[ | [ Investigation of a quadratic trinomial ] |
| :---: | :---: |

For what values of the parameter $a$ are both roots of the equation $(1+a) x^{2}-3 a x+4 a=0$ greater than 1?
Note that $a \neq -1$.

The discriminant $9a^2 - 16a(1 + a) > 0$, therefore $7a^2 + 16a - 1 > 0$. Solving this inequality narrows the range of values for $a$ to $(-\frac{16}{7}, -1)$. For such values of $a$, the branches of the parabola are directed downwards. Therefore, the value of the quadratic polynomial at the point 1 must be negative: $(1 + a) - 3a + 4a = 2a + 1 < 0$. This condition is satisfied over the entire specified interval.

## Answer

For $-\frac{16}{7} < a < -1$.