Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 579

а) х² − 2х − 9 = 0 D = b² − 4ac = (−2)² - 4 ⋅ 1 ⋅ (−9) = 4 + 36 = 40 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{2+\sqrt{40}}{2 ⋅ 1}=\frac{2+2 \sqrt{10}}{2} $$ = 1 + $$\sqrt{10}$$
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{2-\sqrt{40}}{2 ⋅ 1}=\frac{2-2 \sqrt{10}}{2} $$ = 1 − $$\sqrt{10}$$
По обратной теореме Виета х₁ + х₂ = 2, х₁х₂ = −9
б) 3t² − 4t − 4 = 0 D = b² − 4ac = (−4)² − 4 ⋅ 3 ⋅ (−4) = 16 + 48 = 64 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{4+\sqrt{64}}{2 ⋅ 3} $$ = 4 + 8/6 = 12/6 = 2
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{4-\sqrt{64}}{2 \sqrt{3}}= $$ = 4 − 8/6 = −4/6 = −2/3
По обратной теореме Виета х₁ + х₂ = −−4/3 = 4/3 = 11/3 х₁ + х₂ = 2 + (−2/3) = 2 − 2/3 = 11/3, верно
х₁х₂ = −4/3 = −11/3 х₁х₂ = 2 ⋅ (−2/3) = −4/3 = −11/3, верно
в) 2z² + 7z − 6 = 0 D = b² − 4ac = 7² − 4 ⋅ 2 ⋅ (−6) = 49 + 48 = 97 > 0, уравнение имеет 2 корня
x₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{-7+\sqrt{97}}{2 ⋅ 2}=\frac{-7+\sqrt{97}}{4} $$
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-7-\sqrt{97}}{2 ⋅ 2}=\frac{-7-\sqrt{97}}{4} $$
По обратной теореме Виета х₁ + х₂ = −7/2 = −3,5 х₁ + х₂ = $$ \frac{-7+\sqrt{97}}{4}+\frac{-7-\sqrt{97}}{4}=\frac{-7+\sqrt{97}-7-\sqrt{97}}{4} $$ = −14/4 = −3,5, верно
х₁х₂ = −6/2 = −3 х₁х₂ = $$ \frac{-7+\sqrt{97}}{4} \frac{-7-\sqrt{97}}{4}=\frac{(-7+\sqrt{97})(-7-\sqrt{97})}{16} $$ = 49 − 97/16 = −−48/16 = −3, верно
г) 2t² + 9t + 8 = 0 D = b² − 4ac = 9² − 4 ⋅ 2 ⋅ 8 = 81 − 64 = 17 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-9+\sqrt{17}}{2 ⋅ 2}=\frac{-9+\sqrt{17}}{4} $$
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-9-\sqrt{17}}{2 ⋅ 2}=\frac{-9-\sqrt{17}}{4} $$
По обратной теореме Виета х₁ + х₂ = −9/2 = −4,5 х₁ + х₂ = $$ =\frac{-9+\sqrt{17}}{4}+\frac{-9-\sqrt{17}}{4}=\frac{-9+\sqrt{17}-9-\sqrt{17}}{4} $$ = −18/4 = −4,5, верно
х₁х₂ = 8/2 = 4 х₁х₂ = $$ \frac{-9+\sqrt{17}}{4} \frac{-9-\sqrt{17}}{4}=\frac{(-9+\sqrt{17})(-9-\sqrt{17})}{16} $$ = 81 − 17/16 = 64/16 = 4, верно