Question

Помогите решить, пожалуйста |x^2+x-5| больше 3х

Answer 1

$|x^2+x-5|\ \textgreater \ 3x\Longleftrightarrow \left[ \ {{x^2+x-5\ \textgreater \ 3x} \atop {x^2+x-5\ \textless \ -3x}} \right. ,\\\\1)\ x^2+x-5\ \textgreater \ 3x,\\x^2+x-5-3x\ \textgreater \ 0,\\x^2-2x-5\ \textgreater \ 0,\\x^2-2x-5=0,\\D=(-2)^2-4\cdot1\cdot(-5)=4+20=24,\\\\x_{1,2}=\frac{2\pm\sqrt{24}}{2}=1\pm\frac{2\sqrt6}{2}=1\pm\sqrt6,\\x_1=1+\sqrt6,\\x_2=1-\sqrt6,\\a\ \textgreater \ 0\ \Longrightarrow\ x\ \textless \ 1-\sqrt6,\ x\ \textgreater \ 1+\sqrt6;$

$2)\ x^2+x-5\ \textless \ -3x,\\x^2+x+3x-5\ \textless \ 0,\\x^2+4x-5\ \textless \ 0,\\x^2+4x-5=0\\a+b+c=0\ \Longrightarrow x_1=1,\ x_2=-5,\\\\a\ \textgreater \ 0\ \Longrightarrow\ -5<x<1,\\\\3) \left[ \ {{x\ \textless \ 1-\sqrt6,\ x\ \textgreater \ 1+\sqrt6} \atop {-5<x<1}} \right..$

Ответ: $x<1$ или $x>1+\sqrt6.$

Answer 2

X²+x-5<-3x U x²+x-5>3x
1)x²+x-5<-3x
x²+4x-5<0
x1+x2=-4 U x1*x2=-5
x1=-5 U x2=1
-5<x<1
2)x²+x-5>3x
x²-2x-5>0
D=4+20=24
x1=(2-2√6)/2=1-√6
x2=1+√6
x<1-√6 U x>1+√6
x∈(-∞;1) U (1+√6;∞)