Minimum possible number of positive root of the quadratic equation $x^2-(1+\lambda)x+\lambda-2=0,\lambda\in R$ is

Question

Minimum possible number of positive root of the quadratic equation $x^2-(1+\lambda)x+\lambda-2=0,\lambda\in R$ is
$(a)2$
$(b)1$
$(c)0$
$(d)3$

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$x^2-(1+\lambda)x+\lambda-2=0$ I changed this equation to $\lambda=\frac{x^2-x-2}{x-1}$.I am stuck now.

Answer 1

The solutions are

$$x=\frac{\lambda+1\pm \sqrt{(\lambda+1)^2-4(\lambda-2)}}{2}.$$

The factor of 2 plays no role so we can just focus on the numerator. The discriminant is

$$(\lambda+1)^2-4(\lambda-2) = \lambda^2-2\lambda+9=(\lambda-1)^2+8.$$

This is positive, so there are two real roots. Working with the root coming from +discriminant,

$$\lambda+1+\sqrt{(\lambda-1)^2+8} \ge \lambda+1+|\lambda-1|. $$

This is never negative since if $\lambda+1$ is negative, $\lambda-1$ is more negative and thus its absolute value is more positive. Therefore there is at least one positive root.

The -discriminant case is more fun.

$$\lambda+1-\sqrt{(\lambda-1)^2+8} \le \lambda+1-\sqrt{8}.$$

Since the right half can be negative, we see that for at least some choices of $\lambda$ ($\lambda\le \sqrt{8}-1$), this is negative, giving that the minimum number of positive roots is one.

Answer 2

The discriminant is positive, so the equation has distinct roots for every $\lambda$.

The arithmetic mean of the roots is $(\lambda+1)/2$. Thus $\lambda\le-1$ implies at least a root is negative.

Can both roots be negative? Hint: Descartes' rule of signs.

Answer 3

If $\lambda = 2 $ then $ x = 0 $ is a non negative solution.