1. Determine $a$ such that the roots of the equation $a x^{2}+x^{2}+9 a x-x-9 a=0$ are reciprocals of each other.
1. The equation is rearranged to the form $x^{2}(a+1)+x(9 a-1)-9 a=0$. The roots of the equation $x_{1}$ and $x_{2}$ are reciprocals if $x_{1}=\frac{1}{x_{2}}$ or $x_{1} \cdot x_{2}=1$. Considering the formula $x_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}$, we have $\frac{-b-\sqrt{D}}{2 a} \cdot \frac{-b+\sqrt{D}}{2 a}=1$. The equation is rearranged to the form $b^{2}-D=4 a^{2}$ and we use the relation $D=b^{2}-4 a c$. We get $a=c$, which means the leading coefficient must be equal to the constant term. For the given equation, this means $a+1=-9 a$, which yields the solution $a=-\frac{1}{10}$.

Rearrangement of the equation: $x^{2}(a+1)+x(9 a-1)-9 a=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Consideration of the formula: $x_{1,2}=\frac{-b \pm \sqrt{D}}{2 a} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Equation set up from $x_{1}=\frac{1}{x_{2}} \Rightarrow \frac{2 a}{-b-\sqrt{D}}=\frac{-b+\sqrt{D}}{2 a} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-10.jpg?height=63&width=1639&top_left_y=1462&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-10.jpg?height=54&width=1639&top_left_y=1521&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-10.jpg?height=109&width=1642&top_left_y=1559&top_left_x=207)