<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] The quadratic polynomial $ \ x^2 + 2(\alpha-1)x - \alpha+7   \ \  $  in "vertex form" is $$ \ ( \  x \ + \ [\alpha - 1] \ )^2 \ + \ (7 - \alpha) - (\alpha - 1)^2 \ \ = \ \ ( \  x \ + \ [\alpha - 1] \ )^2 \ - \ ( \ \alpha^2 \ - \ \alpha \ - \ 6 \ ) $$
$$ = \ \ ( \  x \ + \ [\alpha - 1] \ )^2 \ - \ ( \ \alpha  \ + \ 2 \ )·( \ \alpha \  - \ 3 \ ) \ \ . $$
So the polynomial has distinct real zeroes only for $ \ \alpha \ < \ -2 \ $ or $ \ \alpha \ > \ 3 \ \ . \ $  Since the vertex must have a negative $ \ x-$coordinate if the two real zeroes $ \ ( \ x-$intercepts) are to be negative, we are only concerned with the case for $ \ \alpha \ > \ 3 \ \ . $  [For $ \ \alpha \ = \ 3 \ \ , \ $ the polynomial becomes $ \ x^2 + 4x + 4 \ = \  ( x + 2 )^2 \ \ . \ ] $
As $ \ \alpha \ $ increases, the two zeroes "spread" farther from the (increasingly negative) vertex, with the larger zero becoming less and less negative.  At $ \ \alpha \ = \ 7 \ \ , \ $ the "constant term" becomes zero (the polynomial is $ \ x^2 + 12x \ ) \ \ , \ $ with one of the zeroes now being $ \ x \ = \ 0 \ \ . \ $  Thus the permissible values are $ \ 3 \ < \ \alpha \ < \ 7 \ \ . $
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