)+1implieslambda^(2)-2lambda lt 0`
`implies lambda in (0,2)`
The other conditions also imply same relation.