B1. Find all real numbers $a \neq-1$, for which the ratio between the solutions of the quadratic equation $(2 a+2) x^{2}+(2 a+3) x+1=0$ is $1: 3$.
B1. We observe that the left side of the equation can be factored to yield

$$
((2 a+2) x+1)(x+1)=0
$$

The solutions are thus $x_{1}=-1$ and $x_{2}=-\frac{1}{2 a+2}$. Therefore, $x_{2}$ must be either $-3$ or $-\frac{1}{3}$. In the first case, we get $2 a+2=\frac{1}{3}$, from which we obtain $a=-\frac{5}{6}$. In the second case, $2 a+2=3$ or $a=\frac{1}{2}$.

2. method. Let the solutions of the equation be $x_{1}=t$ and $x_{2}=3 t$. By Vieta's formulas, we have

$$
\begin{gathered}
4 t=x_{1}+x_{2}=-\frac{2 a+3}{2 a+2} \\
3 t^{2}=x_{1} x_{2}=\frac{1}{2 a+2}
\end{gathered}
$$

From the first equation, we express $t=-\frac{2 a+3}{4(2 a+2)}$ and substitute it into the second equation to get

$$
\frac{1}{2 a+2}=\frac{3(2 a+3)^{2}}{16(2 a+2)^{2}}
$$

We eliminate the fractions: $16(2 a+2)=3(2 a+3)^{2}$ and simplify the equation to $12 a^{2}+4 a-5=0$. The solutions to this quadratic equation are $a=\frac{1}{2}$ and $a=-\frac{5}{6}$.

## 3. method.

The solutions to the quadratic equation are $x_{1,2}=\frac{-2 a-3 \pm \sqrt{(2 a+3)^{2}-8(a+1)}}{4(a+1)}$. The discriminant is a perfect square $(2 a+3)^{2}-8(a+1)=(2 a+1)^{2}$, so the roots simplify to $x_{1}=-\frac{1}{2(a+1)}, x_{2}=-1$. The ratio of the roots is then $x_{1} / x_{2}=3^{\pm 1}$, leading to the pair of solutions $a=-5 / 6$ and $a=1 / 2$.

1. method:

- factorization of the expression $((2 a+2) x+1)(x+1)=0$ 4 points
- two possible ratios: $x_{1} / x_{2}=1 / 3$ or $x_{2} / x_{1}=1 / 3 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ points
- correct final results $a=1 / 2$ and $a=-5 / 6 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

## 2. method:

- roots $x_{1}=t, x_{2}=3 t$. .1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-10.jpg?height=74&width=1733&top_left_y=568&top_left_x=184)

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-10.jpg?height=71&width=1733&top_left_y=661&top_left_x=184)

- Formulated into a quadratic equation $12 a^{2}+4 a-5=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-10.jpg?height=74&width=1728&top_left_y=831&top_left_x=187)
3. method:

- equation for the roots of the quadratic function .1 point
- writing the discriminant as a perfect square 2 points
- simplified form of the roots .1 point
- two possible ratios: $x_{1} / x_{2}=1 / 3$ or $x_{2} / x_{1}=1 / 3$ .2 points
- correct final results $a=1 / 2$ and $a=-5 / 6$ .1 point

If only one possible ratio of the roots is written, 1 point is deducted.