Question

Решите неравенство методом интервалов:
$sqrt{x+3}$* $sqrt{4x+5}$ < 6

Answer 1

$sqrt{x+3} * sqrt{4x+5} textless 6$; _________________________ $sqrt{(x+3)(4x+5)} textless sqrt{36}$; _______________________ $(x+3)(4x+5) textless 36$; ________________________________ $4x^2+5x+12x+15-36 textless 0$; ____________________________ $4x^2+17x-21 textless 0$; _________________________________ $4x^2+17x-21=0; D=17^2-4*4*(-21)=289+336=625=25^2$; ____________ $x_1= frac{-17-25}{2*4} =-5,25; x_2= frac{-17+25}{2*4} =1$; ___________ _______ + ____________________ - ____________________ + ________ -----------------------------(-5,25)----------------------------------------(1)--------------------------------- x∈(-5,25;1)