
Problem 9.1. Let $f(x)=x^{2}+6 a x-a$ where $a$ is a real parameter.

a) Find all values of $a$ for which the equation $f(x)=0$ has at least one real root.

b) If $x_{1}$ and $x_{2}$ are the real roots of $f(x)=0$ (not necessarily distinct) find the least value of the expression

$$
A=\frac{9 a-4 a^{2}}{\left(1+x_{1}\right)\left(1+x_{2}\right)}-\frac{70 a^{3}+1}{\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right)}
$$


Solution: a) The discriminant of $f(x)$ is $D=4\left(9 a^{2}+a\right)$. Therefore the equation $f(x)=0$ has at least one real root iff $9 a^{2}+a \geq 0$, so giving $a \in\left(-\infty,-\frac{1}{9}\right] \cup[0,+\infty)$.

b) Since $f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)$ we obtain $f(-1)=(-1-$ $\left.x_{1}\right)\left(-1-x_{2}\right)=\left(1+x_{1}\right)\left(1+x_{2}\right)=(-1)^{2}+6 a(-1)-a=1-7 a$ and $f(1-6 a)=\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right)=(1-6 a)^{2}+6 a(1-6 a)-a=$ $1-7 a$. Therefore the denominators of the two fractions of $A$ are equal to $1-7 a$ and

$$
A=\frac{-70 a^{3}-4 a^{2}+9 a-1}{1-7 a}=\frac{(1-7 a)\left(10 a^{2}+2 a-1\right)}{1-7 a}=10 a^{2}+2 a-1
$$

The quadratic function $g(a)=10 a^{2}+2 a-1$ attains its minimal value for $a=-\frac{1}{10} \notin\left(-\infty,-\frac{1}{9}\right] \cup[0,+\infty)$. Therefore the minimal value of $A$ equals to the smallest of the numbers $g(0)$ and $g\left(-\frac{1}{9}\right)$, i.e. this value is $g\left(-\frac{1}{9}\right)=-\frac{89}{81}$.
