Question

выполните замену переменной и решите уравнение (8х-3)²-10|8х-3|-24=0​ помогите пожалуйста

Answer 1

(8х - 3)² - 10 |8х - 3| - 24 = 0

Пусть t = 8x - 3, тогда:

t² - 10 |t| - 24 = 0, имеем систему:

$\left \{ {{t^{2}-10t - 24 = 0, t\geq 0 } \atop {t^{2}+10t - 24 = 0, t <0} \right.$

1.

t² - 10t - 24 = 0, t ≥ 0

D = b² - 4ac = 100 - 4 · 1 · (-24) = 196

$t_{1} = \frac{-b + \sqrt{D}}{2a} = \frac{10 + 14}{2} = 5 + 7 = 12$

$t_{2} = \frac{-b + \sqrt{D}}{2a} = \frac{10 - 14}{2} = 5 - 7 = -2$ - не устраивает условие t ≥ 0

2.

t² + 10t - 24 = 0, t < 0

D = b² - 4ac = 100 - 4 · 1 · (-24) = 196

$t_{1} = \frac{-b + \sqrt{D}}{2a} = \frac{-10 + 14}{2} = -5 + 7 = 2$ - не устраивает условие t < 0

$t_{2} = \frac{-b + \sqrt{D}}{2a} = \frac{-10 - 14}{2} = -5 - 7 = -12$

t = 12, -12;

$\left \{ {{8x - 3 = 12} \atop {8x - 3 = -12} \right. \\\left \{ {{8x = 15} \atop {8x = -9}} \right. \\\left \{ {{x = \frac{15}{8} } \atop {x=-\frac{9}{8}}} \right. \\\left \{ {{x = 1\frac{7}{8} } \atop {x=-1\frac{1}{8} }} \right.$

Ответ: $1\frac{7}{8}, -1\frac{1}{8}$ .