Rozwiązanie:
[tex]x^{2}+5px+20p-8=0[/tex]
Warunki:
[tex]\Delta>0\\\Delta=25p^{2}-4(20p-8)=25p^{2}-80p+32\\25p^{2}-80p+32>0\\\Delta_{p}=6400-4*25*32=6400-3200=3200\\p_{1}=\frac{80-40\sqrt{2} }{50} =\frac{8-4\sqrt{2} }{5} \\p_{2}=\frac{80+40\sqrt{2} }{50} =\frac{8+4\sqrt{2} }{5}\\p \in (-\infty, \frac{8-4\sqrt{2} }{5}) \cup (\frac{8+4\sqrt{2} }{5}, \infty)[/tex]
[tex]x_{1}^{2}+x_{2}^{2}=400\\(x_{1}+x_{2})^{2}-2x_{1}x_{2}=400\\25p^{2}-2(20p-8)=400\\25p^{2}-40p+16=400\\25p^{2}-40p-384=0\\\Delta_{p}=1600-4*25*(-384)=1600+38400=1600+38400=40000\\p_{1}=\frac{40-200}{50}=-\frac{16}{5} \\p_{2}=\frac{40+200}{50}=\frac{24}{5}[/tex]
Oba rozwiązania spełniają warunki zadania, więc ostatecznie mamy:
[tex]p \in [-\frac{16}{5},\frac{24}{5} ][/tex]
