Question

решите срочно пожалуйста!!

Answer 1

$\displaystyle\bf\\x^{2} +3(b+2)x-5b=1-b\\\\1)\\\\x=0\\\\0^{2} +3(b+2)\cdot 0-5b=1-b\\\\-5b=1-b\\\\-5b+b=1\\\\-4b=1\\\\\boxed{b=-0,25}$

$\displaystyle\bf\\x^{2} +3(b+2)x-5b=1-b\\\\2)\\\\x=-1\\\\b\cdot(-1)^{2} +3(b+2)\cdot (-1)-5b=1-b\\\\b-3b-6-5b=1-b\\\\-7b-6=1-b\\\\-7b+b=1+6\\\\-6b=7\\\\b=-\frac{7}{6} =-1\frac{1}{6}$