Question

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

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

1)

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

2)

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

3)

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

4)

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

5)

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

6)

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