Question

Один из корней квадратного уравнения x2 - x + q = 0 на 4 больше другого.​

Answer 1

Ответ:

${x}^{2} - x + q = 0$

$x_{2} = x _{1} + 4$

$\left \{ {{{x_{1}}^{2} - x_{1} + q = 0 } \atop {{(x_{1} + 4)}^{2} - (x_{1} + 4) + q = 0 }} \right . \: = > \left \{ {{{x_{1}}^{2} - x_{1} + q = 0 } \atop {{(x_{1} + 4)}^{2} - x_{1} - 4 + q = 0 }} \right . \: = > \left \{ {{{x_{1}}^{2} - x_{1} + q = 0 } \atop {q = - {(x_{1} + 4) }^{2} + x_{1} + 4 }} \right .$

${x_{1}}^{2} - x_{1} - {(x_{1} + 4)}^{2} + x_{1} + 4 = 0 \\ {x_{1}}^{2} - ( {x_{1}}^{2} + 8x_{1} + 16) + 4 = 0 \\ {x_{1}}^{2} - {x_{1}}^{2} - 8x_{1} - 16 + 4 = 0 \\ - 8x_{1} - 12 = 0 \\ 2x_{1} + 3 = 0 \\ 2x_{1} = - 3 \\ x_{1} = - \frac{3}{2}$

${( - \frac{3}{2}) }^{2} - ( - \frac{3}{2} ) + q = 0 \\ \frac{9}{4} + \frac{6}{4} + q = 0 \\ \frac{15}{4} + q = 0 \\ q = - \frac{15}{4}$

$x_{2} = x_{1} + 4 = - \frac{3}{2} + 4 = - \frac{3}{2} + \frac{8}{2} = \frac{5}{2}$