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revocs
@revocs
November 2021
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МНЕ НУЖНО ОЧЕНЬ СРОЧНО
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zinaidazina
1.
1) Решаем уравнение:
2х² - 14х + 20 = 0
D = b² - 4ac
D = 196 - 4 ·2·20 = 196 - 160 = 36
√D = √36 = 6
x₁ = (14-6)/4=12/4=3
x₂ = (14+6)/4=20/4=5
2) Каждое из полученных корней х₁ = 3 и х₂ = 5 умножим на 4.
Х₁ = 4х₁ = 3·4 = 12
Х₂ = 4х₂ = 5·4 = 20
3) Составим уравнение х² + рх + q = 0.
По теореме Виета находим р и q
{X₁ + X₂ = - р
{X₁ · X₂ = q
{12 + 20 = - p
{12 ·20 = q
p = - 32; q = 240
Уравнение :
х² - 32х + 240 = 0
2.
1) m²+8m-9=0
D = 64 -4·1·(-9) = 64 +36 = 100
√D = √100 = 10
x₁ = (-8-10)/2 = -18/2 = - 9
x₂ = (-8+10)/2 = 2/2 = 1
m²+8m-9= (m+9)(m-1)
2) m²+12m+27=0
D = 144 -4·1·27 = 144 -108 = 36
√D = √36 = 6
x₁ = (-12-6)/2 = -18/2 = - 9
x₂ = (-12+6)/2 = -6/2 = - 3
m² + 12 m + 27 = (m+9)(m+3)
3)
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Answers & Comments
1) Решаем уравнение:
2х² - 14х + 20 = 0
D = b² - 4ac
D = 196 - 4 ·2·20 = 196 - 160 = 36
√D = √36 = 6
x₁ = (14-6)/4=12/4=3
x₂ = (14+6)/4=20/4=5
2) Каждое из полученных корней х₁ = 3 и х₂ = 5 умножим на 4.
Х₁ = 4х₁ = 3·4 = 12
Х₂ = 4х₂ = 5·4 = 20
3) Составим уравнение х² + рх + q = 0.
По теореме Виета находим р и q
{X₁ + X₂ = - р
{X₁ · X₂ = q
{12 + 20 = - p
{12 ·20 = q
p = - 32; q = 240
Уравнение :
х² - 32х + 240 = 0
2.
1) m²+8m-9=0
D = 64 -4·1·(-9) = 64 +36 = 100
√D = √100 = 10
x₁ = (-8-10)/2 = -18/2 = - 9
x₂ = (-8+10)/2 = 2/2 = 1
m²+8m-9= (m+9)(m-1)
2) m²+12m+27=0
D = 144 -4·1·27 = 144 -108 = 36
√D = √36 = 6
x₁ = (-12-6)/2 = -18/2 = - 9
x₂ = (-12+6)/2 = -6/2 = - 3
m² + 12 m + 27 = (m+9)(m+3)
3)