6.251. For what positive $p$ are the roots of the equation

$5 x^{2}-4(p+3) x+4=p^{2}$

of opposite signs? Find these roots.
Solution.

Consider the quadratic equation $5 x^{2}-4(p+3) x+4-p^{2}=0$.

The roots of the quadratic equation are of opposite signs when the following system of inequalities is satisfied:

1) $\left\{\begin{array}{l}16(p+3)^{2}-4 \cdot 5\left(4-p^{2}\right)>0, \\ p>0, \\ (p+3)>0, \\ 4-p^{2}<0 ;\end{array}\right.$
2) $\left\{\begin{array}{l}16(p+3)^{2}-4 \cdot 5\left(4-p^{2}\right)>0 \\ p>0, \\ (p+3)<0, \\ 4-p^{2}>0 ;\end{array}\right.$

from which $p>2$. Solving the quadratic equation:

$$
\begin{aligned}
& x_{1,2}=\frac{2(p+3) \pm \sqrt{4(p+3)^{2}-5\left(4-p^{2}\right)}}{5}=\frac{2(p+3) \pm \sqrt{9 p^{2}+24 p+16}}{5}= \\
& =\frac{2(p+3) \pm \sqrt{9(p+4)^{2}}}{5}=\frac{2(p+3) \pm 3(p+4)}{5} \\
& x_{1}=\frac{2 p+6-3 p-4}{5}=\frac{-p+2}{5}, x_{2}=\frac{2 p+6+3 p+4}{5}=p+2
\end{aligned}
$$

Answer: $p>2 ; x_{1}=\frac{-p+2}{5}, x_{2}=p+2$.