If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions...

Question

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 ?

Answer 1

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, $(2\alpha-2)^2-4(7-\alpha)>0\\\implies (\alpha-3)(\alpha+2)>0\\\implies \alpha>3\ or\ \alpha<-2\\ \therefore 3<\alpha<7$

Answer 2

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, $\alpha\in(3,7)$.

Answer 3

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 \ \ . $

Answer 4

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, $3 < \alpha < 7$.