Question

Найдите произведение корней уравнения lg^2(10x) +lg(x)=19​

Answer 1

lg^2(10x) +lg(x)=19​

x > 0

lg(10x) = lg(10) + lg(x) = 1 + lg(x)

(1 + lg(x))^2 + lg(x) = 19

lg^2(x) + 2lg(x) + 1 + lg(x) - 19 = 0

lg^2(x) + 3lg(x) - 18 = 0

lg(x) = y

y² + 3y - 18 = 0

D = 9 + 72 = 81

y12 = (-3 +- 9)/2 = 3  -6

1. lg(x) = 3

x1 = 10^3

2. lg(x) = -6

x2 = 10^(-6)

x1*x2 =10^3 * 10^(-6) = 10^(-3) = 1/1000

ответ 1/1000

Answer 2

Ответ:

x₁ ·x₂=0.001

Объяснение:

$lg^2(10x)+lgx=19;\;\;\;\ x>0\\\\(lg (10x))^2+lgx=19\\\\(lg10+lgx)^2+lgx=19\\\\(1+lgx)^2+lgx=19\\\\lg^2x+2lgx+1+lgx=19\\\\lg^2x+3lgx-18=0\\\\lgx=t\\t^2+3t-18=0\\\\t=\frac{-3\frac{+}{} \sqrt{9+72} }{2} =\frac{-3\frac{+}{} 9 }{2} \\\\t_1=-6;\;\ t_2=3 \\\\lgx=-6\\\\x_1=10^{-6}\\\\lgx=3\\\\x_2=10^3\\\\x_1*x_2=10^3*10^{-6}=10^{3+(-6)} =10^{-3} =\frac{1}{1000} =0.001\\\\x_1*x_2=0.001$