1. ## Roots of a quadratic equation

Given that the roots of the equation $x^2+ax+(a+2)=0$ differ by two, find the possible values of the constant a?

I used the symmetry axis of the parabola $\frac{-b}{2a}=1$to find one of the possible values =-2, but my teacher said, when we were checking this that 6 is also a possible answer. How is this possible? Is my method of solving this correct?

Thank you!

2. Originally Posted by Coach

Given that the roots of the equation $x^2+ax+(a+2)=0$ differ by two, find the possible values of the constant a?
Let $r_1,r_2$ be roots. Then $r_1+r_2 = a$ and $r_1r_2 = (a+2)$. Thus, $r_1r_2 = r_1+r_2+2$. But we also know that $r_1-r_2=2\implies r_1 = r_2+2$ thus $(r_2+2)r_2 = (r_2+2)+r_2+2$. Now solve.

3. I solved for $r$and I got 2 or -2, why is only -2 a possible solution?

I wonder how my teacher then said that also six was a possible answer?

I know this is probably a stupid question, and that I am breaking all internet etiquette rules by pleading for an answer, but I really need this since I have a test tomorrow.

Thank you.

4. Originally Posted by Coach
Given that the roots of the equation $x^2+ax+(a+2)=0$ differ by two, find the possible values of the constant a?
Hello,

the text of your problem indicates that there are real roots:
$x_1=s~\vee~x_2=t$
Then you can write the equation as:

$(x-s)(x-t)=x^2+ax+(a+2)=0$ Expand

$x^2-(s+t)x+st=x^2+ax+(a+2)$
Use the property that the roots should differ by 2. You'll get a system of equations:

$\left \{\begin{array}{l}I:\ -(s+t)=a \\ II: s-t=2 \\III:\ s \cdot t = a+2\end{array} \right.$

Plug a from I into III:

$\left \{\begin{array}{l}II: s-t=2~\iff~s=t+2 \\III:\ s \cdot t = -(s+t)+2\end{array} \right.$

Plug s from II into III:

$(t+2)t=-t-2-t+2~\iff~ t^2+2t=-2t~ \iff~t^2+4t=0~ \iff~t=0~\vee~t=-4$

Then $s = 2~\vee~ s= -2$ and $a=-2~\vee~a=6$

5. Thank you so much!