4A. Let $x_{1}>0$ and $x_{2}=4 x_{1}$ be the roots of the quadratic equation $a x^{2}+b x+c=0$, for which the coefficients satisfy $3 a=2(c-b)$. Find the roots of the equation.
Solution. Note that $a \neq 0$, otherwise the equation is not quadratic. Using the formula for factoring a quadratic trinomial when the roots of the corresponding quadratic equation are known, we can write

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

Substituting the conditions of the problem, we get

$$
a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-4 x_{1}\right)=a x^{2}-5 a x_{1} x+4 a x_{1}^{2}
$$

Comparing the coefficients of the quadratic trinomials, we have $b=-5 a x_{1}$, $c=4 a x_{1}^{2}$. Substituting these values into the condition for the coefficients, we get $3 a=2\left(4 a x_{1}^{2}+5 a x_{1}\right)$. If the equation is divided by $a \neq 0$, it follows that $x_{1}$ is a root of the quadratic equation $8 x_{1}^{2}+10 x_{1}-3=0$, with solutions $x_{1_{1}}=\frac{1}{4}$ and $x_{1_{2}}=-\frac{3}{2}$. Given that $x_{1}>0$, we are left with $x_{1}=\frac{1}{4}, x_{2}=1$.