Nature of roots of a quadratic.

Question

Question: If the equation $x^2+2x+1+\lambda=0$ has real and unequal roots, determine the nature of the roots of the equation

$(\lambda+2)(x^2+2x+1+\lambda)=2\lambda(x^2+1)$.

My attempt: Taking $(\lambda+2)$ to the other side we have $\dfrac{2\lambda}{(\lambda+2)}(x^2+1)$ which must have real and unequal roots.

Let $a=\dfrac{2\lambda}{(\lambda+2)}$.

Then $ax^2+a$ has real and unequal roots so $\sqrt{4a^2}>0$ and so $a>\dfrac{1}{2}$ which means that $\lambda>\dfrac{2}{3}$. (I'm not sure how helpful this is.)

Besides this I have no idea what else can be done as I am sure that expanding the brackets given will lead nowhere.

Answer 1

HINT:

The discriminant of $x^2+2x+1+\lambda=0$ is $2^2>4(1+\lambda)\iff\lambda<0$

Now rearrange the second equation as $Ax^2+Bx+C=0$ and find the sign of the discriminant with $\lambda<0$

Answer 2

Expanding and collecting terms, $$(\lambda-2)x^2-2(\lambda+2)x-(\lambda^2+\lambda+2)=0\ .$$ The given equation has unequal real roots if and only if $\lambda<0$; in this case the second quadratic has discriminant $$4(\lambda+2)^2+4(\lambda-2)(\lambda^2+\lambda+2)=4\lambda(\lambda^2+4)<0\ .$$ So the quadratic has (non-real) complex conjugate roots.

Answer 3

Since the equation $x^2+2x+1+\lambda=0$ has real and unequal roots, we have $$4-4(1+\lambda)>0$$ Hence $\lambda<0$.

Write $(\lambda+2)(x^2+2x+1+\lambda)=2\lambda(x^2+1)$ as $(-\lambda+2)x^2+2(\lambda+2)x+(\lambda^2+\lambda+2)=0$.

After computation, you can find the discriminant is $4\lambda(\lambda^2+4)<0$.

And $(-\lambda+2)>0$, so no real root.