Problem 7. For what values of the parameter $a$ does the equation

$$
\log _{2}^{2} x+(a-6) \log _{2} x+9-3 a=0
$$

have exactly two roots, one of which is four times the other?
Answer: $-2.2$

Solution. Let $t=\log _{2} x$, then the equation becomes $t^{2}+(a-6) t+(9-3 a)=0$. Notice that $3 \cdot(3-a)=9-3 a, 3+(3-a)=6-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -3 and $3-a$. We make the reverse substitution: $\log _{2} x=3$ or $\log _{2} x=3-a$, i.e., either $x=8$, or $x=2^{3-a}$. We get two cases: $8=4 \cdot 2^{3-a}$ or $2^{3-a}=4 \cdot 8$. In the first case $a=2$, in the second $a=-2$.