A Common Root

Observe that $b = 0$ is not a solution. This will allow us to divide by $b$ subsequently.

We have

$2x^3 + 7x^2 + 5x + 1 - (x+1) ( 2x^2 + 5x + 2b) = -2bx - 2b + 1$

Hence, the common root must satisfy $0 = -2bx -2b + 1 \Rightarrow x = \frac{ 1-2b}{2b}$ .

Substituting this into the first equation (either equation works) and multiplying by $(2b)^2$ we get

$0 = 8b^3 - 12b^2 + 2b + 2 = 2 (b-1) ( 4b^2 - 2b -1 )$

which has 3 real roots (check the discriminant).

For each of these values of $b$ , we get a corresponding value of $\alpha = \frac {1-2b}{b}$ which is the common root. Hence, by Vieta, the sum of these values are $\frac{12}{8} = \frac{3}{2}$ .

2 x^3 + 7 x^2 + 5 x + 1 = (2 x + 1)(x^3 + 3 x + 1) = 0 means this is not a genuine cubic equation. x = {-0.5, (- 3 + sqrt(5))/ 2, (- 3 - sqrt(5))/ 2)}.

2 x^2 + 5 x + 2 b = 0 => b = - x^2 - 2.5 x.

Hence, b = {1, -0.309016694, 0.809016994}

Therefore, sum = 1.5 = 3/ 2.

3 + 2 = 5.

First, choosing two more general equations:

${ x }^{ 2 }+ax+b=0$ and ${ x }^{ 3 }+(a+1){ x }^{ 2 }+qx+r=0$ .

Take $\alpha$ to be a common root of the two equations. From now on, these two equations are referred to as (1) and (2) respectively.

From (1), we have that: ${ \alpha }^{ 2 }=-b\alpha -b$ ,

and substituting this into (2), it follows that: $\alpha (-b\alpha -b)+(a+1)(-b\alpha -b)+q\alpha +r=0$ .

Rearranging this to make $\alpha$ the subject yields:

$\alpha =-\frac { b-r }{ a+b-q }$ .

Substituting this into (1), we have that: ${ \frac { { (b-r) }^{ 2 } }{ { (a+b-q) }^{ 2 } } }-a\frac { (b-r) }{ (a+b-q) } +b=0$ .

Tidying up gives an equation now referred to as $(\ast )$ :

${ (b-r) }^{ 2 }-a(b-r)(a+b-q)+b{ (a+b-q) }^{ 2 }=0$ ,

which is satisfied if and only if (1) and (2) have a common root.

Halving both sides of each of the equations given in the question, and denoting these as (i) and (ii) respectively, we have that:

${ x }^{ 2 }+\frac { 5 }{ 2 } x+b=0$ and

${ x }^{ 3 }+\frac { 7 }{ 2 } { x }^{ 2 }+\frac { 5 }{ 2 } x+\frac { 1 }{ 2 } =0$ .

Comparing the coefficients of these equations with the forms of the general equations that we initially stated, we have that:

$a=\frac { 5 }{ 2 }$ , $b=b$ , $q=\frac { 5 }{ 2 }$ , and $r=\frac { 1 }{ 2 }$ .

Putting these into $(\ast )$ , then expanding and simplifying to a monic polynomial yields:

${ b }^{ 3 }-\frac { 3 }{ 2 } { b }^{ 2 }+\frac { 1 }{ 4 } b+\frac { 1 }{ 4 } =0$ .

Assuming that all three roots of the cubic are real (though this is implied in the wording of the question), by Vieta's Formula, the sum of the values of b that satisfy the cubic equation in b is:

$\frac { -\frac { 3 }{ 2 } }{ -1 } =\frac { 3 }{ 2 }$ .

More rigorously, a factor of $(b-1)$ should be taken out of the cubic, and then the resolving quadratic factorized, showing that all three values of b are indeed real.

Hence, the final result is: $3+2=\boxed { \quad 5\quad }$