Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 603

а) 10х² + 5х − 5 D = b² − 4ac = 5² − 4 ⋅ 10 ⋅ (−5) = 25 + 200 = 225 > 0, имеет 2 корня x₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{-5+\sqrt{225}}{2 ⋅ 10} $$ = −5 + 25/20 = 20/20 = 1 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-5-\sqrt{225}}{2 ⋅ 10} $$ = −5 − 25/20 = −30/20 = −1,5
x₁ = 1, x₂ = −1,5
б) −2х² + 12х − 18 D = b² − 4ac = 12² − 4 ⋅ (−2) ⋅ (−18) = 144 − 144 = 0 = 0, имеет 1 корень x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-12+\sqrt{0}}{2(-2)} $$ = −12/−4 = 3
x₁ = 3
в) х² − 2х − 4 D = b² − 4ac = (−2)² − 4 ⋅ 1 ⋅ (−4) = 4 + 16 = 20 > 0, имеет 2 корня x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{2+\sqrt{20}}{2 ⋅ 1}=\frac{2+2 \sqrt{5}}{2}= $$ = 1 + $$\sqrt{5}$$ x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{2-\sqrt{20}}{2 ⋅ 1}=\frac{2-2 \sqrt{5}}{2}= $$ = 1 − $$\sqrt{5}$$
x₁ = 1 + $$\sqrt{5}$$, x₂ = 1 − $$\sqrt{5}$$
г) 12х² − 12 12х² − 12 = 0 12х(х − 1) = 0 12х = 0, х₁ = 0 х − 1 = 0, х₂ = 1