Question

50 ! х1 і х2 – корені рівняння х²-2х-7=0 знайдіть: 5х+х2-х1*х2

Answer 1

$x^{2} - 2x -7 = 0\\D = (-2)^{2} - 4 \cdot 1 \cdot (-7) = 4 + 28 = 32\\x_{1} = \dfrac{-(-2) + \sqrt{32} }{2 \cdot 1} = \dfrac{2 + 4\sqrt{2}}{2} = \dfrac{2(1 + 2\sqrt{2})}{2} = 1 + 2\sqrt{2} \\x_{2} = \dfrac{-(-2) - \sqrt{32} }{2 \cdot 1} = \dfrac{2 - 4\sqrt{2}}{2} = \dfrac{2(1 - 2\sqrt{2})}{2} = 1 - 2\sqrt{2}$

$5x + x_{2} - x_{1}x_{2} = 5x + 1 - 2\sqrt{2} - (-7) = 5x + 8 - 2\sqrt{2}$

Есть второй вариант решения, — если $x_{1} =1 - 2\sqrt{2}; \ x_{2} = 1 + 2\sqrt{2}$. Тогда $5x + x_{2} - x_{1}x_{2} = 5x + 8 + 2\sqrt{2}$