Решить уравнение [tex]x^2-4|x|+3=0[/tex]. Created by yugolovin. algebra-ru

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1) x^2 - 4x + 3 = 0
2) x^2 + 4x + 3 = 0

(1) x^2 - 4x + 3 = 0
D = 16 - 12 = 4
x1 = ( 4 + 2)/2 = 6/2 = 3
x2 = (4 - 2)/2 = 2/2 = 1

(2) x^2 + 4x+ 3 = 0
D = 16 - 12 = 4
x3 = ( - 4 + 2)/2 = - 2/2 = - 1
x4 = ( - 4 - 2)/2 = - 6/2 = - 3

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Luluput

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$x^2-4|x|+3=0$
Замена: $|x|=a,$ $a \geq 0$
$a^2-4a+3=0$
$D=(-4)^2-4*1*3=4$
$a_1= \frac{4+2}{2} =3$
$a_2= \frac{4-2}{2} =1$

$|x|=3$ или $|x|=1$
$x=3$ или $x=-3$ или $x=1$ или $x=-1$

Ответ: $б3;$ $б1$

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