Question

Срочно!!!!! Даю 30 балов
$1) {x}^{2} - 5x + 6 = 0 \\ 2) {x}^{2} + 3x + 2 = 0 \\ 3)7 {x}^{2} + 7x + 5 = 0 \\ 4)2 {x}^{2} - 7x - 4 = 0 \\ 5)3 {x}^{2} - 17x + 10 = 0 \\ 6) {x}^{2} + 3x - 18 = 0$
​

Answer 1

Для квадратного уравнения вида ax²+bx+c=0 решение следующее: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

1)

$x^2-5x+6=0\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x_{1,\:2}=\frac{-(-5)\pm \sqrt{(-5)^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{5\pm \sqrt{1} }{2}\\\\x_1 = \frac{5+1}{2}=\frac{6}{2}=3 \\\\\:\:x_2= \frac{5-1}{2} = \frac{4}{2} =2$

2)

$x^2+3x+2=0\\\\x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-3 \pm \sqrt{1}}{2}\\\\x_{1}=\frac{-3 +1}{2} = -\frac{2}{2} =-1\\\\\:\:x_2=\frac{-3-1}{2}= -\frac{4}{2}=-2$

3)

$7x^2+7x+5=0\\\\D=49-140 <0 \Rightarrow x \in\varnothing$

4)

$2x^2-7x-4=0\\\\x_{1,\:2}=\frac{-(-7)\pm \sqrt{(-7)^2-4\cdot 2 (-4)}}{2\cdot \:2}\\\\x_{1,\:2}=\frac{7\pm \sqrt{64} }{4}\\\\x_{1}=\frac{7+9}{4} = \frac{16}{4}=4 \\\\x_{2}=\frac{7-9}{4} = -\frac{2}{4}=-\frac{1}{2}=-0,5$

5)

$3x^2-17x+10=0\\\\x_{1,\:2}=\frac{-(-17)\pm \sqrt{(-17)^2-4\cdot \:3\cdot \:10}}{2\cdot \:3}\\\\x_{1,\:2}=\frac{17 \pm \sqrt{169}}{6}\\\\x_{1}=\frac{17 +13 }{6} = \frac{30}{6}=5 \\\\x_{1}=\frac{17 -13 }{6} = \frac{4}{6}=\frac{2}{3}$

6)

$x^2+3x-18=0\\\\x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\left(-18\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-3\pm \sqrt{81}}{2}\\\\x_{1}=\frac{-3+ 9}{2} = \frac{6}{2} =3\\\\x_{2}=\frac{-3-9}{2} = -\frac{12}{2} =-6$