If the equation ` ( lambda^2-5lambda+6) x^2+ ( lambda^2_3lambda+2) x+ (lambda^2_4)=0` is satisfied by more then two values of ` x` then `lambda =`

If the equation (lambda^(2)-5 lambda+6)x^(2)+(lambda^(2)-3 lambda+2)x+(lambda^(2)-4)=0 is satisfied by more then two values of x then lambda=

the values of lambda for which (lambda^(2)-3 lambda+2)x^(2)+(lambda^(2)-5 lambda+6)+lambda^(2)-4=0 is identity in x is

Let alpha and beta be the values of x obtained form the equation lambda^(2) (x^(2)-x) + 2lambdax +3 =0 and if lambda_(1),lambda_(2) be the two values of lambda for which alpha and beta are connected by the relation alpha/beta + beta/alpha = 4/3 . then find the value of (lambda_(1)^(2))/(lambda_(2)) + (lambda_(2)^(2))/(lambda_(1)) and (lambda_(1)^(2))/lambda_(2)^(2) + (lambda_(2)^(2))/(lambda_(1)^(2))

The equation lambda^(2)x^(2)+(lambda^(2)-5 lambda+4)xy+(3lambda-2)y^(2)-8x+12y-4=0 will represent a circle, if lambda=

The equation lambda^(2)x^(2)+(lambda^(2)-5 lambda+4)xy+(3lambda-2)y^(2)-8x+12y-4=0 will represent a circle, if lambda=

If the following equations x+y-3=0,(1+lambda)x+(2+lambda)y-8=0,x-(1+lambda)y+(2+lambda)=0 are consistent then the value of lambda is

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

If alpha,beta be the roots of the equation lambda^(2)(x^(2)-x)+2 lambda x+3=0 and lambda_(1),lambda_(2) be the two values of lambda for which alpha and beta are connected by the relation (alpha)/(beta)+(beta)/(alpha)=(4)/(3), then find the equation whose roots are ((lambda_(1))^(2))/(lambda_(2)) and ((lambda_(2))^(2))/(lambda_(1))