The equation `x^2-6x + 8 + lambda(x^2-4x+3) = 0, AA lambda in R-{-1}` has

The equation (x^(2)-6x+8)+lambda (x^(2)-4x+3)=0, lambda in R has

the equation (x^(2)-6x+8)+lambda(x^(2)-4x+3)=0 for lambda varepsilon R-{-1} has

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

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

The number of real roots of the equation 4x^(3)-x^(2)+lambda^(2)x-mu=0 , mu in R, lambda > 1 is

If alpha,beta be the roots of the equation 4x^2-16x+lambda=0 , lambda in R such that 1

If alpha,beta be the roots of the equation 4x^(2)-16x+lambda=0 where lambda in R such that 1

For all lambda in R , The equation ax^2+ (b - lambda)x + (a-b-lambda)= 0, a != 0 has real roots. Then

If the equation of a circle is lambda x^(2) + (2 lambda - 3)y^(2) - 4x + 6y - 1 = 0 , then the coordinates of centre are -

The values of lambda for which the system of equations x + y - 3 = 0 (1+lambda)x+(2+lambda)y-8=0 x-(1 +lambda) y + (2 + lambda) = 0 is consistent are a) -5//3,1 b) 2//3,-3 c) -1//3,-3 d)0,1