[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameters ]

For which values of the parameter $a$ are both roots of the equation $(2-a) x^{2}-3 a x+2 a=0$ greater than $1 / 2$?
When $a2$ the coefficient of $x^{2}$ and the constant term have different signs, meaning one of the roots is negative. When $a=0$ one of the roots is 0, and when $a=2$ the equation has only one root.

Consider the interval $016 / 17$. For these values of $a$

$f(1 / 2)=1 / 4(2-a)-3 / 2 a+2 a=1 / 4 a+1 / 2>0$ and the x-coordinate of the vertex $\frac{3 a}{2(2-a)}>\frac{\frac{48}{17}}{\frac{36}{17}}=\frac{3}{2}>\frac{1}{2}$, therefore, both roots are greater than $1 / 2$.

## Answer

For $16 / 17<a<2$