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6. $x^{2} +2x +q =0 \\$

Пусть х1 и х2 - корни, тогда по теореме Виета

$x_{1} + x_{2} = -2 \\ x_{1} x_{2} =q \\$
и по условию
x_{1} = 6 x_{2} \\ " alt=" \frac{ x_{1} }{ x_{2} } = 6 => x_{1} = 6 x_{2} \\ " align="absmiddle" class="latex-formula">
Тогда
x_{1} = 6 * (- \frac{2}{7} )= - \frac{12}{7}= -1 \frac{5}{7}\\ q = x_{1} x_{2} = (-1 \frac{5}{7})*(- \frac{2}{7} ) = \frac{12}{7}*\frac{2}{7}=\frac{24}{49} " alt="x_{1} + x_{2} = -2 \\ 6 x_{2} + x_{2} = -2 \\ 7x_{2} = -2 \\ x_{2} = - \frac{2}{7} => x_{1} = 6 * (- \frac{2}{7} )= - \frac{12}{7}= -1 \frac{5}{7}\\ q = x_{1} x_{2} = (-1 \frac{5}{7})*(- \frac{2}{7} ) = \frac{12}{7}*\frac{2}{7}=\frac{24}{49} " align="absmiddle" class="latex-formula">