Question

РЕШИТЕ ПОЖАЛУЙСТА НОМЕР 2

Answer 1

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Answer 2

$x^2+2px-3q=0\; ,\; \; x_1=p\; ,\; \; x_2=4q\\\\\left \{ {{x_1\cdot x_2=-3q} \atop {x_1+x_2=-2p}} \right. \; \left \{ {{p\cdot 4q+3q=0} \atop {p+4q=-2p}} \right. \left \{ {{q(4p+3)=0} \atop {4q+3p=0}} \right. \; \left \{ {{q=0\; ,\; \; p=-\frac{3}{4}} \atop {3p+4q=0}} \right. \\\\a)\; \; \left \{ {{q=0} \atop {3p=-4q}} \right. \; \left \{ {{q=0} \atop {p=0}} \right.\\\\b)\; \; \left \{ {{p=-\frac{3}{4}} \atop {4q=-3p}} \right. \; \left \{ {{p=-\frac{3}{4}} \atop {q=-\cdot \frac{3p}{4}}} \right. \; \left \{ {{p=-\frac{3}{4}} \atop {q=\frac{9}{16}}} \right. \\\\Otvet:\; \; p=0\; ,\; q=0\; \; \; ili\; \; \; p=-\frac{3}{4}\; ,\; \; q=\frac{9}{16}\; .$