If $f\left(x\right)$  is a quadratic polynomial such that $\mathrm{f}\left(1\right)+\mathrm{f}\left(2\right)=0$  and “-2” is a root of $\mathrm{f}\left(\mathrm{x}\right)=0$. Then the other root of $\mathrm{f}\left(\mathrm{x}\right)=0$ is

Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$

$\begin{array}{l}\therefore \mathrm{f}\left(1\right)=\mathrm{a}+\mathrm{b}+\mathrm{c};\mathrm{f}\left(2\right)=4\mathrm{a}+2\mathrm{b}+\mathrm{c}\\ \therefore \mathrm{f}\left(1\right)+\mathrm{f}\left(2\right)=0\Rightarrow 5\mathrm{a}+3\mathrm{b}+2\mathrm{c}=0........\left(1\right)\end{array}$

since -2 is a root of f(x)=0

$\therefore \mathrm{a}{(-2)}^2+\mathrm{b}(-2)+\mathrm{c}=0\Rightarrow 4\mathrm{a}-2\mathrm{b}+\mathrm{c}=0........\left(2\right)$

From (1) and (2), we get  $3\mathrm{a}-7\mathrm{b}=0\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{3}{7}$
If other root of f(x)=0 is $\alpha$ , then sum of roots
$\mathrm{\alpha }+(-2)=-\frac{\mathrm{b}}{\mathrm{a}}\Rightarrow \mathrm{\alpha }-2=-\frac{3}{7}\Rightarrow \mathrm{\alpha }=\frac{11}{7}$.