Problem 6: Symmetrical Properties of Roots

Let the roots be $r$ and $r+5$

By Vieta the genius, we have $r+r+5=-(4-a)$ and $r(r+5)=-(a+1)$

Rearrange theses two equations to get $a=9+2r$ and $a=-r^2-5r-1$ and solve for $r$ , which is $-2,-5$

Using these two values of $r$ , solve for $a$ by substitution. It turns out $a=5,-1$ so the sum is $\boxed{4}$

We know $\alpha +\beta$ = a-4

$\alpha -\beta$ = 5

$\alpha \beta$ = -(a+1)

Now, we will evaluate

${ (\alpha -\beta ) }^{ 2 }\quad =\quad { \alpha }^{ 2 }{ +\beta }^{ 2 }-2\alpha \beta \\ \quad \quad \quad { 5 }^{ 2 }\quad =\quad { (\alpha +\beta ) }^{ 2 }-2\alpha \beta -2\alpha \beta \quad \\ \quad \quad 25\quad \quad =\quad { (a-4) }^{ 2 }\quad -\quad 4(-(a+1))\\ \quad \quad 25\quad \quad =\quad { a }^{ 2 }+16-8a+4a+4\\ \quad \quad 0\quad \quad \quad =\quad { a }^{ 2 }-4a+20-25\\ \quad \quad 0\quad \quad \quad =\quad { a }^{ 2 }-4a-5\\ \quad \quad 0\quad \quad \quad =\quad { a }^{ 2 }-5a+a-5\\ \quad \quad 0\quad \quad \quad =\quad (a+1)(a-5)\quad$

Hence,

a = -1 or 5

Therefore, Sum = 5-1 = 4

Let the roots be k & k+5. Now sum of roots= -(4-a). So k+(k+5)=a-4. 2k+5=a-4. So k=(a-9)/2. Again product of roots=-(a+1). So k*(k+5)=-a-1. k^2 +5k= -a-1. Substituting the value of k here, we get a quadratic equation in a, a^2 -4a -5=0 (after simplyfying)...so sum of possible values of k=4.