Question

Определите количество корней уравнения |х^2-4х|=а в зависимости от значения параметра а.

Answer 1

$|x^2-4x|=a, \\ 1) \ a\ \textless \ 0, x\in\varnothing; \\ 2) \ a=0, |x^2-4x|=0, \\ x^2-4x=0, \\ x(x-4)=0, \\ x_1=0, x_2=4; \\ 3) \ a\ \textgreater \ 0, \left [ {{x^2-4x=-a,} \atop {x^2-4x=a;}} \right. \left [ {{x^2-4x+a=0,} \atop {x^2-4x-a=0;}} \right. \\ \left [ {{D_{/4}=(-2)^2-1\cdot a,} \atop {D_{/4}=(-2)^2-1\cdot(-a);}} \right. \\ 3.1) \ \left [ {{4-a\ \textless \ 0,} \atop {4+a\ \textless \ 0;}} \right. \left [ {{a\ \textgreater \ 4,} \atop {a\ \textless \ -4;}} \right. \\ a\ \textgreater \ 4, x\in\varnothing;$
$3.2) \ \left [ {{4-a=0,} \atop {4+a=0;}} \right. \left [ {{a=4,} \atop {a=-4;}} \right. a=4, \\ x^2-4x+4=0, \ (x-2)^2=0, \\ x_1=x_2=2; \\ 3.3) \ \left [ {{4-a\ \textgreater \ 0,} \atop {4+a\ \textgreater \ 0;}} \right. \left [ {{a\ \textless \ 4,} \atop {a\ \textgreater \ -4;}} \right. 0\ \textless \ a\ \textless \ 4, \\ x_1=2-\sqrt{4-a}, \ x_2=2+\sqrt{4-a}. \\ \\ a\in(-\infty;0)\cup(4;+\infty) \ - \ x\in\varnothing, \\ a\in\{0,4\} \ - \ x_1=x_2, \\ a\in(0;4) \ - \ x_1 \neq x_2.$