7. (4 points) The leading coefficient of the quadratic trinomial $f(x)$ is 1. All three coefficients in some order form an arithmetic progression of three elements with a common difference $d$. Find all possible values of $d$, given that it is a rational number and the roots of $f(x)$ differ from each other by a factor of $d$.
Answer: $-1,-\frac{1}{2}$.

Solution: The coefficients of the trinomial have the following form: $a, a+d, a+2 d$ in some order. The roots of the trinomial can be represented as $b$ and $b d$.

Case 1: $a=1$, the other two coefficients: $1+d$ and $1+2 d$.

Subcase 1.1: $b+b d=-1-d$, so $b=-1$ or $d=-1$. If $b=-1$, we also write the equality for the constant term: $b^{2} d=1+2 d \Rightarrow d=1+2 d \Rightarrow d=-1$.

Subcase $1.2: b+b d=-1-2 d \Rightarrow b=-\frac{1+2 d}{1+d}$. Further, $b^{2} d=1+d \Rightarrow d(1+2 d)^{2}=(1+d)^{3} \Rightarrow 3 d^{3}+d^{2}-2 d-1=0$. The rational roots of this equation can only be $\pm 1, \pm \frac{1}{3}$, it is easy to verify that they do not fit.

Second case: $a+d=1$, the other two coefficients: $1-d, 1+d$

Subcase $2.1: b+b d=-1+d \Rightarrow b=-\frac{1-d}{1+d}$. Further, $b^{2} d=1+d=\Rightarrow d(1-d)^{2}=(1+d)^{3} \Rightarrow 5 d^{2}+2 d+1=0$. The rational roots of this equation can only be $\pm 1, \pm \frac{1}{5}$, it is easy to verify that they do not fit.

Subcase $2.2: b+b d=-1-d$, so $b=-1$ or $d=-1$. If $b=-1$, we also write the equality for the constant term: $b^{2} d=1-d \Rightarrow d=\frac{1}{2}$.

Third case: $a+2 d=1$, the other two coefficients: $1-2 d, 1-d$.

Subcase $3.1: b+b d=-1+2 d \Rightarrow b=-\frac{1-2 d}{1+d}$. Further, $b^{2} d=1-d \Rightarrow d(1-2 d)^{2}=(1-d) *(1+d)^{2} \Rightarrow$ $5 d^{3}-3 * d^{2}-1=0$. The rational roots of this equation can only be $\pm 1, \pm \frac{1}{5}$, it is easy to verify that they do not fit.

Subcase 3.2: $b+b d=-1+d \Rightarrow b=\frac{1-d}{1+d}$. Further, $b^{2} d=1-2 d \Rightarrow d(1-d)^{2}=(1-2 d)(1+d)^{2} \Rightarrow 3 d^{3}+d^{2}+d-1=0$ The rational roots of this equation can only be $\pm 1, \pm \frac{1}{5}$, it is easy to verify that they do not fit.