Be carefulll

Let $ax^2+bx+c=0$ have roots $x_1,x_2$ and $dx^2+ex+f=0$ have roots $x_2,x_3$ . Then, by Vieta's formulas:

$x_1+x_2=-\dfrac{b}{a} ... (1) \\ x_1x_2=\dfrac{c}{a} ... (2) \\ x_2+x_3=-\dfrac{e}{d} ... (3) \\ x_2x_3=\dfrac{f}{d} ... (4)$

Then with the identity $(A-B)^2=(A+B)^2-4AB$ we can obtain:

$x_1-x_2=\dfrac{\sqrt{b^2-4ac}}{a} ... (5)\\ x_2-x_3=\dfrac{\sqrt{e^2-4df}}{d} ... (6)$

Subtract $(3)$ from $(1)$ :

$x_1-x_3=-\dfrac{b}{a}+\dfrac{e}{d} ... (7)$

Sum $(5)$ and $(6)$ :

$x_1-x_3=\dfrac{\sqrt{b^2-4ac}}{a}+\dfrac{\sqrt{e^2-4df}}{d} ... (8)$

Equate $(7)$ and $(8)$ :

$-\dfrac{b}{a}+\dfrac{e}{d}=\dfrac{\sqrt{b^2-4ac}}{a}+\dfrac{\sqrt{e^2-4df}}{d}$

Simplify and square both sides:

$ae-bd=d\sqrt{b^2-4ac}+a\sqrt{e^2-4df}$

$(ae-bd)^2=(d\sqrt{b^2-4ac}+a\sqrt{e^2-4df})^2$

$\cancel{a^2e^2}-2abde+\cancel{b^2d^2}=\cancel{b^2d^2}-4acd^2+\cancel{a^2e^2}-4a^2df+2ad\sqrt{(b^2-4ac)(e^2-4df)}$

Divide both sides by $2ad$ :

$2cd-be+2af=\sqrt{(b^2-4ac)(e^2-4df)}$

Square again:

$(2cd-be+2af)^2=(b^2-4ac)(e^2-4df)$

$4c^2d^2+\cancel{b^2e^2}+4a^2f^2-4bcde+8acdf-4abef=\cancel{b^2e^2}-4b^2df-4ace^2+16acdf$

$4a^2f^2-8acdf+4c^2d^2=4bcde+4abef-4b^2df-4ace^2$

$\cancel{4}(af-cd)^2=\cancel{4}(bcde+abef-b^2df-ace^2)$

$(af-cd)^2=bd(ce-bf)+ae(bf-ce)$

$(af-cd)^2=(ae-bd)(bf-ce)$

In this case the equations are $2x^2+bx+1=0$ and $5x^2+x+b=0$ , hence the condition is:

$(2b-5)^2=(2-5b)(b^2-1)$

$4b^2-20b+25=2b^2-2-5b^3+5b$

$5b^3+2b^2-25b+27=0$

The discriminant of that equation is less tha cero, hence there are no repeated roots. By Vieta's formulas, the sum of all values of $b$ is $-\dfrac{2}{5}=\boxed{-0.4}$ .

If $ax^2 + bx + c = 0$

$dx^2 + ex + f = 0$

then condition for common roots is :

$(dc - af)^2 = (bf - ec)(ae - db)$

Applying it here, we get :

$5b^3 + 2b^2 - 25b + 27 = 0$

Using vieta, sum of all possible values of $b$ is -0.4

There is a direct formula for the condition of two quadratics to have a common root. But I would like to present a solution that shows where the answer came from. Apologies for the lack of knowledge of LaTeX.

Let's say the two quadratics have a common root. Multiply a constant factor to one of them and make their leading coefficients equal. Subtract them. If two terms are equal to zero, their difference is zero as well. And because the difference is a linear term, it's root is the common root!

But we need to put this thing in to any one of the equations and make it equal to zero, as it was a mere assumptions that the quadratics ACTUALLY had a common root. We get a cubic in this case in terms of b. What are you looking at now? Put Vieta to some use!