Question

Х^4-11х^2+30=0
Х*(х+1)-(х+2)(х+3)+8=х*(х+4)-(х+5)(х+2)

Answer 1

x²=y
y²-11y+30=0
D=121-4*30=1
y₁=(11-1)/2=5
y₂=(11+1)/2=6

x₁ ₂=+-√5
x₃ 4=+-√6


2) Х*(х+1)-(х+2)(х+3)+8=х*(х+4)-(х+5)(х+2)
x²+x-(x²+2x+3x+6)+8=x²+4x-(x²+5x+2x+10)
x²+x-x²-5x-6+8=x²+4x-x²-7x-10
-4x+2=-3x-10
-3x+4x=2+10
x=12

Answer 2

$x^4-11x^2+30=0\\t=x^2 geq 0; ,; ; t^2-11t+30=0\\t_1=5; ,; t_2=6; ; (teorema; Vieta)\\x^2=5; ,; ; x=pm sqrt5\\x^2=6; ,; ; x=pm sqrt6\\Otvet:; ; -sqrt6,; -sqrt5; ,; sqrt5; ,; sqrt6; .$

$2); ; x(x+1)-(x+2)(x+3)+8=x(x+4)-(x+5)(x+2)\\x^2+x-x^2-5x-6+8=x^2+4x-x^2-7x-10\\-4x+2=-3x-10\\x=12$