Question

Найдите сумму квадратов корней уравнения (x^2 + x - 12) √(x+1) = 3x^2 + 3x -36

Answer 1

$( {x}^{2} + x - 12) \sqrt{x + 1} = 3 {x}^{2} + 3 - 36 \\ ( {x}^{2} + x - 12) \sqrt{x + 1} - 3({x}^{2} + x - 12) = 0 \\ ( {x}^{2} + x - 12)( \sqrt{x + 1} - 3) = 0 \\ x1 = 3 \\ x2 = - 4 \\ x3 = 8$

${( - 4)}^{2} + {3}^{2} + {8}^{2} = \\ 16 + 9 + 64 = 89$