Find the quadratic polynomial, the sum of whose zeros is

Let $\alpha$ and $\beta$ be the zeros of the required polynomial $f(x)$.

Then $(\alpha+\beta)=\frac{5}{2}$ and $\alpha \beta=1$

$\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$

$=>f(x)=x^{2}-\frac{5}{2} x+1$

$=>f(x)=2 x^{2}-5 x+2$

Hence, the required polynomial is $f(x)=2 x^{2}-5 x+2$.

$\therefore f(x)=0=>2 x^{2}-5 x+2=0$

$=>2 x^{2}-(4 x+x)+2=0$

$=>2 x^{2}-4 x-x+2=0$

$=>2 x(x-2)-1(x-2)=0$

$=>(2 x-1)(x-2)=0$

$=>(2 x-1)=0$ or $(x-2)=0$

$=>x=\frac{1}{2}$ or $x=2$

So, the zeros of $f(x)$ are $\frac{1}{2}$ and 2 .