Question

Найдите корни квадратного трехчлена: 1) x^2-5x+6; 2) 2b^2-18; в) 0,3x^2+0,1x

Answer 1

$1.\\\\x^2 - 5x + 6 = 0$

По теореме Виета:

$\left \{\begin{array}{lcl} {{x_1 + x_2=5} \\ {x_1\cdot x_2 =6}}\end{array} \right. \Rightarrow \;\;x_1 = 3,\;\;x_2 = 2$

$2.\\\\2b^2 - 18 = 0 \quad \Big|:2\\\\b^2 = 9 \\\\b = \pm3$

$3.\\\\0,3x^2 + 0,1x = 0\\\\x (0,3x + 0,1) = 0\\\\x = 0 \qquad x = -\frac{1}{3}$