Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 594

а) (3x + 1)² = 3x + 1 (3х)² + 6х + 1 − 3х − 1 = 0 9х² + 3х = 0 3х(3х + 1) = 0 3х = 0 х₁ = 0 3х + 1 = 0 3х = −1 х₂ = −1/3
б) (3x + 1)² = 3(x + 1) (3х)² + 6х + 1 − 3х − 3 = 0 9х² + 3х − 2 = 0 D = b² − 4ac = 3² − 4 ⋅ 9 ⋅ (−2) = 9 + 72 = 81 > 0, имеет 2 корня х₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{-3+\sqrt{81}}{2 ⋅ 9} $$ = −3 + 9/18 = 6/18 = 1/3 х₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-3-\sqrt{81}}{2 ⋅ 9} $$ = −3 − 9/18 = −12/18 = −2/3
в) (3x + 1)² = (2x − 5)² (3х)² + 6х + 1 = (2х)² − 20х + 25 9х² + 6х + 1 − 4х² + 20х − 25 = 0 5х² + 26х − 24 = 0 D = b² − 4ac = 26² − 4 ⋅ 5 ⋅ (−24) = 676 + 480 = 1156 > 0, имеет 2 корня х₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-26+\sqrt{1156}}{2 ⋅ 5} $$ = −26 + 34/10 = 8/10 = 0,8 х₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-26-\sqrt{1156}}{2 ⋅ 5} $$ = −26 − 34/10 = −60/10 = −6
г) (3x + 4)² = 4(x + 3) (3х)² + 24х + 16 − 4х − 12 = 0 9х² + 20х + 4 = 0 D = b² − 4ac = 20² − 4 ⋅ 9 ⋅ 4 = 400 − 144 = 256 > 0, имеет 2 корня х₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-20+\sqrt{256}}{2 ⋅ 9} $$ = −20 + 16/18 = −4/18 = −2/9 х₂ = $$ =\frac{-b-\sqrt{D}}{2 a}=\frac{-20-\sqrt{256}}{2 ⋅ 9} $$ = −20 − 16/18 = −36/18 = −2
д) 4(x + 3)² = (2x + 6)² 4(х² + 6х + 9) = (2х)² + 24х + 36 4х² + 24х + 36 = 4х² + 24х + 36 4х² + 24х + 36 - 4х² − 24х − 36 = 0 0 = 0 х − любое число
е) (6x + 3)² = (x − 4)² (6х)² + 36х + 9 = х² − 8х + 16 36х² + 36х + 9 − х² + 8х − 16 = 0 35х² + 44х - 7 = 0 D = b² − 4ac = 44² − 4 ⋅ 35 ⋅ (−7) = 1936 + 980 = 2916 > 0, имеет 2 корня х₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-44+\sqrt{2916}}{2 ⋅ 35} $$ = −44 + 54/70 = 10/70 = 1/7 х₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-44-\sqrt{2916}}{2 ⋅ 35} $$ = −44 − 54/70 = −98/70 = −1,4