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Quadratic

Posted June 3rd, 2009 by alpha
in

The equation $ x^2-6x+8+\lambda(x^2-4x+3)=0 , \lambda\in R $ has ,
a)real and unequal roots for all $ \lambda $
b)real roots for $ \lambda < 0 $ only
c)real roots for $ \lambda >0 $ only
d)real and unequal roots for $ \lambda=0 $ only

‹ Quadratic Factor or square root equation ›

Comments

answer

On June 5th, 2009 Structure says:

$ x^2-6x+8+\lambda(x^2-4x+3)=0 , \lambda\in R $

$$(1+\lambda)x^2-2(2\lambda+3)x+3\lambda+8=0$$
$$\Delta=4(\lambda^2+\lambda+1)=(2\lambda+1)^2+3\geq3>0$$

So the equation has real distinct roots for all $ \lambda \in R $

THANK YOU.

On June 6th, 2009 alpha says:

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