Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 619

а) 2х² + 12х − 14 D = b² − 4ac = 12² − 4 ⋅ 2 ⋅ (−14) = 144 + 112 = 256 > 0 имеет 2 корня x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-12+\sqrt{256}}{2 ⋅ 2} $$ = −12 + 16/4 = 4/4 = 1 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-12-\sqrt{256}}{2 ⋅ 2}= $$ = −12 − 16/4 = −28/4 = −7 x₁ = 1, x₂ = −7
б) −m² + 5m − 6 D = b² − 4ac = 5² − 4 ⋅ (−1) ⋅ (−6) = 25 − 24 = 1 > 0 имеет 2 корня m₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-5+\sqrt{1}}{2(-1)}= $$ = −5 + 1/−2 = −4/−2 = 2 m₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-5-\sqrt{1}}{2(-1)}= $$ = −5 − 1/−2 = −6/−2 = 3 m₁ = 2, m₂ = 3 −m² + 5m − 6 = −(х − 2)(х − 3)
в) 3х² + 5х − 2 D = b² − 4ac = 5² − 4 ⋅ 3 ⋅ (−2) = 25 + 24 = 49 > 0 имеет 2 корня x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-5+\sqrt{49}}{2 ⋅ 3} $$ = −5 + 7/6 = 2/6 = 1/3 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-5-\sqrt{49}}{2 ⋅ 3} $$ = −5 − 7/6 = −12/6 = −2 x₁ = 1/3, x₂ = −2 3х² + 5х − 2 = 3(x − 1/3)(x + 2) = (3х − 1)(х + 2)
г) 6х² − 12х + 6 D = b² − 4ac = (−12)² − 4 ⋅ 6 ⋅ 6 = 144 − 144 = 0 = 0 имеет 1 корень x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{12+\sqrt{0}}{2 ⋅ 6} $$ = 12 + 0/12 = 12/12 = 1 x₁ = 1 6х² − 12х + 6 = 6(х − 1)(х − 1) = 6(х − 1)²