Question

Помогите с 27 заданием​

Answer 1

$\displaystyle\bf\\x^{2} +(3-b)x-6b=0\\\\Teorema \ Vieta:\\\\x_{1} +x_{2} =-(3-b)\\\\x_{1} \cdot x_{2} =-6b\\\\\\x_{1} ^{2}+x_{2} ^{2} =\Big(\underbrace{x_{1} + x_{2} }_{b-3} \Big)^{2} -2\cdot \underbrace{x_{1} \cdot x_{2} }_{-6b}=\Big(b-3\Big)^{2} -2\cdot\Big(-6b\Big)=\\\\\\=b^{2} -6b+9+12b=b^{2} +6b+9=\Big(b+3\Big)^{2} \\\\\\\Big(b+3\Big)^{2} =16\\\\\\\left[\begin{array}{ccc}b+3=-4\\b+3=4\end{array}\right\\\\\\\left[\begin{array}{ccc}b=-4-3\\b=4-3\end{array}\right$

$\displaystyle\bf\\\left[\begin{array}{ccc}b_{1} =-7\\b_{2} =1\end{array}\right\\\\\\Otvet \ : \ -7 \ ; \ 1$