Question

Lg^2x+lgx-12=0 1)-10 2)12 3)-12 4)1\10

Answer 1

lg² x+lgx-12=0    ( x > 0)
lgx = t
t² +t -12 = 0
по т. Виета корни  t₁ = - 4  и  t₂ = 3
a) lgx = - 4                         б) lgx = 3
    x = 10⁻⁴= 0,0001                x = 10³ = 1000
Ответ: 0,0001;   1000
 

Answer 2

В приложении .....,......

Answer 3

To solve this equation, we can use a substitution. Let y = lg(x), then the equation becomes:

y^2 + y - 12 = 0

This is a quadratic equation that can be factored as:

(y + 4)(y - 3) = 0

So, either y + 4 = 0 or y - 3 = 0.

If y + 4 = 0, then y = -4. Substituting back y = lg(x), we get:

lg(x) = -4

x = 10^-4 = 1/10000

If y - 3 = 0, then y = 3. Substituting back y = lg(x), we get:

lg(x) = 3

x = 10^3 = 1000

Therefore, the solutions of the equation are x = 1/10000 and x = 1000. None of the given options matches any of the solutions, so the answer is none of the above.