Question

Прошу помогите с алгеброй на картинке задание

Answer 1

Ответ:

3.

$\boxed{x_{2} = 3,4}$

$\boxed{q = -34}$

4.

а) $\boxed{ \dfrac{1}{x_{1}} + \dfrac{1}{x_{2}} = -\dfrac{9}{17} }$

б) $\boxed{ \dfrac{x_{1}}{x_{2}} + \dfrac{x_{2}}{x_{1}} = -\dfrac{115}{17} }$

в) $\boxed{ x_{1}^{2} + x_{2}^{2} = 115 }$

Объяснение:

$\boxed{x \in \mathbb R}$

$ax^{2} + bx +c = 0$

По теореме Виета для квадратного уравнения:

$\displaystyle \left \{ {{ x_{1} + x_{2} = -\dfrac{b}{a} } \atop { x_{1} \cdot x_{2} = \dfrac{c}{a}}} \right.$

3.

$5x^{2} - 7x + q =0$

$x_{1} = -2$

По теореме Виета:

$\displaystyle \left \{ {{ -2 + x_{2} = -\dfrac{-7}{5} } \atop {-2 x_{2} = \dfrac{q}{5}}} \right$  

$\displaystyle \left \{ {{ x_{2} = 1,4 + 2 = 3,4 } \atop {q = -10x_{2} = -10 \cdot 3,4 = -34} \right$

4.

$x^{2} - 9x - 17 = 0$

$D = 81 - 4\cdot 1 \cdot (-17) = 81 + 68 = 149 > 0$

$D > 0 \Longrightarrow x \in \mathbb R$

По теореме Виета:

$\displaystyle \left \{ {{ x_{1} + x_{2} = -\dfrac{-9}{1} = 9 } \atop { x_{1} \cdot x_{2} = \dfrac{-17}{1}} = -17} \right.$

а)

$\dfrac{1}{x_{1}} + \dfrac{1}{x_{2}} = \dfrac{x_{1}+x_{2}}{x_{1}x_{2}} = \dfrac{9}{-17} = -\dfrac{9}{17}$

б)

$\dfrac{x_{1}}{x_{2}} + \dfrac{x_{2}}{x_{1}} = \dfrac{x_{1}^{2} + x_{2}^{2}}{x_{1}x_{2}} = \dfrac{(x_{1} + x_{2})^{2} - 2x_{1}x_{2}}{x_{1}x_{2}} = \dfrac{9^{2} - 2 \cdot(-17)}{-17} = \dfrac{81 + 34}{-17} = -\dfrac{115}{17}$

в)

$x_{1}^{2} + x_{2}^{2} = (x_{1} + x_{2})^{2} - 2x_{1}x_{2} =9^{2} - 2 \cdot(-17) =81 + 34 =115$