<s>[INST] If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then

*

*$ \ \displaystyle{ \alpha<-2 }$  ;

*$ \ \displaystyle{ 3<\alpha<7 }$  ;

*it is impossible  ;

*none of (a)-(c).

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I have done the following :
The value $\alpha$ is real and such that the equation $x^2+2(\alpha-1)x-\alpha+7=0$ has two different real negative solutions.
The solutions of the equation are given from the quadratic formula \begin{align*}x_{1,2}&=\frac{-2(\alpha-1)\pm \sqrt{[2(\alpha-1)]^2-4\cdot 1\cdot (-\alpha+7)}}{2}\\ & =\frac{-2(\alpha-1)\pm \sqrt{4(\alpha^2-2\alpha-1)-4\cdot (-\alpha+7)}}{2}\\ & =-(\alpha-1)\pm \sqrt{2(\alpha^2-2\alpha-1)-2\cdot (-\alpha+7)} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-4\alpha-2+2\alpha-14} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}\end{align*}
So that we have two different solutions the discriminant must be non-zero.
So that we have two negative solutions, it must hold $-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}<0$.
So that we have real solutions the expression under the square root must be non negative.
The expression under the square root has the sign of the coefficient of $x^2$, i.e. positive, outside the roots.
We have that \begin{equation*}2\alpha^2-2\alpha-16=0 \Rightarrow \alpha_{1,2}=\frac{1}{2}\pm \frac{\sqrt{33}}{2}\end{equation*}
So we have that the expression under the square root if $\alpha<\frac{1}{2}- \frac{\sqrt{33}}{2}$ and if $\alpha>\frac{1}{2}+ \frac{\sqrt{33}}{2}$.
Is my attempt correct so far? Now do we check if the first two intervals of $\alpha$ can hold for all these conditions? Or how do we continue? Or is there a better way to solve that exercise ?
 [/INST] You made a slight algebra error.
$$a = 1; b = 2(\alpha - 1); c=-\alpha + 7$$
$$x = \frac{-2(\alpha - 1) \pm \sqrt{(2(\alpha - 1))^2 - 4(-\alpha + 7)}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4(\alpha^2 - 2\alpha + 1) + 4(\alpha - 7)}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4\alpha^2 - 8\alpha + 4 + 4\alpha - 28}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4\alpha^2 - 4\alpha - 24}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4}\sqrt{\alpha^2 - \alpha - 6}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4}\sqrt{\alpha^2 - \alpha - 6}}{2}$$
$$x = -\alpha + 1 \pm \sqrt{\alpha^2 - \alpha - 6}$$
For the roots to be real, the radicand must be nonnegative.  For the roots to be distinct, the radicand must be nonzero.  Thus,
$$\alpha^2 - \alpha - 6 > 0$$
$$(\alpha + 2)(\alpha - 3) > 0$$
$$\alpha < -2 \text{ or } \alpha > 3$$
But we also need both roots to be negative.  This is equivalent to the greater of the two roots being negative.
$$-\alpha + 1 + \sqrt{\alpha^2 - \alpha - 6} < 0$$
$$\sqrt{\alpha^2 - \alpha - 6} < \alpha - 1$$
But $\sqrt{\alpha^2 - \alpha - 6} > 0$, so by transitivity we must also have $\alpha - 1 > 0$, or $\alpha > 1$.  This rules out the $\alpha < -2$ possibility from earlier, so at this point we know $\alpha > 3$.  Anyhow, since both sides of the inequality are positive, we can square it without changing their order.
$$\alpha^2 - \alpha - 6 < (\alpha - 1)^2$$
$$\alpha^2 - \alpha - 6 < \alpha^2 - 2\alpha + 1$$
$$-6 < -\alpha + 1$$
$$-7 < -\alpha$$
$$7 > \alpha$$
Therefore, $3 < \alpha < 7$.
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