Roots of the equation `(x^2-4x+3)+lamda(x^2-6x+8)=0` , `lamda in R` will be

Options

A. always real
B. real only when `lambda` is positive
C. real only when `lambda` is negative
D. always is magnary

Answer

A

Explanation

Roots of the quadratic …………..
`(x^(2)-4x+3)+lambda(x^(2)-6x+8)=0`
`x^(2)(1+lambda)-2x(2+3lambda)+(3+8lambda) = 0`
Discriminant.
`D = 4(2+3lambda)^(2)-4(1+lambda)(3+8lambda)`
`D = 4(lambda^(2)+lambda+1)`
If `lambda epsilon R` then `D gt 0`
so root of given quadratic always real