1. Let $x_{1}$ and $x_{2}$ be the distinct roots of the polynomial $p(x)=x^{2}+a x+b, x_{1}^{2}-\frac{1}{2}$ and $x_{2}^{2}-\frac{1}{2}$ are the roots of the polynomial $q(x)=x^{2}+\left(a^{2}-\frac{1}{2}\right) x+b^{2}-\frac{1}{2}$. Determine $a$ and $b$.
IV /1. The roots of the polynomial $p(x)=x^{2}+a x+b$ are $x_{1}=\frac{-a+\sqrt{a^{2}-4 b}}{2}$ and $x_{2}=\frac{-a-\sqrt{a^{2}-4 b}}{2}$. Since they must be different, $a^{2}-4 b \neq 0$. From this, we calculate

$$
x_{1}^{2}-\frac{1}{2}=\frac{\left(a^{2}-2 b-1\right)-a \sqrt{a^{2}-4 b}}{2}
$$

and

$$
x_{2}^{2}-\frac{1}{2}=\frac{\left(a^{2}-2 b-1\right)+a \sqrt{a^{2}-4 b}}{2}
$$

Since these are the roots of the polynomial $q(x)$, they must satisfy the equation $x^{2}+\left(a-\frac{1}{2}\right) x+b^{2}-\frac{1}{2}=0$. When we substitute both into this equation and rearrange, we get

$$
\left(4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b\right)-\left(4 a^{3}-4 a b-3 a\right) \sqrt{a^{2}-4 b}=0
$$

and

$$
\left(4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b\right)+\left(4 a^{3}-4 a b-3 a\right) \sqrt{a^{2}-4 b}=0
$$

Thus, it must be that

$$
4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b=0
$$

and

$$
\left(4 a^{3}-4 a b-3 a\right) \sqrt{a^{2}-4 b}=0
$$

Since $a^{2}-4 b \neq 0$, from the second equation it follows that $\left(4 a^{3}-4 a b-3 a\right)=a\left(4 a^{2}-4 b-3\right)=0$. If $a=0$, then from the first equation it follows that $8 b^{2}+6 b=2 b(4 b+3)=0$. Since $a^{2}-4 b \neq 0$, $b \neq 0$, so $b=-\frac{3}{4}$. If $a \neq 0$, then it must be that $4 a^{2}-4 b-3=0$ or $b=a^{2}-\frac{3}{4}$. When we substitute this into the first equation, we get $-2 a^{2}=0$ or $a=0$, which is a contradiction. Therefore, $a=0$ and $b=-\frac{3}{4}$.

2. method. By Vieta's formulas, $x_{1}+x_{2}=-a, x_{1} x_{2}=b, x_{1}^{2}+x_{2}^{2}-1=\frac{1}{2}-a^{2}$ and $\left(x_{1}^{2}-\frac{1}{2}\right)\left(x_{2}^{2}-\frac{1}{2}\right)=b^{2}-\frac{1}{2}$. First, let's rearrange the third equation, taking into account the first two,

$$
\frac{3}{2}-a^{2}=x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=a^{2}-2 b
$$

from which we get $b=a^{2}-\frac{3}{4}$. Now, let's rearrange the fourth equation,

$$
b^{2}-\frac{1}{2}=\left(x_{1}^{2}-\frac{1}{2}\right)\left(x_{2}^{2}-\frac{1}{2}\right)=x_{1}^{2} x_{2}^{2}-\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}\right)+\frac{1}{4}=b^{2}-\frac{1}{2}\left(\frac{3}{2}-a^{2}\right)+\frac{1}{4}
$$

from which it follows that $a^{2}=0$ and thus $a=0$ and $b=-\frac{3}{4}$.

## 1. method

Expression of $x_{1}$ and $x_{2}$ in terms of $a$ and $b$ using the quadratic equation 1 point

Expression of $x_{1}^{2}-\frac{1}{2}$ and $x_{2}^{2}-\frac{1}{2}$ in terms of $a$ and $b$

Substitution of these roots, expressed in terms of $a$ and $b$, into the polynomial $q(x)$

Writing the system of equations $4 a^{4}-12 a^{2} b+8 b^{2}-5 a^{2}+6 b=0$ and $\left(4 a^{3}-4 a b-3 a\right) \sqrt{a^{2}-4 b}=0$

Correct solution of the system of equations

Writing the solution $a=0, b=-\frac{3}{4}$

## 2. method

Use of Vieta's formulas

1 point

Correct writing of all four equations with Vieta's formulas $\left(x_{1}+x_{2}=-a, x_{1} x_{2}=b\right.$,

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-19.jpg?height=69&width=1637&top_left_y=274&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-19.jpg?height=62&width=1633&top_left_y=326&top_left_x=220)
Expression of the equation $\left(x_{1}^{2}-\frac{1}{2}\right)\left(x_{2}^{2}-\frac{1}{2}\right)=b^{2}-\frac{1}{2}$ in terms of $a$ and $b \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$.

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-19.jpg?height=65&width=1610&top_left_y=430&top_left_x=220)