Решите уравнение: а) (x² + 3)² – 11⋅ (x² + 3) + 28 = 0; б) (x² – 4x)² + 9 ⋅ (x² – 4x) + 20 = 0; в) (x² + x) ⋅ (x² + x – 5) = 84.

Решение:

а) (x² + 3)² – 11⋅ (x² + 3) + 28 = 0

Замена x² + 3 = t

t² – 11t + 28 = 0

D = (-11)² – 4 ⋅ 1 ⋅ 28 = 121 – 112 = 9 больше 0

√D = √9 = √3² = 3

t₁ = -(-11) – 3 / 2 = 11 – 3 / 2 = 8 / 2 = 4

t₂ = -(-11) + 3 / 2 = 11 + 3 / 2 = 14 / 2 = 7

При t = 4

x² + 3 = 4

x² = 4 – 3

x² = 1

При t = 7

x² + 3 = 7

x² = 7 – 3

x² = 4

x = +-2

б) (x² – 4x)² + 9 ⋅ (x² – 4x) + 20 = 0

Замена x² – 4x = t

t² + 9t + 20 = 0

D = 9² – 4 ⋅ 1 ⋅ 20 = 81 – 80 = 1 больше 0

√D = √1 = 1

t₁ = – 9 – 1 / 2 = -(9 + 1) / 2 = – 10 / 2 = – 5

t₂ = – 9 + 1 / 2 = -(9 – 1) / 2 = 8 / 2 = – 4

При t = – 5

x² – 4x = – 5

x² – 4x + 5 = 0

D = (-4)² – 4 ⋅ 1 ⋅ 5 = 16 – 20 = -4 меньше 0 нет корней

При t = -4

x² – 4x = -4

x² – 4x + 4 = 0

x – 2² = 0

x – 2 = 0

x = 2

в) (x² + x) ⋅ (x² + x – 5) = 84

Замена x² + x = t

t(t – 5) = 84

t² – 5t – 84 = 0

D = (-5)² – 4 ⋅ 1 ⋅ (-84) = 25 + 336 = 361 больше 0

√D = √361 = √19² = 19

t₁ = – (-5) – 19 / 2 = 5 – 19 / 2 = – 14 / 2 = – 7

t₂ = – (-5) + 19 / 2 = 5 + 19 / 2 = 24 / 2 = 12

При t = – 7

x² + x = – 7

x² + x + 7 = 0

D = 1² – 4 ⋅ 1 ⋅ 7 = 1 – 28 = -27 меньше 0 корней нет

При t = 12

x² + x = 12

x² + x – 12 = 0

D = 1² – 4 ⋅ 1 ⋅ (-12) = 1 + 48 = 49 больше 0

√D = √ 49 = √7² = 7

x₁ = – 1 – 7 / 2 = – (1 + 7) / 2 = – 8/2 = -4

x₂ = – 1 + 7 / 2 = 7 – 1 / 2 = 6/2 = 3

Источник : Учебник по Алгебре 9 класса авторов Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова, 2014г.