Roots of functions f(x) and g(x)

Question

Two distinct polynomial f(x) and g(x) are defined as follows: $\ f(x)=x^2+ax+2$ and $\ g(x)=x^2+2x+a$. If the equation $\ f(x)=0$ and $\ g(x)=0$ have a common root then the sum of the roots of the equation $\ f(x)+g(x)=0$ is
(A)-1/2
(B)0
(C)1/2
(D)1
Whatever I could do:
they have a common root so
$\ x^2+ax+2=x^2+2x+a$
$\ ax+2=2x+a$
$\ (a-2)x=(a-2)$
$\ (a-2)x=(a-2)$
$\ x=1$
but I don't know what to do with this...

Answer 1

It's important to note that it's safe to divide both sides of the equation by $a - 2$ because if $a = 2$, then $f$ and $g$ are not distinct.

Continuing your work, since $x = 1$ is a root of $f$, we have that: $$ 0 = f(1) = 1^2 + a + 2 = a + 3 \implies a = -3 $$

Now let's analyze $f + g = 2x^2 - x - 1$. If it's two roots are $p$ and $q$, then we know that: $$ 2x^2 - x - 1 = 2(x - p)(x - q) = 2(x^2 - (p + q)x + pq) = 2x^2 - 2(p + q)x + 2pq $$ Comparing the coefficient of $x$, we see that $-1 = -2(p + q)$ so that the desired sum is: $$ p + q = \frac{1}{2} $$

Answer 2

Hint:

Exactly 1 common root

Let ${\alpha}$ be the common root of equation. $$a_1{\alpha}^2 + b_1{\alpha} +c_1=0$$ $$a_2{\alpha}^2 + b_2{\alpha} +c_2=0$$ $$\frac{{\alpha}^2}{b_1c_2-b_2c_1}=\frac{{\alpha}}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$$

Answer 3

Hint $\,f,\,g\,$ have same coeff's so $\,f(1)=g(1)\ (= $ coeff sum), so $\,x=1\,$ is a root of $\,f-g.\,$ But $\,0\ne f-g\,$ is linear so it has a unique root, which is necessarily the common root of $\,f,\,g.$

The rest is straightforward, e.g. evaluate $\,f\,$ at its root $\,x=1\,$ to solve for $\,a,\,$ then use Vieta find the root sum of $\,f+g.$