# 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).



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 ?

• Unfortunately you expanded the $(\alpha-1)^2$ incorrectly inside the square root. Commented Aug 13, 2022 at 0:15
• Also when you divide $\sqrt{4(\alpha^2-2\alpha+1)-4(-\alpha+7)}$ by $2$ the result is $\sqrt{(\alpha^2-2\alpha+1)-(-\alpha+7)}$, not $\sqrt{2(\alpha^2-2\alpha+1)-2(-\alpha+7)}$. Commented Aug 13, 2022 at 1:32
• Once you know that the x-coordinate of the vertex must be less than 0 which gives $-\frac{2(\alpha-1)}{2} < 0 \implies 1 < \alpha$, then you can check what happens when $\alpha = 3, 7$. Using the options is what you would do if you were in a competition. Commented Aug 13, 2022 at 4:55
• Thank you!! :-) Commented Aug 13, 2022 at 15:20

Sum of roots$$=-2(\alpha-1)<0\implies \alpha>1$$

Product of roots$$=(7-\alpha)>0\implies\alpha<7$$

with the condition that the discriminant, $$(2\alpha-2)^2-4(7-\alpha)>0\\\implies (\alpha-3)(\alpha+2)>0\\\implies \alpha>3\ or\ \alpha<-2\\ \therefore 3<\alpha<7$$

• Nearly there... Commented Aug 12, 2022 at 23:06
• @Suzu Hirose...Thanks, I fixed it now. Commented Aug 12, 2022 at 23:11
• Thanks a lot!! :-) Commented Aug 13, 2022 at 15:20

If $$r$$ and $$s$$ are the roots of $$x^2+2(\alpha-1)x-\alpha+7$$, then$$\left\{\begin{array}{l}r+s=2-2\alpha\\rs=7-\alpha.\end{array}\right.$$Therefore$$\left\{\begin{array}{l}2-2\alpha<0\\7-\alpha>0;\end{array}\right.$$in other words, $$\alpha\in(1,7)$$. But, in fact, we cannot have $$\alpha\in(1,3]$$, because $$(r-s)^2>0$$, and\begin{align}(r-s)^2>0&\iff(r+s)^2-4rs>0\\&\iff(2-2\alpha)^2-4(7-\alpha)>0\\&\iff4\alpha^2-4\alpha-24>0\\&\iff\alpha^2-\alpha+6>0\\&\iff(\alpha+2)(\alpha-3)>0\\&\iff\alpha<-2\vee\alpha>3.\end{align}So, $$\alpha\in(3,7)$$.

• Thank you very much!! :-) Commented Aug 13, 2022 at 15:20

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 $$\ 3 \ < \ \alpha \ < \ 7 \ \ .$$

• Thanks a lot!! :-) Commented Aug 13, 2022 at 15:20

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, $$3 < \alpha < 7$$.

• Thank you very much for your answer!! :-) Commented Aug 13, 2022 at 15:20